Class 11 Commerce Applied Mathematics - Number Theory

If $Z_{1}$ and $Z_{2}$ are two complex numbers and $c> 0,$ then prove that
$\left | z_{1}+z_{2} \right |^{2}\leq \left ( 1+c \right )\left | z_{1} \right |^{2}+\left ( 1+c^{-1} \right )\left | z_{2} \right |^{2}$

We have to prove:

$\left | z_{1}+z_{2} \right |^{2}\leq \left ( 1+c \right )\left | z_{1} \right |^{2}+\left ( 1+c^{-1} \right )\left | z_{2} \right |^{2}$

i.e.

$\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}+z_{1}\bar{z}_{2}+\bar{z}_{1}z_{2}\leq \left ( 1+c \right )\left | z_{1} \right |^{2}+\left ( 1+c^{-1} \right )\left | z_{2} \right |^{3}$

$z_{1}\bar{z}_{2}+\bar{z}_{1}z_{2}\leq c\left | z_{1} \right |^{2}+c^{-1}\left | z_{2} \right |^{2}$

$\displaystyle c\left | z_{1} \right |^{2}+\frac{1}{c}\left | z_{2} \right |^{2}-z_{1}\bar{z}_{2}-\bar{z}_{1}z_{2}\geq 0$ (using Re

$\left ( z_{1}\bar{z}_{2} \right )\leq \left | z_{1}\bar{z}_{2} \right |$ )

$\displaystyle \left ( \sqrt{c}\left | z_{1} \right |-\frac{1}{\sqrt{c}}\left | z_{2} \right | \right )^{2}\geq 0$ which is always true.

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1}\neq z_{2}$ and $\left | z_{1} \right |=\left | z_{2} \right |.$ If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then show that $\displaystyle \frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is purely imaginary.

$\displaystyle z_{1}=r\left ( \cos \theta +i\sin \theta \right ), -\frac{\pi }{2}< \theta < \frac{\pi }{2}$

$z_{2}=r\left ( \cos \phi +i\sin \phi \right ), -\pi < \phi < 0$

$\Rightarrow$

$\displaystyle \frac{z_{1}+z_{2}}{z_{1}-z_{2}}=-i\cot \left ( \frac{0-\phi }{2} \right ), -\frac{\pi }{4}< \frac{\theta -\phi }{2}< \frac{3\pi }{4}$
Hence purely imaginary

Multiply $3\sqrt { -7 } -5\sqrt { -2 }$ by $3\sqrt { -7 } +5\sqrt { -2 }$

$(3\sqrt{-7}-5\sqrt{-2})\times(3\sqrt{-7}+5\sqrt{-2})$

$=(3\sqrt{7}\text i-5\sqrt{2}\text i)\times(3\sqrt{7}\text i+5\sqrt{2}\text i)$

$=-(3\sqrt{7}-5\sqrt{2})\times(3\sqrt{7}+5\sqrt{2})$

$=-(63+15\sqrt{14}-15\sqrt{14}-50)$

$=-13$

Find the modulus of complex number $-2+2\sqrt{3}i$ .

Let

$z=-2+2\sqrt{3}i$

Then, modulus of z =

$|z|=\sqrt{(-2)^2+(2\sqrt{3})^2}=\sqrt{16}=4$ .

Compute :
$(1 \, + \, i)^{-1}$

$(1 \, + \, i)^{-1} \, = \, \dfrac{1}{1 \, + \, i} \, = \, \dfrac{1 \, - \, i}{1 \, - \, i^2} \ = \, \dfrac{1 \, - \, i}{2}$

Locate the points representing the complex number for which ${ \log }_{ 1/2 }\dfrac { \left| z-1 \right| +4 }{ \left| z-1 \right| -2 } >1$

Put

$\left| z-1 \right| =t=+ive$
Since its base is less than

$1$ , the given inequality
implies that $$\dfrac { t+4 }{ t-2 } <{ \left( \dfrac { 1 }{ 2 } \right)
}^{ 1 }=\dfrac { 1 }{ 2 }$$
or

$\dfrac { 2t+8 }{ t-2 } <1$ or

$\dfrac { 2t+8 }{ t-2 } -1<0$
or

$\dfrac { t+10 }{ t-2 } <0$ where

$t$ is

$+ive$
We cannot multiply by

$t-2$ as we do not know whether it is $$
+ive

$or$ -ive$$.

$\therefore t-2<0$ or

$t<2$
Again

${ \log }_{ 1/2 }\left( \dfrac { t+4 }{ t-2 } \right) >1$ (given)
Now order relation occurs only in real numbers and we know that

$\log \ x$ is real only when

$x$ is

$+ive$

$i.e.,>0$

$\therefore \dfrac { t+4 }{ t-2 }$ is

$+ive\Rightarrow t>2$
Since results

$(1)$ and

$(2)$ cannot hold together and hence
no such

$z$ exists which satisfies the given inequality.

compute.
$[ (cos \theta \, + \, i \, sin \, \theta ) (\, cos \theta \, - \, i sin \theta ) ] = ?$

$[(cos \, \theta \, + \, i \, sin \, \theta) (cos \, \theta \, - \, i \, sin \, \theta)]$

$= \, (cos^2 \, \theta \, - \, i^2 \, sin^2 \, \theta)$

$= \, (cos^2 \, \theta \, + \, sin^2 \, \theta)\, = \, 1.$

Prove the identity, $|1+z_1\bar{z_2}|^2+|z_1-z_2|^2=(1+|z_1|^2)(1+|z_2|^2)$

Given,

$|1-z_1\bar{z}_2|^2-|z_1-z_2|^2$

$=(1+z_1 ^2z_2 ^2-2 z_1 z_2)-(z_1 ^2+z_2^2-2z_1z_2)$

$=1+z_1^2z_2^2-2z_1z_2-z_1^2-z_2^2+2z_1z_2$

$=1+z_1^2z_2^2-z_1^2-z_2^2$ .......(1)

$(1-|z_1|^2)(1-|z_2|^2)$

$=1-|z_2|^2-|z_1|^2+|z_1|^2|z_2|^2$ .......(2)

Comparing (1) and (2), it's clear that,

$|1-z_1\bar{z}_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

Solve $\left(\dfrac{1+i}{1-i}\right)^{24}$

$\displaystyle{{ \left( \frac { 1+i }{ 1-i } \right) }^{ 24 }\\ { \left( \frac { 1+i }{ 1-i } \times \frac { 1+i }{ 1+i } \right) }^{ 24 }\\ { \left( \frac { { (i+i) }^{ 2 } }{ 1-{ i }^{ 2 } } \right) }^{ 24 }\\ { \left( \frac { 1+{ i }^{ 2 }+2i }{ 2 } \right) }^{ 24 }\\ { \left( \frac { 1-1+2i }{ 2 } \right) }^{ 24 }\\ { i }^{ 24 }\\ { { (i }^{ 2 }) }^{ 12 }\\ 1}$

Prove the identity $|1-z_1{z_2}|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

Given,

$|1-z_1\bar{z}_2|^2-|z_1-z_2|^2$

$=(1+z_1 ^2z_2 ^2-2 z_1 z_2)-(z_1 ^2+z_2^2-2z_1z_2)$

$=1+z_1^2z_2^2-2z_1z_2-z_1^2-z_2^2+2z_1z_2$

$=1+z_1^2z_2^2-z_1^2-z_2^2$ .......(1)

$(1-|z_1|^2)(1-|z_2|^2)$

$=1-|z_2|^2-|z_1|^2+|z_1|^2|z_2|^2$ .......(2)

Comparing (1) and (2), it's clear that,

$|1-z_1\bar{z}_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

The HCF of two or more prime numbers is always __________

HCF of two or more prime number is always One.

If $\left( {a + ib} \right)\left( {c + id} \right) = A + iB$ , then show that $\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {A^2} + {B^2}$ .

$\left( {a + ib} \right)\left( {c + id} \right) = A + iB\,\,\,\,\,\left( 1 \right)$

$\left( {a - ib} \right)\left( {c - id} \right) = A - iB\,\,\,\,\,\left( 2 \right)$

$Multiply.$

$\left( {a + ib} \right)\left( {a - ib} \right)\left( {c + id} \right)(c-id) = \left( {A + iB} \right)\left( {A - iB} \right)$

$\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {A^2} + {B^2}$

If $(a+ib)(c+id)=A+iB$ , then show that $({a^2} + {b^2})({c^2} + {d^2}) = {A^2} + {B^2}$

$\left( {a + ib} \right)\left( {c + id} \right) = A + iB{\text{ }} \to {\text{ }}\left( 1 \right)$

${\text{ Replace i by - i,}}$

${\text{A = iB = }}\left( {a - ib} \right)\left( {c - id} \right)\,{\text{ }} \to \left( 2 \right)$

$\left( {A + iB} \right)\left( {A - iB} \right) = \left( {a + ib} \right)\left( {c + id} \right)\left( {a - ib} \right)\left( {c - id} \right)$

${\text{ = (a - ib)}}\left( {a + ib} \right)\left( {c + id} \right)\left( {c - id} \right)$

${\text{ = }}\left( {{a^2} - {{\left( {ib} \right)}^2}} \right)\left( {{c^2} - {{\left( {id} \right)}^2}} \right)$

${\text{ = }}\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)$

${\text{ (}}\because {i^2} = - 1)$

Define and Give one example of each :
a) Parallel lines
b) Intersecting lines
c) Perpendicular line
d) Prime Number
e) Twin Prime
f) Even Number

(a) Parallel lines are two lines that are always the same distance apart and never touch each other.

Ex :

$3x+2y=10,\;3x+2y=4$

(b) Intersecting lines are two lines that meet exactly at one point

Ex :

$3x+2y=10,\;4x+5y=6$

Ex :

$3x-2y=10,\;2x+3y=-5$

(d) The numbers which are divisible only by $$144 or itself are called prime numbers.

Ex :

$3,7,11,....$

(e) Twin prime numbers are those which have difference of

$2$

Ex :

$11,13$

(f) A prime number which is also an even number is called even prime number.

Ex :

$2$ .

Find the values of the following:
$(1+i\sqrt {3})^{3}$

${(1 + i\sqrt 3 )^3}$

$=(1)^3+(i\sqrt3)^3+3(1)^2i\sqrt3+3(1)(i\sqrt3)^2$

$=1-3\sqrt3i+3\sqrt3i-9=-8$

Find the value of ${(-1+\sqrt{-3})}^{2}+{(-1-\sqrt{-3})}^{2}$ .

$\begin{array}{l} { \left( { -1+\sqrt { -3 } } \right) ^{ 2 } }+{ \left( { -1-\sqrt { -3 } } \right) ^{ 2 } } \\ ={ \left( { -1 } \right) ^{ 2 } }+{ \left( { \sqrt { 3 } } \right) ^{ 2 } }+2\left( { -1 } \right) \left( { \sqrt { 3 } } \right) +{ \left( { -1 } \right) ^{ 2 } }+{ \left( { -\sqrt { 3 } } \right) ^{ 2 } } \\ =1-3-2\sqrt { 3 } +1-3+2\sqrt { -3 } \\ =2-6 \\ =-4 \end{array}$

Find the modules and the argument of the complex no $Z=1+i$

Given,

$z=1+i$

modulus:

$|z|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$

$\theta =\tan^{1}\left ( \dfrac{1}{1} \right )=\tan^{1}(1)$

argument:

$\therefore \theta =\dfrac{\pi}{4}$

If $a = \frac{{ - 1 + \sqrt {3i} }}{2}$ , $b = \frac{{ - 1 - \sqrt {3i} }}{2}$ then show that $\,\,{a^2} = b$ and $\,\,{b^2} = a$ .

Given,

$a=\dfrac{-1+\sqrt{3}i}{2}$

$a^2=\left(\dfrac{-1+\sqrt{3}i}{2}\right)^2$

$=\dfrac{-2-2\sqrt{3}i}{2^2}$

$=-\dfrac{2\left(1+\sqrt{3}i\right)}{2^2}$

$=\dfrac{-1-\sqrt{3}i}{2}$

$=b$

----------------------------------

$b=\dfrac{-1-\sqrt{3}i}{2}$

$b^2=\left(\dfrac{-1-\sqrt{3}i}{2}\right)^2$

$=\dfrac{-2+2\sqrt{3}i}{2^2}$

$=-\dfrac{2\left(-1+\sqrt{3}i\right)}{2^2}$

$=\dfrac{-1+\sqrt{3}i}{2}$

$=a$

$z=2+3i$ find $|z|$

$z=2+3i$

$|z|=\sqrt {x^2+y^2}\\|z|=\sqrt {2^2+3^2}=\sqrt {4+9}=\sqrt {13}$

Find the principal arguments of the following Complex Number
(i) $5 + 5i$

Given,

$z=5+5i=a+ib$

Principle argument,

$\theta =\tan ^{-1}\left (\dfrac{b}{a} \right )=\tan ^{-1}\left (\dfrac{5}{5} \right )=\tan ^{-1}(1)$

$\therefore \theta =\dfrac{\pi}{4}$

If ${z_1},{z_2} \in C,$ then $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right|$ .

Let

$z_{1}=x_{1}+iy_{1}, z_{2}=x_{2}+iy_{2}$

where

$x_{1},x_{2},y_{1},y_{2}\in R$

$\left | \dfrac{z_{1}}{z_{2}} \right |=\dfrac{\left | z_{1} \right |}{\left | z_{2} \right |}$

$(z_{2}\neq 0)$

$=\dfrac{\sqrt{x_{1}^{2}+y_{1}^{2}}}{\sqrt{x_{2}^{2}+y_{2}^{2}}}$

Let $Z = x + iy$ and $\omega = \dfrac {1 - iZ}{Z - i}$ . If $|\omega| = 1$ , show that $Z$ is purely real.

$|w|=1$

$|\dfrac{1-iz}{z-i}|=1$

$|1-iz|=|z-i|$

$|1-i(x+iy)|=|x+i(y-1)|$ , where

$z=x+iy$ .

$\sqrt{(1+y)^2+(-x)^2}=\sqrt{x^2+(y-1)^2}$

$(1+y)^2+x^2=x^2+(y-1)^2$

$y=0$

$z=x+i0=x$ , which is purely real.

Find the modulus of $\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}$

$\\z= (\dfrac{1+i}{1-i})-(\dfrac{1-i}{1+i})$

$\\=(\dfrac{(1+i)^2-(1-i)^2}{(1-i)(1+i)})$

$\\=(\dfrac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2})$

$\\=(\dfrac{4i}{2})$

$\\=2i$

$\\\therefore |z|=|2i|=2$

Solve: $x^2 - (3\sqrt{2} - 2i)x - \sqrt{2}i = 0$

Given,

$x^2-\left(3\sqrt{2}-2i\right)x-\sqrt{2}i=0$

$\left(a+bi\right)^2-\left(3\sqrt{2}-2i\right)\left(a+bi\right)-\sqrt{2}i=0$

$\left(a^2-b^2-3\sqrt{2}a-2b\right)+i\left(-\sqrt{2}+2a+2ab-3\sqrt{2}b\right)=0+0i$

$\begin{bmatrix}a^2-b^2-3\sqrt{2}a-2b=0\\ -\sqrt{2}+2a+2ab-3\sqrt{2}b=0\end{bmatrix}$

solving the above 2 equations, we get,

$\begin{pmatrix}a=\dfrac{3\sqrt{2}-4}{2},&b=-\dfrac{-\sqrt{2}+2}{2}\\ a=\dfrac{3\sqrt{2}+4}{2},&b=-\dfrac{2+\sqrt{2}}{2}\end{pmatrix}$

$x=\dfrac{3\sqrt{2}-4}{2}-\dfrac{-\sqrt{2}+2}{2}i,\dfrac{3\sqrt{2}+4}{2}-\dfrac{2+\sqrt{2}}{2}i$

State whether the given statement is true or false $\overline{(z^{-1})} \, = \, (\bar{z})^{-1}$

We have,

$\overline {(z^{-1}})=(\overline z)^{-1}$

$\overline {\left |z^{-1}\right|}=|\overline z|^{-1}$

$\overline {\left |\dfrac{1}{z}\right|}=\dfrac{1}{|\overline z|}$

$\dfrac{1}{|\overline {z}|}=\dfrac{1}{|\overline z|}$

Hence, this is the answer.

Express $\dfrac{{5 + i\sqrt 2 }}{{2i}}$ in the form of $x + iy$ .

$\dfrac{5+i\sqrt{2}}{2i}=x+iy$

$\dfrac{5}{2i}+\dfrac{i\sqrt{2}}{2i}=x+iy$

$\dfrac{5i}{2{i}^{2}}+\dfrac{\sqrt{2}}{2}=x+iy$

$\dfrac{5i}{-2}+\dfrac{\sqrt{2}}{2}=x+iy$ since

${i}^{2}=-1$

$\dfrac{\sqrt{2}}{2}-i\dfrac{5}{2}=x+iy$

Comparing both the sides, we have

$x=\dfrac{\sqrt{2}}{2}$ and

$y=-\dfrac{5}{2}$

$\therefore\,x+iy=\dfrac{\sqrt{2}}{2}+i\left(-\dfrac{5}{2}\right)$

If $\dfrac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x - iy,$ then find the value of $x + y$ .

$\dfrac{(1+i)^{2}}{2-i}=x-iy$

$=\dfrac{1+i^{2}+2i}{2-i}$

$=\left ( \dfrac{2i(2+i)}{2-i(2+i)} \right )$

$=\dfrac{4i-2}{4+1}$

$=\dfrac{4i}{5}-\dfrac{2}{5}$

$x=\dfrac{-2}{5},y=\dfrac{-4}{5}$

$x+y=\dfrac{-6}{5}$

Find the real and imaginary parts of the complex number z= $\dfrac{3i^{20}-i^{19}}{2i-1}$

$z=\dfrac{3i^{20}-i^{19}}{2i-1}=\dfrac{3(1)-(-i)))}{2i-1}$ =

$\dfrac{3+i}{-1+2i}$

$=\dfrac{3+i}{-1+2i}\times \dfrac{-1-2i}{-1-2i}$

$=\dfrac{-3-6i-i-2i^{2}}{(-1^{2})+2^{2}}$

$=\dfrac{-1-7i}{5} =\dfrac{-1}{5}-\dfrac{7}{5}i$

$Re(z)=-\dfrac{1}{5}$ and

$Im(z)=\dfrac{-7}{5}$

If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\frac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$ , then find $(x,\,y)$ ,

$\left ( \dfrac{1+i}{1-i} \right )^{3} - \left ( \dfrac{1-i}{1+i} \right )^{3} = x+iy$

Rationalizing, we get,

$\left ( \dfrac{(1+i)^{2}}{2} \right )^{3} - \left ( \dfrac{(1-i)^{2}}{2} \right )^{3} = x + iy$

$\left ( \dfrac{1+2i-1}{2} \right )^{3}-\left ( \dfrac{1-2i-1}{2} \right )^{3}=x+iy$

$(i)^{3} - (-i)^{3} = x + iy$

$-i - (+i) = x + iy$

$- 2i = x + iy$

so,

$y = - 2$

$x = 0$

For the complex number $\sqrt { 37 } + \sqrt { - 19 }$ .
Real part is $\sqrt 37$ and imaginary is $\sqrt 19$
Enter 1 if true or 0 for false

$\\\sqrt{37}+\sqrt{-19}\\=\sqrt{37}+\sqrt{-1}\sqrt{19}$

$\\=\sqrt{37}+\sqrt{19}i$

$\therefore$ Real Part =

$\sqrt{37}$

& Imaginary Part =

$\sqrt{19}$

What are twin-primes ?

Pair of prime numbers whose difference is

$2$ called twin-primes. For example,

$(3,5), (5,7)$ etc.

Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(1 + 2i)(-2 + i)$

$(1 + 2i)(-2 + i)\\-2+i-4i-2\\-4-3i\\a=-4,b=-3$

If ABC are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find value of xyz

Given:-

$x = cis{A}$

$y = cis{B}$

$z = cis{C}$

To find:-

$xyz = ?$

Solution:-

Since

$A, B$ and

$C$ are angles of a triangle.

Therefore,

$A + B + C = 180°$

$xyz$

$= cis{A} \cdot cis{B} \cdot cis{C}$

$= cis{\left( A + B + C \right)}$

$= cis{180°}$

$= \cos{180°} + i \sin{180°}$

$= -1 + i \left( 0 \right) = -1$

Hence the value of

$xyz$ is

$-1$ .

Give three pairs of prime numbers whose difference is $2$ . [Remark: Two prime numbers whose difference is $2$ are called twin primes].

As we know,

Two prime numbers whose difference is

22 are called twin primes.

examples are as:

$3, 5$

$41, 43$

$71, 73$ .

If $z \neq 0$ is a complex number,then prove that $Re(z) = 0 \Longrightarrow Im(z^2)=0$ .

$z \neq 0$ , let

$z = x + iy$ .
Then,

$z^2=(x+iy)(x+iy)$

$\Rightarrow z^2= x^2-y^2+i(2xy)$

$\because Re(z)= 0$

$\Longrightarrow x=0$

$\Longrightarrow Im(z^2)=2xy=0$

If $(a + ib) (c + id) (e + if)(g +ih) = A + iB$ , then prove that $(a^2 + b^2) (c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$ .

$(a + ib) (c + id) (e + if)(g +ih) = A + iB$

$\therefore \left|(a + ib) (c + id) (e + if)(g +ih)\right| = \left|A + iB\right|$
or

$\sqrt{a^2 + b^2} \times \sqrt{c^2 + d^2} \times \sqrt{e^2 + f^2} \times \sqrt{g^2 + h^2} = \sqrt{A^2 + B^2}$
On squaring both sides, we get

$(a^2 + b^2) (c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$

If $z = x + iy$ and $w = \dfrac{(1 - iz)}{(z - i)}$ and $\left|w\right| = 1$ , then prove that $z$ is purely real.

We have,

$\left|w\right| = 1 \Longrightarrow \left|\dfrac{1 - iz}{z - i}\right| = 1$
or

$\dfrac{\left|1 - iz\right|}{\left|z - i\right|} =1$
or

$\left|1 - iz\right| = \left|z - i\right|$ (1)
or

$\left|1 - i (x + iy) \right| = \left|x + iy - i\right|$ , where

$z = x + iy$
or

$\left|1 + y - ix\right| = \left|x + i(y - 1)\right|$
or

$\sqrt{(1+y)^2 + (-x)^2} = \sqrt{x^2 + (y - 1)^1}$
or

$(1 + y)^2 + x^2 = x^2 + (y-1)^2$
or

$y = 0$

$\Longrightarrow z = x + i0 = x,$ which is purely real.

Ans: 1

If $(x+iy)^3=u+iv$ , then prove that $\displaystyle \frac{u}{x}+\frac{v}{y} =4(x^2 - y^2)$ .

Given,

$(x+iy)^3=u+iv$

$x^3+(iy)^3 + 3. x. iy(x+iy) = u+iv$

$x^3+i^3y^3+3x^2yi+3xy^2 i^2=u+iv$

$x^3-iy^3+3x^2yi-3xy^2=u+iv$

$(x^3-3xy^2)=i(3x^2y-y^3)=u+iv$
On equating real and imaginary parts, we get

$u=x^3-3xy^2, v=3x^2y-y^3$

$\displaystyle \therefore \frac{u}{x}+\frac{v}{y} =\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}$

$\displaystyle=\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}$

$=x^2-3y^2+3x^2-y^2 =4x^2-4y^2 =4(x^2-y^2)$

If $z_1$ and $z_2$ are complex numbers and $u = \sqrt{z_1z_2}$ , then prove that
$\left|z_1\right| + \left|z_2\right| = \left|\displaystyle \frac{z_1 + z_2}{2}+u\right| + \left|\displaystyle \frac{z_1 + z_2}{2}-u\right|$ .

$\left|\dfrac{z_1 + z_2}{2}+u\right| + \left|\dfrac{z_1 + z_2}{2}-u\right|$

$=\left|\dfrac{z_1 + z_2}{2}+\sqrt{z_1z_2}\right| + \left|\dfrac{z_1 + z_2}{2}-\sqrt{z_1z_2}\right|$

$= \left|\dfrac{(\sqrt{z_1} + \sqrt{z_2})^2}{2}\right| + \left|\dfrac{(\sqrt{z_1} - \sqrt{z_2})^2}{2}\right|$

$= \left|\dfrac{(p+q)^2}{2}\right| + \left|\frac{(p-q)^2}{2}\right|$ (where

$p = \sqrt{z_1}$ and

$q = \sqrt{z_2}$ )

$= \dfrac{1}{2}\left[\left|p+q\right|^2 + \left|p-q\right|^2\right]$

$= \dfrac{1}{2}\left[2\left|p\right|^2 +2\left|q\right|^2 \right]$

$= \left|p\right|^2 + \left|q\right|^2$

$= \left|p^2\right| + \left|q^2\right|$

$= \left|z_1\right| + \left|z_2\right|$

Ans: 1

If $\left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right|$ , then show that $arg(z_1) = arg(z_2)$ .

$\left|z_1 + z_2\right| =$

$\left|z_1\right| + \left|z_2\right|$
or

$\left|z_1\right|^2 + \left|z_2\right|^2 + 2Re(z_1\overline{z}_2) = \left|z_1\right|^2 + \left|z_2\right|^2 + 2\left|z_1\right| \left|z_2\right|$
or

$2Re(z_1\overline{z}_2) = 2\left|z_1\right| \left|z_2\right|$

$\Longrightarrow cos (\theta_1 - \theta_2) = 1$

$\Longrightarrow \theta_1 - \theta_2 = 0$

$\Longrightarrow arg(z_1) = arg(z_2)$

Ans: 1

If $\left|z_1 - z_2\right| = \left|z_1\right| + \left|z_2\right|$ , then prove that $arg(z_1) - arg (z_2) = \pi$ . I

$\left|z_1\right|^2 + \left|z_2\right|^2 - 2Re(z_1\overline{z}_2) = \left|z_1\right|^2 + \left|z_2\right|^2 + 2\left|z_1\right| \left|z_2\right|$
or

$-2Re(z_1\overline{z}_2) = 2\left|z_1\right| \left|z_2\right|$

$\Longrightarrow cos (\theta_1 - \theta_2) = -1$

$\Longrightarrow \theta_1 - \theta_2 = \pi$

Ans: 1

Find the minimum value of $\left | 1+z \right |+\left | 1-z \right |.$

$\Rightarrow$

$\left | 1+z \right |+\left | 1-z \right |\geq 2$

$\therefore$ minimum value of

$\left ( \left | 1+z \right |+\left | 1-z \right | \right )=2$
Geometrically

$z$ from

$1$ and

$-1$
it can be seen easily that minimum

$\left ( PA+PB \right )=AB=2$
1032289_332835_ans_372ec579868a4122a5e9dc36ef764fd3.png

$2^{i}=e^{i(lnx)}$ .

$2^{i}=e^{i\ln 2}=\left [ \cos \left ( \ln 2 \right )+i\sin \left ( \ln 2 \right ) \right ]$

$\cos \left ( \ln 2 \right )+i\sin \left ( \ln 2 \right )=e^{i\left ( \ln 2 \right )}$

Compute the modulus of complex number $z=\sqrt5-2i$

$|z|=\sqrt{\sqrt5^2+2^2}=\sqrt{9}=3$ .

If $\displaystyle a+ib=\dfrac { { \left( x+i \right) }^{ 2 } }{ { 2x }^{ 2 }+1 }$ , prove that $\displaystyle { a }^{ 2 }+{ b }^{ 2 }=\dfrac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( 2x^2+1 \right) }^{ 2 } }$ .

$\displaystyle a+ib=\frac { { \left( x+i \right) }^{ 2 } }{ { 2x }^{ 2 }+1 }$

$\displaystyle =\frac { { x }^{ 2 }+{ i }^{ 2 }+2xi }{ 2{ x }^{ 2 }+1 }$

$\displaystyle =\frac { { x }^{ 2 }-1+2xi }{ 2{ x }^{ 2 }+1 }$

$\displaystyle =\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } +i\left( \frac { 2x }{ 2{ x }^{ 2 }+1 } \right)$

On comparing real and imaginary parts, we obtain

$\displaystyle a=\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } and\quad b=\frac { 2x }{ 2{ x }^{ 2 }+1 }$

$\displaystyle \therefore \quad { a }^{ 2 }+{ b }^{ 2 }={ \left( \frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \right) }^{ 2 }+{ \left( \frac { 2x }{ 2{ x }^{ 2 }+1 } \right) }^{ 2 }$

$\displaystyle =\frac { { x }^{ 4 }+1-{ 2x }^{ 2 }+4{ x }^{ 2 } }{ { \left( 2{ x }^{ 2 }+1 \right) }^{ 2 } }$

$\displaystyle =\frac { { x }^{ 4 }+1+{ 2x }^{ 2 } }{ { \left( 2{ x }^{ 2 }+1 \right) }^{ 2 } }$

$\displaystyle =\frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( 2{ x }^{ 2 }+1 \right) }^{ 2 } }$

$\displaystyle \therefore \quad { a }^{ 2 }+{ b }^{ 2 }=\frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( 2{ x }^{ 2 }+1 \right) }^{ 2 } }$
Hence, proved.

Express $\cfrac{2+i}{(1+i)(1-2i)}$ in the form of $a+ib$ . Find its modulus and argument.

Here

$\cfrac{2+i}{(1+i)(1-2i)}=\cfrac{2+i}{1+i-2i-2{i}^{2}}=\cfrac{2+i}{3-i}$

$=\cfrac { 2+i }{ 3-i } \times \cfrac { 3+i }{ 3+i } =\cfrac { 6+3i+2i+{ i }^{ 2 } }{ 9+1 }$

$=\cfrac { 5+5i }{ 10 } =\cfrac{1}{2}+\cfrac{1}{2}i=\left( \cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 2 } \right)$

$\therefore$ Modulus

$=\displaystyle \sqrt { { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 } } =\cfrac { 1 }{ \sqrt { 2 } }$
Since

$\left( \cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 2 } \right)$ lies in the first quadrant

$\therefore$ Argument (

$\alpha$ )

$=\tan ^{ -1 }{ \left| \cfrac { 1/2 }{ 1/2 } \right| } =\tan ^{ -1 }{ (-1) } =\cfrac { \pi }{ 4 }$

Find the real and imaginary parts of the complex number $\dfrac {a + ib}{a - ib}$

$=\dfrac {a + ib}{a - ib}\cdot \dfrac{a+ib}{a+ib}$ , [rationalize the denominator]

$=\dfrac{(a+ib)^2}{a^2-i^2b^2}=\dfrac{a^2+i^2b^2+2iab}{a^2+b^2}$

$=\dfrac{a^2-b^2+(2ab)i}{a^2+b^2}$

$=\dfrac{a^2-b^2}{a^2+b^2}+\dfrac{2ab}{a^2+b^2}i$

Hence real part is

$\dfrac{a^2-b^2}{a^2+b^2}$ and imaginary part is

$\dfrac{2ab}{a^2+b^2}$

Find the number of integral solution of $(1-i)^x=2^x$ .

Let first find the value of

$|1-i|$

$1-i$ is the complex no in the form of

$x+i y=0$

where

$x=1$ and

$y=-1$

$|1-i|=\sqrt2$

Given

$|1-i|^x=2^x$

$\Rightarrow (\sqrt2)^x=2^x$

$\Rightarrow (\sqrt2)^x=(\sqrt2)^{2x}$

compair power

$\Rightarrow x=2x$

$\Rightarrow x=0$

So the answer is

$0$

Hence, there is only

$1$ integral solution.

If $z$ and $\alpha$ complex numbers such that $|z| = |\alpha| = r, r > 0$ and $\omega = \dfrac {z - \overline {\alpha}}{r^{2} + z\overline {\alpha}}$ , find $Re(\omega)$ .

$|z=|\alpha|=r$

$z=a_1 +ib_1 \ \Rightarrow z=\sqrt {a_1^2 +b_1^2}=r \Rightarrow a_1^2 +b_1^2 =a_2^2 +b_2^2$

$x=a_2+ib_2 \Rightarrow \ x=\sqrt {a_2^2 +b_2^2} =r \ a_1^2 -a_2^2 =b_2^2 -b_1^2$

$w=\dfrac {a_1 +ib_1 -a_2 -ib_2}{r^2 +a_1 a_2 -b_1 b_2} \Rightarrow \dfrac {a_1 -a_2 +ib_1 -ib_2}{r^2 +a_1 a_2 -b_1 b_2}$

$Re(w)=\dfrac {a_1 -a_2}{r^2 +a_1 a_2 -b_1 b_2}$

$a_1^2 +b_1^2$ or

$a_2^2 +b_2^2$

If the ratio $\left (\dfrac {1 - z}{1 + z}\right )$ is purely imaginary, then find value of $|z|$ .

$z=x+cy\\ \Rightarrow\dfrac{1-z}{1+z}=\left[ \cfrac { 1-x-cy }{ 1+x+cy } \right] \\ =\cfrac { \left( 1-x-cy \right) \left( 1+x-cy \right) }{ \left( 1+x+cy \right) \left( 1+x-cy \right) } \\ =\cfrac { 1+x-cy-x-{ x }^{ 2 }+cxy-cy-cxy+{ x }^{ 2 }{ y }^{ 2 } }{ { \left( 1+x \right) }^{ 2 }+{ y }^{ 2 } } \\ =\cfrac { 1-{ x }^{ 2 }-{ y }^{ 2 }-2xy }{ { \left( 1+x \right) }^{ 2 }+{ y }^{ 2 } } \\$

So, $=\cfrac { 1-{ x }^{ 2 }-{ y }^{ 2 } }{ { \left( 1+x \right) }^{ 2 }+{ y }^{ 2 } } =0\\ 1-{ x }^{ 2 }-{ y }^{ 2 }=0\\ { x }^{ 2 }+{ y }^{ 2 }=1\\ \left| z \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 1 } =1$

If $a=\cfrac { 1+i }{ \sqrt { 2 } }$ , find the value of ${ a }^{ 6 }+{ a }^{ 4 }+{ a }^{ 2 }+1$

${ a }^{ 6 }+{ a }^{ 4 }+{ a }^{ 2 }+1={ a }^{ 4 }\left( { a }^{ 2 }+1 \right) +1\left( { a }^{ 2 }+1 \right) \\ =\quad \left( { a }^{ 2 }+1 \right) \left( { a }^{ 4 }+1 \right) \quad ..............eq\quad \left( 1 \right) \\ Here,\\ \quad \quad \quad a=\cfrac { 1+i }{ \sqrt { 2 } } \\ \Rightarrow \quad \sqrt { 2a } =1+i\\ Squaring\quad both\quad sides,\\ { \left( \sqrt { 2 } a \right) }^{ 2 }={ \left( 1+i \right) }^{ 2 }\\ \Rightarrow 2{ a }^{ 2 }=1+2i+i^{ 2 }\quad \left( { i }^{ 2 }=-1 \right) \\ \Rightarrow 2{ a }^{ 2 }=2i\\ \quad { a }^{ 4 }=-1\\ \left( { a }^{ 2 }+1 \right) \left( { a }^{ 4 }+1 \right) =\left( 1+1 \right) \left( -1+1 \right) =0$

If $\left| { z }_{ 1 } \right| =2,\left| { z }_{ 2 } \right| =3,\left| { z }_{ 3 } \right| =4$ and $\left| 2{ z }_{ 1 }+3{ z }_{ 2 }+4{ z }_{ 3 } \right| =10$ , then absolute value $8{ z }_{ 2 }{ z }_{ 3 }+27{ z }_{ 3 }{ z }_{ 1 }+64{ z }_{ 1 }{ z }_{ 2 }$ must be equal to-

$|z_1|=2,z_1\bar{z_1}=4$

$|z_2|=3,z_2\bar{z_2}=9$

$|z_3|=4,z_3\bar{z_3}=16$

$|8z_2z_3+27z_3z_1+64z_1z_2|$

$\implies|2z_1\bar{z_1}z_2z_3+3z_2\bar{z_2}z_3z_1+4z_3\bar{z_3}z_1z_2|$

$\implies|z_1z_2z_3||2\bar{z_1}+3\bar{z_2}+4\bar{z_3}|$

$\implies 2\times 3\times 4|2{z_1}+3{z_2}+4{z_3}|$

$\implies 24\times 10=240$

Find the points of the plane which satisfy the following equations.
$\displaystyle \left | \, \frac{z \, - \, 2}{z \, + \, 3} \, \right | \, = \, 1.$

1148852_889382_ans_df2109de7ef7424db5aeb8ae9e7e415d.jpg

Find the points of the plane which satisfy the following equations.
$\displaystyle \left | \, \frac{z \, + \, i}{z \, - \, 3i} \, \right | \, = \, 1.$

1148847_889386_ans_088cb905849a48bdb195466f369c37b2.pdf

Simplify the following :
$\dfrac{(1 \, - \, i)^3}{1 \, - \, i^3}$

$\dfrac{(1 \, - \, i)^3}{1 \, - \, i^3}$

$= \, \dfrac{(1 \, - \, i)^2}{1 \, + \, i^2 \, + \, i} \, = \, \dfrac{-2i}{i} \, = \, -2 \,$

Express the following in the form A + iB :
$\dfrac{1}{1 \, - \, \cos \theta \, + \, 2i \, \sin \theta}$

$\dfrac{1}{1 \, - \, cos \theta \, + \, 2i \, sin \theta} \, = \, \dfrac{1}{2 \, sin^2 \dfrac{\theta}{2} \, + \, 4i \, sin \dfrac{\theta}{2} \, cos \dfrac{\theta}{2}}$

$= \, \dfrac{1}{2sin (\theta / 2)} \, \cdot \, \dfrac{1}{[sin(\theta / 2) \, + \, 2i \, cos \, (\theta / 2)]}$

$= \, \dfrac{1}{2sin (\theta / 2)} \, \cdot \, \dfrac{sin(\theta / 2) \, - \, 2i \, cos \, (\theta / 2)}{sin^2(\theta / 2) \, + \, 4 \, cos^2 \, (\theta / 2)}$

$=\,\dfrac{1 \, - \, 2i \, cot \, (\theta /2)}{2sin^2(\theta / 2) \, + \, 4[ 2cos^2 \, (\theta / 2)]}$

Change the double angle.

$= \,\dfrac{1 \, - \, 2i \, cot \, (\theta /2)}{5 \, + \, 3 \, cos \theta}$

Put the following in the form A + iB :
$\dfrac{(a \, + \, ib)^2}{(a \, - \, ib)} \, - \, \dfrac{(a \, - \, ib)^2}{(a \, + \, ib)}$

$\dfrac{(a \, + \, ib)^2}{(a \, - \, ib)} \, - \, \dfrac{(a \, - \, ib)^2}{(a \, + \, ib)} \, = \, \dfrac{(a \, + \, ib)^3 \, - \, (a \, - \, ib)^3}{(a \, - \, ib)(a \, + \, ib)}$

$= \, \dfrac{2(3a^2 \, . \, ib \, + \, i^3b^3)}{a^2 \, + \, b^2} \, = \, \dfrac{2(3a^2b \, - \, b^3)}{a^2 \, + \, b^2}i$

$= \, \dfrac{2b(3a^2 \, - \, b^2)}{a^2 \, + \, b^2}i$

If $Z_r \, = \, \left ( \cos\dfrac{r\pi }{10} \, + \, i \, \sin \dfrac{r\pi }{10}\right ).$ Then, find the value of $Z_1 \cdot Z_2 \cdot Z_3 \cdot Z_4$

$Z_1 \, = \, \cos\dfrac{\pi }{10} \, + \, i \, \sin \dfrac{\pi }{10}=e^{i\pi/10}.$

$Z_2 \, = \, \cos\dfrac{2\pi }{10} \, + \, i \, \sin \dfrac{2\pi }{10}=e^{i2\pi/10}.$

$Z_3 \, = \, \cos\dfrac{3\pi }{10} \, + \, i \, \sin \dfrac{3\pi }{10}=e^{i3\pi/10}.$

$Z_4 \, = \, \cos\dfrac{4\pi }{10} \, + \, i \, \sin \dfrac{4\pi }{10}=e^{i4\pi/10}.$

So,

$Z_1 \cdot Z_2 \cdot Z_3 \cdot Z_4$

$=e^{\dfrac {\pi}{10}}e^{\dfrac {2\pi}{10}}e^{\dfrac {3\pi}{10}}e^{\dfrac{4\pi}{10}}$

$=e^{ {\dfrac{\pi} {10}} \cdot (1+2+3+4)}$

$=e^{i\pi}=cos\pi+isin\pi=-1$

If $i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0$ , then show that $\left| z \right| =1$ .

Dividing throughout by

$i$ and knowing that

$\cfrac { 1 }{ i } =-i$
we get

${ z }^{ 3 }-i{ z }^{ 2 }+iz+1=0$
or

${ z }^{ 2 }\left( z-i \right) +i\left( z-i \right) =0 \qquad as\ 1=-{ i }^{ 2 }$
or

$\left( z-i \right) \left( { z }^{ 2 }+i \right) =0$

$\therefore z=i$ or

${ z }^{ 2 }=-i$

$\therefore \left| z \right| =\left| i \right| =1$

$\left| { z }^{ 2 } \right| ={ \left| z \right| }^{ 2 }=\left| -i \right| =1$

$\therefore \left| z \right| =1$
Hence in either case

$\left| z \right| =1$ .

$The\quad complex\quad number\quad \cfrac { { 2 }^{ n } }{ { (1+i) }^{ 2n } } +\cfrac { { (1+i) }^{ 2n } }{ { 2 }^{ n } } ,\quad n\quad \in Z,\quad is\quad equal\quad to\quad$

$\cfrac { { 2 }^{ n } }{ { (1+i) }^{ 2n } } +\cfrac { { (1+i) }^{ 2n } }{ { 2 }^{ n } } ,\quad n\quad \in I\\ { (1+p) }^{ 2 }=1-1+2i=2i\\ \cfrac { { 2 }^{ n } }{ { (2i) }^{ n } } +\cfrac { { (2i) }^{ n } }{ { 2 }^{ n } } =\quad \cfrac { { 2 }^{ n } }{ i^{ n }x{ 2 }^{ n } } +\cfrac { i^{ n }x{ 2 }^{ n } }{ { 2 }^{ n } } \\ =\cfrac { 1 }{ 4 } +{ i }^{ n }\\ =\cfrac { 1+{ 2 }^{ n } }{ { i }^{ n } } \\ =\cfrac { 1+{ (-1) }^{ n } }{ { i }^{ n } } \\ if\quad n\quad is\quad even\quad { (-1) }^{ n }=1=\cfrac { 2 }{ { i }^{ n } } \\ n\quad is\quad odd\quad { (-1) }^{ n }=-1=0$

Find all the values of $z$ which satisfy the equation
exp $\left| \dfrac { { \left| z \right| }^{ 2 }-\left| z \right| +4 }{ { \left| z \right| }^{ 2 }+1 } \log2 \right| ={ \log }_{ \sqrt { 2 } }\left| 3\sqrt { 15 } +11i \right|$

1317004_1001092_ans_b7a910b38c994aefa8a1257d359c45fb.jpg

Find the modulus and argument of the complex number:
$\dfrac{1}{1+i}$

Consider,

$z=\dfrac{1}{1+i}$

$z=\dfrac{1}{1+i}\ \times\ \dfrac{1-i}{1-i}$

$z=\dfrac{1-i}{(1+i)(1-i)}$

$z=\dfrac{1-i}{1^2-i^2}$

$z=\dfrac{1-i}{1-(-1)}$ -------- We know that

$i^2=-1$

$z=\dfrac{1-i}{2}$

$z=\dfrac{1}{2}-\dfrac{1}{2}i$

This is the complex number of the form

$z=x+iy$

$x=\dfrac{1}{2}$ and

$y=-\dfrac{1}{2}$

We know that,

$Modulus=\sqrt{x^2+y^2}$

$=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}$

$=\sqrt{\dfrac{2}{4}}$

$=\dfrac{1}{\sqrt{2}}$

Argument:

$\dfrac{1}{2}+\left(-\dfrac{1}{2}\right)i=r\cos \theta+i\ r\sin \theta$

Equating the real parts,

$\dfrac{1}{2}=r\cos \theta$

$\dfrac{1}{2}=\dfrac{1}{\sqrt{2}}\cos \theta$ ------ Above,

$modulus=r=\dfrac{1}{\sqrt{2}}$

$\cos \theta=\dfrac{1}{\sqrt{2}}$

Similarly, equating the imaginary parts we get,

$-\dfrac{1}{2}=r\sin \theta$

$\implies \sin \theta=-\dfrac{1}{\sqrt{2}}$

Here,

$\sin \theta$ is

$-ve$ and

$\cos \theta$ is

$+ve$ . Hence,

$\theta$ is in 4th quadrant.

So,

$Argument=-45^\circ=-\dfrac{\pi}{4}\ radians$

For any two complex number $z_1$ and $z_2$ prove that $Re(z_1z_2)= Rez_1Rez_2-I mz_1Imz_1$

Let

${ z }_{ 1 }=a+ib,{ z }_{ 2 }=x+iy$

${ z }_{ 1 }{ z }_{ 2 }=\left( a+ib \right) \left( x+iy \right) =ax+aiy+ibx+{ i }^{ 2 }by$

$Re\left( { z }_{ 1 }{ z }_{ 2 } \right) =ax-by$

$Re\left( { z }_{ 1 }{ z }_{ 2 } \right) =Re\left( { z }_{ 1 } \right) Re\left( { z }_{ 2 } \right) -Im({ z }_{ 1 })Im({ z }_{ 2 })$

Hence proved

let $z_1$ and $z_2$ be two complex number such that $|1-{z_1}z_2|^2-|z_1-z_2|^2=k\left(1-|z_1|^2\right)\left(1-|z_2|^2\right)$ find the value of $k$

Given,

$|1-\bar{z}_1{z}_2|^2-|z_1-z_2|^2$

$=(1+z_1 ^2z_2 ^2-2 z_1 z_2)-(z_1 ^2+z_2^2-2z_1z_2)$

$=1+z_1^2z_2^2-2z_1z_2-z_1^2-z_2^2+2z_1z_2$

$=1+z_1^2z_2^2-z_1^2-z_2^2$ .......(1)

$(1-|z_1|^2)(1-|z_2|^2)$

$=1-|z_2|^2-|z_1|^2+|z_1|^2|z_2|^2$ .......(2)

Comparing (1) and (2), it's clear that,

$|1-z_1\bar{z}_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

$\therefore k=1$

If $\left( {a + ib} \right) = \frac{{1 + i}}{{1 - i}}$ , then prove that $\left( {{a^2} + {b^2}} \right) = 1$

According to question,

$a+ib=\dfrac{1+i}{1-i}$

On taking modulus on both side we get,

$|a+ib|=\dfrac{|1+i|}{|1-i|}$

=>

$\sqrt{a^2+b^2}=\dfrac{1}{1}$

=>

$a^2+b^2=1$

Hence proved

How solve and what is its meaning
$z_{1}=2+3i$
$z_{2}=3+4i$
$\left|z_{1}+z_{2}\right|=?$

$z_1 = 2+3i$

$z_2 = 3+4i$

$\implies z_1 + z_2 = 5+7i$

$\therefore |z_1 + z_2| = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}$

It is modulus of a complex number

Let $\dfrac{1}{a+ib}=c+id$ for non-zero $a,b$ , then find $c,d$ .

Given,

$\dfrac{1}{a+ib}=c+id$

or,

$\dfrac{a-ib}{a^2+b^2}=c+id$

or,

$\dfrac{a}{a^2+b^2}+i\dfrac{-b}{a^2+b^2}=c+id$

Comparing the real and imaginary part both sides we get,

$c=\dfrac{a}{a^2+b^2},d=\dfrac{-b}{a^2+b^2}$ .

Find the modulus and the arguments of each of the complex numbers is $z=-\sqrt{3}+i$

$z = - \sqrt 3 + i$

$\left| z \right| = \sqrt {{{\left( { - \sqrt 3 } \right)}^2} + {1^2}} = \sqrt {3 + 1} = 2$

$\tan \phi = \left| {\dfrac{y}{x}} \right| = \left| { - \dfrac{1}{{\sqrt 3 }}} \right| = \dfrac{1}{{\sqrt 3 }}$

$= \phi = \dfrac{\pi }{6}$

Since

$z$ lies in second quadrant so, argument of

$z$ =

$z = \pi - \phi$

$= \pi - \dfrac{\pi }{6}$

$= \dfrac{{5\pi }}{6}$

Find real values of $\theta$ for which $\left( {\dfrac{{4 + 3i\;\sin \theta }}{{1 - 2i\;\sin \theta }}} \right)$ is purely real.

Given:

$\Rightarrow\dfrac{4+3i \sin(\theta)}{1-2i\sin{\theta}}$ .......(i)

As the function is purely real, the imaginary part of eqn.(i) would be 0

rationalizing eqn.(i) we get,

$\Rightarrow\dfrac{4+3i \sin(\theta)}{1-2i\sin{\theta}} \times \dfrac{1+2i\sin{\theta}}{1+2i\sin{\theta}}$

$\Rightarrow\dfrac{(4+3i \sin(\theta))\times (1+2i\sin(\theta))}{1^2-(2i\sin{\theta})^2}$

.......{

$since,(a+b)(a-b) = a^2 - b^2$ }

$\Rightarrow\dfrac{4+8i\sin\theta+3i\sin\theta+6i^2\sin^2\theta}{1-4i^2\sin^2\theta}$

$\Rightarrow\dfrac{4+11i\sin\theta-6\sin^2\theta}{1+4\sin^2\theta}$

.......{

$since, i^2=-1$ }

$\Rightarrow \dfrac{1}{1+4\sin\theta} \times [(4-6\sin^2\theta)+ i (11\sin\theta)]$

the imainary part is equal to zero for eqn. (i) to be purely real.

$\therefore$

$\Rightarrow\dfrac{11\sin\theta}{1+4\sin^2\theta} = 0$

$\Rightarrow\sin\theta = 0$

the general solution for

$\sin\theta = 0$ is

$\theta \in n\pi \quad where,n \in Z$

Hence, for

$\theta \in n\pi \quad where,n \in Z$

$\Rightarrow\dfrac{4+3i \sin(\theta)}{1-2i\sin{\theta}}$ is purely real.

Prove that ${\left( {1 + i} \right)^4}{\left( {1 + \dfrac{1}{i}} \right)^4} = 16$

$LHS$

${{\left( 1+i \right)}^{4}}{{\left( 1+\dfrac{1}{i} \right)}^{4}}$

$={{\left( 1+i \right)}^{2}}{{\left( 1+\dfrac{1}{i} \right)}^{2}}{{\left( 1+i \right)}^{^{2}}}{{\left( 1+\dfrac{1}{i} \right)}^{2}}$

$={{\left( 1+i \right)}^{2}}{{\left( 1+i \right)}^{^{2}}}{{\left( 1+\dfrac{1}{i} \right)}^{2}}{{\left( 1+\dfrac{1}{i} \right)}^{2}}$

$=\left( {{1}^{2}}+{{i}^{2}}+2i \right)\left( {{1}^{2}}+{{i}^{2}}+2i \right)\left( {{1}^{2}}+\dfrac{1}{{{i}^{2}}}+\dfrac{2}{i} \right)\left( {{1}^{2}}+\dfrac{1}{{{i}^{2}}}+\dfrac{2}{i} \right)$

$=\left( 1-1+2i \right)\left( 1-1+2i \right)\left( {{1}^{2}}+\dfrac{1}{-1}+\dfrac{2}{i} \right)\left( {{1}^{2}}+\dfrac{1}{-1}+\dfrac{2}{i} \right)$

$=2i\times 2i\times \dfrac{2}{i}\times \dfrac{2}{i}$

$=16$

$RHS$

Hence, proved.

$\dfrac{4+3i}{(2+3i)(4-3i)}$ .

$\cfrac{4+3i}{(2+3i)(4-3i)}$

$\cfrac{(4+3i)(4+3i)}{25(2+3i)}$

$\cfrac{7+24i}{25(2+3i)}$

$\cfrac{(7+24i)(2-3i)}{25 \times 13}$

$\cfrac{14-21i+48i+72}{325}$

$\cfrac{86+27i}{325}$

For any two complex numbers $z_1,z_2$ and any two real numbers a and b,
$|az_1-bz_2|^2+|bz_1+az_2|^2=.....$ .

$|az_1-bz_2|^2+|bz_1+az_2|^2$

$=|az_1|^2|+|bz_2|^2-2|abz_1z_2|+|bz_1|^2+|az_2|^2+2|az_1bz_2|$

$=a^2|z_1|^2+b^2|z_2|^2+b^2|z_1|^2+a^2(|z_2|)^2$

$=a^2(|z_1|^2+|z_2|^2)+b^2(|z_1|^2+|z_2|^2)$

$=(a^2+b^2)(|z_1|^2+|z_2|^2)$ .

Among the complex numbers $Z$ satisfying the condition $|z+3-\sqrt{3}i| = \sqrt{3}$ , find the number having the least positive argument.

1360189_1133996_ans_70c21869723a47d590bbd8257bd57be8.jpg

Show that:
$\left(\dfrac{1+i}{\sqrt{2}}\right)^{8}+\left(\dfrac{1-i}{\sqrt{2}}\right)^{8}=2$

$\left(\dfrac{1+i}{\sqrt{2}}\right)^8+\left(\dfrac{1-i}{\sqrt{2}}\right)^2=2$

$\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}$

$\Rightarrow \left(\cos \dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$

$(r\cos \theta +i\sin\theta)^n=r^n(\cos n\theta)+i\sin n\theta)$

$\left(\dfrac{1+i}{\sqrt{2}}\right)^8=(1)^8(\cos(2\pi)+i\sin (2\pi))$

$=(1)$

$\left(\dfrac{1-i}{\sqrt{2}}\right)=1^8(\cos(2\pi +ii\sin(2\pi)))$

$=(1)$ .

1198036_1158330_ans_c356a9d748da4b17bb1b5133adaa00b1.jpg

Find the value of ${ x }^{ 3 }+2{ x }^{ 2 }-3x+21$ if $x=1+2i$

$x^3+2x^2-3x+21$ if

$x=1+2i$

$(x-1)^2=(2i)^2$

$x^2-2x+1=-4$

$x^2-2x+5=0$

$\therefore x^3+2x^2-3x+21$

$=x(x^2-2x+5)+4x^2-8x+21$

$=0+4x^2-8x+21$

$=4(x^2-2x+5)+1$

$=0+1$

$=1$

Value is

$1$ .

Let $z_{1}=2-i,z_{2}=-2+i$ . Find $Re\left(\dfrac {z_{1} z_{2}}{ \vec {z_{1}}}\right)$

1118068_1163939_ans_35ede8965a19452e8350e7c9978d94dd.jpeg

Find the expanded form when $z$ is divided by $z-0.5$

1050648_1177652_ans_6de383a760154f10a4537949f7654bf1.png

Show that if $\left| \dfrac { z - 3 i } { z + 3 i } \right| = 1 ,$ then $z$ is a real number.

$|(\dfrac{z-3i}{z+3i})|=1\\|z-3i|=|z+3i|\\|x+iy-3i|=|x+iy+3i|\\|x+i(y-3)|=|x+i(y+3)|\\\sqrt{x^2+(y-3)^2}=\sqrt{x^2+(y+3)^2}\\x^2+(y-3)^2=x^2+(y+3)^2\\(y-3)^2=(y+3)^2\\y-3=\pm (y+3)\\\therefore y-3 = y+3 or -3 = 3 \>which \>is\> not\> possible\\then\>y-3 = -(y+3)or \>y =0\\so\>z = x+iy = x+0 = x\\\$

z is real number.

If $|z-1|+|z-3| \le 8$ , then find the range of $|z-4|$ .

Given :

$|z+1|+|z-3| \le 8$

To find : Range of

$|z-4|$

Triangle Inequality :

$|z_1 + z_2| \le |z_1| + |z_2|$

$\to |z +1 + 2 -3| \le 8$

$\to |2z - 2| \le 8$

$\to |z -1| \le 4$

$\to |z| \le 3$

Now,

$\to ||z| - 4| \le |z - 4| \le ||z| + 4|$

$\to 1 \le |z - 4| \le 7$

$\therefore |z-4| \in [1,7]$

Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(1+i)(1-i)^{-1}$

$(1+i)(1-i)^{-1}\\\dfrac{1+i}{1-i}\\\dfrac{(1+i)^2}{1-i^2}\\\dfrac{1-1+2i}2=i\\a=0,b=1$

Show that $(-1 + \sqrt 3 i)^3$ is a real number.

$\Rightarrow (\displaystyle -1+\sqrt 3 i)^3$

$\Rightarrow -1+3\sqrt 3i-3(-3)-3\sqrt 3i$

$[\because (a-b)^3=a^3-b^3-3ab(a-b)]$

$\Rightarrow-1+9=8$

It is a real number.

Express $(5-3i)^{3}$ in the form $a+ib$ .

$\textbf{Use} \mathbf{(a-b)^3}\ \textbf{expansion}$

$(5-3i)^3=5^3-3\times5^2\times3i+3\times5\times(3i)^2-(3i)^3$

$[\because (a-b)^3=a^3-3a^2b+3ab^2-b^3]$

$=125-3\times25\times3i+3\times5\times9i^2-27i^3$

$=125-225i-135+27i$

$[\because i^2=-1]$

$=-10-198i$

$\textbf{Hence,} \mathbf{(5-3i)^3}\ \textbf{in}\ \mathbf{(a+ib)}\ \textbf{form is}\ \mathbf{-10-198i}$

Simplify: $\left( -\sqrt { 3 } +\sqrt { -2 } \right) \left( 2\sqrt { 3 } -i \right)$ =(a + ib) $\left( -\sqrt { 3 } +\sqrt { -2 } \right)$ . Find value of a and b.

$\left( -\sqrt { 3 } +\sqrt { -2 } \right) \left( 2\sqrt { 3 } -i \right) =\left( a+ib \right) \left( -\sqrt { 3 } +\sqrt { -2 } \right)$

$\Rightarrow \left( -\sqrt { 3 } +i\sqrt { 2 } \right) \left( 2\sqrt { 3 } -i \right) =\left( a+ib \right) \left( -\sqrt { 3 } +i\sqrt { 2 } \right)$

$\Rightarrow \left( 2\sqrt { 3 } -i \right) =\left( a+ib \right)$

$\Rightarrow a=2\sqrt { 3 } ,\quad b=-1$

Simplify and express the result in the form of $a+ib$
$(\sqrt{3}+i)^6$

2042820_1534994_ans_94566cd3c27148a5affcbadad03d4f90.jpg

Find the possible missing twins for the following number so that they become twin primes.
$89$ .

Twin primes: Two prime numbers are said to be twin primes if there is only one composite number between them.

So, twin prime number can be obtained by adding or subtracting 2 to the given number.

$89 + 2=91$ , or

$89−2=87$

but

$87$ is not a prime number.

$∴89,91$ -Twin primes

Express the following number as the sum of twin primes.
$36$ .

1504807_1658867_ans_0fb7167f97ba41a1bc56ed0cdab5fcbe.jpg

Find the possible missing twins for the following number so that they become twin primes.
$101$ .

Twin primes: Two prime numbers are said to be twin primes if there is only one composite number between them.

So, twin prime number can be obtained by adding or subtracting 2 to the given number.

$101 + 2=103$ , or

$101−2=99$

but

$99$ is not a prime number.

$∴101,103$ -Twin primes

Find the possible missing twins for the following number so that they become twin primes.
$29$ .

Twin primes: Two prime numbers are said to be twin primes if there is only one composite number between them.

So, twin prime number can be obtained by adding or subtracting 2 to the given number.

$29 + 2=31$ or

$29-2=27$

but

$27$ is not a prime number.

$\therefore29, 31$ -Twin primes

Evaluate $(\sqrt {-36}\times \sqrt{-25})$ .

$(\sqrt{-36}\times \sqrt{-25})$

$\Rightarrow \sqrt{36} \ . \sqrt{-1} \ \times \sqrt{25} \ . \sqrt{-1}$

$\Rightarrow(6i\times 5i)=30i^{2}=-30$

Define twin prime numbers.

$A \space pair\space of\space prime\space numbers\space is\space said\space to\space be\space twin\space primes\space it\space they\space differ\space by \space$

$2$ .

$Eg:$

$(3,5)$ &

$(11,13)$

Given the output of the following,
when num1 = 4, num2 = 3, num3 = 2
1. num1 + = num2 + num3
print (num1)
2. num1 = num1 * (num2 + num3)
print (numl)
3. num1 = num2* (num2 + num 3)
print(num1)
4. num1 = '5' + '5'
print (numl)

undefined

1985814_1940777_ans_f57c5ad14e9b476cbe5def461827586e.jpg

Prove that if $\displaystyle x \, + \,\frac{1}{x} \, = \, 2 \, cos \, \alpha, \, then \, x^n \, + \, \frac{1}{x^n} \, = \, 2 \, cos \, n\alpha.$

$x+\cfrac{1}{x}=2\cos\alpha$ (Given)

To Prove:

$x^{n}+\cfrac{1}{2^:{n}}=2\cos n\alpha$
Proof:

$x^{2}+1=2x\cos\alpha$ -----------------(1)

then,

$x^{2n}+1=2x^{k}\cos(n\theta)$ ----------------------(2)

By Induction:

$n=1: x^{2}+1=2\cos\theta$ --------------------------This is just Equation 1 is so true.

From Pythagoras and trigonometric relations, this also means:

$\sqrt{4x^{2}-(x^{2}+1)^{2}}=2x\sin\theta=(x^{2}+1)\tan\theta$

Assume that,

$n=k:\,\,\,x^{2k}+1=2x^{k}\cos(k\theta)$ is true.

Then,

$n=k+1:\,\,x^{2k+1}\cos(k\theta+\theta)$

Now,

$x=\cos\theta+i\sin\theta$

$x^{n}+x^{-n}=\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta=2\cos n\theta$

$\therefore\,\,x^{n}+\cfrac{1}{x^{n}}=2\cos n\alpha$

Given: ${ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=A;{ z }_{ 1 }+{ z }_{ 2 }w+{ z }_{ 3 }{ w }^{ 2 }=B;{ z }_{ 1 }+{ z }_{ 2 }{ w }^{ 2 }+{ z }_{ 3 }{ w }^{ }=C$ where $w$ is cube rott of unity
Prove: ${ \left| A \right| }^{ 2 }+{ \left| B \right| }^{ 2 }+{ \left| C \right| }^{ 2 }=3\left( { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 3 } \right| }^{ 2 } \right)$

${ \left| A \right| }^{ 2 }+{ \left| B \right| }^{ 2 }+{ \left| C \right| }^{ 2 }=A\overline { A } +B\overline { B } +C\overline { C } ....(1)$

$A\overline { A } =\left( { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right) \left( \overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 3 } } \right) ={ z }_{ 1 }{ z }_{ 1 }+{ z }_{ 2 }{ z }_{ 2 }+{ z }_{ 3 }{ z }_{ 3 }+\overline { { z }_{ 1 } } \left( { z }_{ 2 }+{ z }_{ 3 } \right) +\overline { { z }_{ 2 } } \left( { z }_{ 3 }+{ z }_{ 1 } \right) +\overline { { z }_{ 3 } } \left( { z }_{ 1 }+{ z }_{ 2 } \right) ....$

$B\overline { B } =\left( { z }_{ 1 }+{ z }_{ 2 }\omega +{ z }_{ 3 }{ \omega }^{ 2 } \right) \left( \overline { { z }_{ 1 } } +\overline { { z }_{ 2 }\omega } +\overline { { z }_{ 3 }{ \omega }^{ 2 } } \right) =\left( { z }_{ 1 }+{ z }_{ 2 }\omega +{ z }_{ 3 }{ \omega }^{ 2 } \right) \left( \overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } { \omega }^{ 2 }+\overline { { z }_{ 3 } } \omega \right) \quad \left[ \because \overline { \omega } ={ \omega }^{ 2 };\left( \overline { { \omega }^{ 2 } } \right) =\omega \right]$

$={ z }_{ 1 }\overline { { z }_{ 1 } } +{ z }_{ 2 }\overline { { z }_{ 2 } } { \omega }^{ 3 }+{ z }_{ 3 }\overline { { z }_{ 3 } } { \omega }^{ 3 }+\overline { { z }_{ 1 } } \left( { z }_{ 2 }\omega +{ z }_{ 3 }{ \omega }^{ 2 } \right) +\overline { { z }_{ 2 } } \left( { z }_{ 3 }{ \omega }^{ 4 }+{ z }_{ 1 }{ \omega }^{ 2 } \right) +\overline { { z }_{ 3 } } \left( { z }_{ 1 }\omega +{ z }_{ 2 }{ \omega }^{ 2 } \right) ={ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 3 } \right| }^{ 2 }+\overline { { z }_{ 1 } } \left( { z }_{ 2 }\omega +{ z }_{ 3 }{ \omega }^{ 2 } \right) +\overline { { z }_{ 2 } } \left( { z }_{ 3 }{ \omega }^{ 4 }+{ z }_{ 1 }{ \omega }^{ 2 } \right) +\overline { { z }_{ 3 } } \left( { z }_{ 1 }\omega +{ z }_{ 2 }{ \omega }^{ 2 } \right) ....(2)\quad$
Similarly

$C\overline { C } ={ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 3 } \right| }^{ 2 }+\overline { { z }_{ 1 } } \left( { { z }_{ 2 }{ \omega }^{ 2 }+z }_{ 3 }\omega \right) +\overline { { z }_{ 2 } } \left( { z }_{ 3 }{ \omega }^{ 2 }+{ z }_{ 1 }{ \omega }^{ \quad } \right) +\overline { { z }_{ 3 } } \left( { z }_{ 1 }{ \omega }^{ 2 }+{ z }_{ 2 }{ \omega }^{ } \right) ....(3)$
Adding (1), (2) and (3), we get

$A\overline { A } +B\overline { B } +C\overline { C } =3\left[ { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 3 } \right| }^{ 2 } \right] +\overline { { z }_{ 1 } } \left[ { z }_{ 2 }\left( 1+\omega +{ \omega }^{ 2 } \right) +{ z }_{ 3 }\left( 1+{ \omega }^{ 2 }+\omega \right) \right] +\overline { { z }_{ 2 } } \left[ { z }_{ 2 }\left( 1+\omega +{ \omega }^{ 2 } \right) +{ z }_{ 1 }\left( 1+{ \omega }^{ 2 }+\omega \right) \right] +\overline { { z }_{ 3 } } \left[ { z }_{ 2 }\left( 1+\omega +{ \omega }^{ 2 } \right) +{ z }_{ 2 }\left( 1+{ \omega }^{ 2 }+\omega \right) \right]$
From (1) and (2) we conclude

${ \left| A \right| }^{ 2 }+{ \left| B \right| }^{ 2 }+{ \left| C \right| }^{ 2 }=3\left[ { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 3 } \right| }^{ 2 } \right]$

Perform the indicated operations.
$\displaystyle (- \, 1 \, + \, \sqrt{3i})^3.$

$(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}$

$(-1+\sqrt{3i})^{3}=(-1)^{3}+(\sqrt{3i})+(\sqrt{3i})+(\sqrt{3i})+3(-1)^{2}(\sqrt{3i})+3(-1)(\sqrt{3i})^{2}$

$=-1+3i\sqrt{3i}+3\sqrt{3i}-3\times 3i$

$=-1+3i\sqrt{3i}+3\sqrt{3i}-9i$

Solve $\left(\dfrac {1}{1-4i}-\dfrac {2}{1+i}\right)\left(\dfrac {3-4i}{5+i}\right)$

$\left( {\dfrac{1}{{1 - 4i}} - \dfrac{2}{{1 + i}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)$

$= \left( {\dfrac{{1 + i - 2 + 8i}}{{1 + i - 4i + 4}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)$

$= \left( {\dfrac{{ - 1 + 9i}}{{5 - 3i}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)$

$= \left( {\dfrac{{ - 3 + 4i + 27i + 36}}{{25 + 5i - 15i + 3}}} \right)$

$= \left( {\dfrac{{33 + 31i}}{{28 - 10i}}} \right)$

Solve $\dfrac {1}{1+i}$

Rationalizing $\dfrac{1}{{1 + i}}$ .

$\dfrac{1}{{1 + i}} = \dfrac{1}{{1 + i}} \times \dfrac{{1 - i}}{{1 - i}}$

$= \dfrac{{1 - i}}{{\left( {{1^2}} \right) - \left( {{i^2}} \right)}}$

$= \dfrac{{1 - i}}{{1 + 1}}$

$= \dfrac{{1 - i}}{2}$

Express the following number given in the form $a+ib$ .
$(5i)\left(-\dfrac{3}{5}i\right)$ .

$\left( 5i \right) \left( -\cfrac { 3i }{ 5 } \right) \\ =-3{ i }^{ 2 }\\ =-3\left( -1 \right) \qquad \left[ \because { i }^{ 2 }=-1 \right] \\ =3\\ =3+i\left( 0 \right)$

Comparing with a+ib,

$a=3,b=0$

Therefore,

$3+i(0)$

If $|z + 1| = z + 2 (1 + i)$ , then find $z$ .

$\left | z+1 \right |=z+2(1+i)$

Let

$z=x+iy$

$\left | x+iy+1 \right |=x+iy+2(1+i)$

$\sqrt{(x+1)^{2}+y^{2}}=x+iy+2+2i$

$\sqrt{(x+1)^{2}+y^{2}}+0i=(x+2)+i(y+2)$

$\Rightarrow y+2=0$

$y=-2$

And,

$x+2=\sqrt{(x+1)^{2}+y^{2}}$

$x+2=\sqrt{(x+1)^{2}+(-2)^{2}}$

Squaring both sides,

$(x+2)^{2}=(x+1)^{2}+4$

$x^{2}+4+4x=x^{2}+1+2x+4$

$2x=1$

$x=\cfrac{1}{2}$

Therefore,

$z=\cfrac{1}{2}-2i$

Define twin primes.Determine twin primes between $50$ and $1000$

2039359_1269195_ans_77bb1d51ed6f447b81c1722dd71ae5f2.jpg

What are twin-primes? Write all pairs of twin-primes between $50$ and $100$ .

Twin primes are pair of prime numbers which differ by 2

So, all the pairs of twin primes between 50 and 100 are - (59,61) and (71,73)

Express the following number as the sum of twin primes.
$120$ .

The twin prime numbers are pair of prime numbers whose difference is equal to 2

Now, for 120, first divide 120 by 2 , we get 60 and then by adding 1 and subtracting 1 , we get two numbers

60-1=59

60+1=61

Here, 59 and 61 are prime numbers which have a difference of 2 and add up to form 120

59+61=120

Express the following number as the sum of twin primes.
$84$ .

1504809_1658868_ans_c1eb8f52e51a4e188aa6f9869c2b27f4.jpg

Number Theory - Class 11 Commerce Applied Mathematics - Extra Questions

If Z1Z_{1} and Z2Z_{2} are two complex numbers and c>0,c> 0, then prove that|z1+z2|2≤(1+c)|z1|2+(1+c−1)|z2|2\left | z_{1}+z_{2} \right |^{2}\leq \left ( 1+c \right )\left | z_{1} \right |^{2}+\left ( 1+c^{-1} \right )\left | z_{2} \right |^{2}

Multiply 3√−7−5√−23\sqrt { -7 } -5\sqrt { -2 } by 3√−7+5√−23\sqrt { -7 } +5\sqrt { -2 }

Find the modulus of complex number −2+2√3i-2+2\sqrt{3}i.

Compute :(1+i)−1(1 \, + \, i)^{-1}

Locate the points representing the complex number for which log1/2|z−1|+4|z−1|−2>1{ \log }_{ 1/2 }\dfrac { \left| z-1 \right| +4 }{ \left| z-1 \right| -2 } >1

compute.[(cosθ+isinθ)(cosθ−isinθ)]=?[ (cos \theta \, + \, i \, sin \, \theta ) (\, cos \theta \, - \, i sin \theta ) ] = ?

Prove the identity,|1+z1¯z2|2+|z1−z2|2=(1+|z1|2)(1+|z2|2)|1+z_1\bar{z_2}|^2+|z_1-z_2|^2=(1+|z_1|^2)(1+|z_2|^2)

Solve (1+i1−i)24\left(\dfrac{1+i}{1-i}\right)^{24}

Prove the identity |1−z1z2|2−|z1−z2|2=(1−|z1|2)(1−|z2|2)|1-z_1{z_2}|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)

The HCF of two or more prime numbers is always __________

If (a+ib)(c+id)=A+iB\left( {a + ib} \right)\left( {c + id} \right) = A + iB, then show that (a2+b2)(c2+d2)=A2+B2\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {A^2} + {B^2}.

If (a+ib)(c+id)=A+iB(a+ib)(c+id)=A+iB, then show that (a2+b2)(c2+d2)=A2+B2({a^2} + {b^2})({c^2} + {d^2}) = {A^2} + {B^2}

Define and Give one example of each :a) Parallel linesb) Intersecting linesc) Perpendicular lined) Prime Numbere) Twin Primef) Even Number

Find the values of the following:(1+i√3)3(1+i\sqrt {3})^{3}

Find the value of (−1+√−3)2+(−1−√−3)2{(-1+\sqrt{-3})}^{2}+{(-1-\sqrt{-3})}^{2}.

Find the modules and the argument of the complex no Z=1+iZ=1+i

If a=−1+√3i2a = \frac{{ - 1 + \sqrt {3i} }}{2} , b=−1−√3i2b = \frac{{ - 1 - \sqrt {3i} }}{2} then show that a2=b\,\,{a^2} = b and b2=a\,\,{b^2} = a.

z=2+3iz=2+3i find |z||z|

Find the principal arguments of the following Complex Number(i) 5+5i5 + 5i

If z1,z2∈C,{z_1},{z_2} \in C, then |z1z2|\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right|.

Let Z=x+iyZ = x + iy and ω=1−iZZ−i\omega = \dfrac {1 - iZ}{Z - i}. If |ω|=1|\omega| = 1, show that ZZ is purely real.

Find the modulus of 1+i1−i−1−i1+i\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}

Solve: x2−(3√2−2i)x−√2i=0x^2 - (3\sqrt{2} - 2i)x - \sqrt{2}i = 0

State whether the given statement is true or false ¯(z−1)=(ˉz)−1\overline{(z^{-1})} \, = \, (\bar{z})^{-1}

Express 5+i√22i\dfrac{{5 + i\sqrt 2 }}{{2i}} in the form of x+iyx + iy.

If (1+i)22−i=x−iy,\dfrac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x - iy, then find the value of x+yx + y.

Find the real and imaginary parts of the complex number z=3i20−i192i−1\dfrac{3i^{20}-i^{19}}{2i-1}

If (1+i1−i)3−(1−i1+i)3=x+iy{\left( {\frac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\frac{{1 - i}}{{1 + i}}} \right)^3} = x + iy, then find (x,y)(x,\,y),

For the complex number√37+√−19\sqrt { 37 } + \sqrt { - 19 }.Real part is √37\sqrt 37 and imaginary is √19\sqrt 19Enter 1 if true or 0 for false

What are twin-primes ?

Express the following in the form of a = ib, a,bϵ\epsilonR i=√−1i = \sqrt{-1}. State the values of a and b. (1+2i)(−2+i)(1 + 2i)(-2 + i)

If ABC are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find value of xyz

Give three pairs of prime numbers whose difference is 22. [Remark: Two prime numbers whose difference is 22 are called twin primes].

If z≠0z \neq 0 is a complex number,then prove that Re(z)=0⟹Im(z2)=0Re(z) = 0 \Longrightarrow Im(z^2)=0.

If (a+ib)(c+id)(e+if)(g+ih)=A+iB(a + ib) (c + id) (e + if)(g +ih) = A + iB, then prove that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2(a^2 + b^2) (c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2.

If z=x+iyz = x + iy and w=(1−iz)(z−i)w = \dfrac{(1 - iz)}{(z - i)} and |w|=1\left|w\right| = 1, then prove that zz is purely real.

If (x+iy)3=u+iv(x+iy)^3=u+iv, then prove that ux+vy=4(x2−y2)\displaystyle \frac{u}{x}+\frac{v}{y} =4(x^2 - y^2).

If z1z_1 and z2z_2 are complex numbers and u=√z1z2u = \sqrt{z_1z_2}, then prove that |z1|+|z2|=|z1+z22+u|+|z1+z22−u|\left|z_1\right| + \left|z_2\right| = \left|\displaystyle \frac{z_1 + z_2}{2}+u\right| + \left|\displaystyle \frac{z_1 + z_2}{2}-u\right|.

If |z1+z2|=|z1|+|z2| \left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right|, then show that arg(z1)=arg(z2)arg(z_1) = arg(z_2).

If |z1−z2|=|z1|+|z2|\left|z_1 - z_2\right| = \left|z_1\right| + \left|z_2\right|, then prove that arg(z1)−arg(z2)=πarg(z_1) - arg (z_2) = \pi . I

Find the minimum value of |1+z|+|1−z|.\left | 1+z \right |+\left | 1-z \right |.

2i=ei(lnx)2^{i}=e^{i(lnx)}.

Compute the modulus of complex number z=√5−2iz=\sqrt5-2i

If a+ib=(x+i)22x2+1\displaystyle a+ib=\dfrac { { \left( x+i \right) }^{ 2 } }{ { 2x }^{ 2 }+1 } , prove that a2+b2=(x2+1)2(2x2+1)2\displaystyle { a }^{ 2 }+{ b }^{ 2 }=\dfrac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( 2x^2+1 \right) }^{ 2 } } .

Express 2+i(1+i)(1−2i)\cfrac{2+i}{(1+i)(1-2i)} in the form of a+iba+ib. Find its modulus and argument.

Find the real and imaginary parts of the complex number a+iba−ib\dfrac {a + ib}{a - ib}

Find the number of integral solution of (1−i)x=2x(1-i)^x=2^x.

If zz and α\alpha complex numbers such that |z|=|α|=r,r>0|z| = |\alpha| = r, r > 0 and ω=z−¯αr2+z¯α\omega = \dfrac {z - \overline {\alpha}}{r^{2} + z\overline {\alpha}}, find Re(ω)Re(\omega).

If the ratio (1−z1+z)\left (\dfrac {1 - z}{1 + z}\right ) is purely imaginary, then find value of |z||z|.

If a=1+i√2a=\cfrac { 1+i }{ \sqrt { 2 } } , find the value of a6+a4+a2+1{ a }^{ 6 }+{ a }^{ 4 }+{ a }^{ 2 }+1

Find the points of the plane which satisfy the following equations.|z−2z+3|=1.\displaystyle \left | \, \frac{z \, - \, 2}{z \, + \, 3} \, \right | \, = \, 1.

Find the points of the plane which satisfy the following equations.|z+iz−3i|=1.\displaystyle \left | \, \frac{z \, + \, i}{z \, - \, 3i} \, \right | \, = \, 1.

Simplify the following :(1−i)31−i3\dfrac{(1 \, - \, i)^3}{1 \, - \, i^3}

Express the following in the form A + iB :11−cosθ+2isinθ\dfrac{1}{1 \, - \, \cos \theta \, + \, 2i \, \sin \theta}

Put the following in the form A + iB :(a+ib)2(a−ib)−(a−ib)2(a+ib)\dfrac{(a \, + \, ib)^2}{(a \, - \, ib)} \, - \, \dfrac{(a \, - \, ib)^2}{(a \, + \, ib)}

If Zr=(cosrπ10+isinrπ10).Z_r \, = \, \left ( \cos\dfrac{r\pi }{10} \, + \, i \, \sin \dfrac{r\pi }{10}\right ). Then, find the value of Z1⋅Z2⋅Z3⋅Z4Z_1 \cdot Z_2 \cdot Z_3 \cdot Z_4

If iz3+z2−z+i=0i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0, then show that |z|=1\left| z \right| =1.

Thecomplexnumber2n(1+i)2n+(1+i)2n2n,n∈Z,isequaltoThe\quad complex\quad number\quad \cfrac { { 2 }^{ n } }{ { (1+i) }^{ 2n } } +\cfrac { { (1+i) }^{ 2n } }{ { 2 }^{ n } } ,\quad n\quad \in Z,\quad is\quad equal\quad to\quad

Find all the values of zz which satisfy the equationexp||z|2−|z|+4|z|2+1log2|=log√2|3√15+11i|\left| \dfrac { { \left| z \right| }^{ 2 }-\left| z \right| +4 }{ { \left| z \right| }^{ 2 }+1 } \log2 \right| ={ \log }_{ \sqrt { 2 } }\left| 3\sqrt { 15 } +11i \right|

Find the modulus and argument of the complex number:11+i\dfrac{1}{1+i}

For any two complex number z1z_1 and z2z_2 prove that Re(z1z2)=Rez1Rez2−Imz1Imz1Re(z_1z_2)= Rez_1Rez_2-I mz_1Imz_1

let z1z_1 and z2z_2 be two complex number such that |1−z1z2|2−|z1−z2|2=k(1−|z1|2)(1−|z2|2)|1-{z_1}z_2|^2-|z_1-z_2|^2=k\left(1-|z_1|^2\right)\left(1-|z_2|^2\right) find the value of kk

If \left( {a + ib} \right) = \frac{{1 + i}}{{1 - i}}\left( {a + ib} \right) = \frac{{1 + i}}{{1 - i}} , then prove that \left( {{a^2} + {b^2}} \right) = 1\left( {{a^2} + {b^2}} \right) = 1

How solve and what is its meaningz_{1}=2+3iz_{1}=2+3iz_{2}=3+4iz_{2}=3+4i\left|z_{1}+z_{2}\right|=?\left|z_{1}+z_{2}\right|=?

Let \dfrac{1}{a+ib}=c+id\dfrac{1}{a+ib}=c+id for non-zero a,ba,b, then find c,dc,d.

Find the modulus and the arguments of each of the complex numbers is z=-\sqrt{3}+iz=-\sqrt{3}+i

Find real values of \theta \theta for which\left( {\dfrac{{4 + 3i\;\sin \theta }}{{1 - 2i\;\sin \theta }}} \right)\left( {\dfrac{{4 + 3i\;\sin \theta }}{{1 - 2i\;\sin \theta }}} \right) is purely real.

Prove that {\left( {1 + i} \right)^4}{\left( {1 + \dfrac{1}{i}} \right)^4} = 16{\left( {1 + i} \right)^4}{\left( {1 + \dfrac{1}{i}} \right)^4} = 16

\dfrac{4+3i}{(2+3i)(4-3i)}\dfrac{4+3i}{(2+3i)(4-3i)}.

For any two complex numbers z_1,z_2z_1,z_2 and any two real numbers a and b,|az_1-bz_2|^2+|bz_1+az_2|^2=.....|az_1-bz_2|^2+|bz_1+az_2|^2=..... .

Among the complex numbers ZZ satisfying the condition |z+3-\sqrt{3}i| = \sqrt{3}|z+3-\sqrt{3}i| = \sqrt{3}, find the number having the least positive argument.

Show that:\left(\dfrac{1+i}{\sqrt{2}}\right)^{8}+\left(\dfrac{1-i}{\sqrt{2}}\right)^{8}=2\left(\dfrac{1+i}{\sqrt{2}}\right)^{8}+\left(\dfrac{1-i}{\sqrt{2}}\right)^{8}=2

Find the value of { x }^{ 3 }+2{ x }^{ 2 }-3x+21{ x }^{ 3 }+2{ x }^{ 2 }-3x+21 if x=1+2ix=1+2i

Let z_{1}=2-i,z_{2}=-2+iz_{1}=2-i,z_{2}=-2+i. Find Re\left(\dfrac {z_{1} z_{2}}{ \vec {z_{1}}}\right)Re\left(\dfrac {z_{1} z_{2}}{ \vec {z_{1}}}\right)

Find the expanded form when zz is divided by z-0.5z-0.5

Show that if \left| \dfrac { z - 3 i } { z + 3 i } \right| = 1 ,\left| \dfrac { z - 3 i } { z + 3 i } \right| = 1 , then zz is a real number.

If |z-1|+|z-3| \le 8|z-1|+|z-3| \le 8, then find the range of |z-4||z-4|.

Express the following in the form of a = ib, a,b\epsilon\epsilonR i = \sqrt{-1}i = \sqrt{-1}. State the values of a and b. (1+i)(1-i)^{-1}(1+i)(1-i)^{-1}

Show that (-1 + \sqrt 3 i)^3(-1 + \sqrt 3 i)^3 is a real number.

If $Z_{1}$ and $Z_{2}$ are two complex numbers and $c> 0,$ then prove that
$\left | z_{1}+z_{2} \right |^{2}\leq \left ( 1+c \right )\left | z_{1} \right |^{2}+\left ( 1+c^{-1} \right )\left | z_{2} \right |^{2}$

Multiply $3\sqrt { -7 } -5\sqrt { -2 }$ by $3\sqrt { -7 } +5\sqrt { -2 }$

Find the modulus of complex number $-2+2\sqrt{3}i$ .

Compute :
$(1 \, + \, i)^{-1}$

Locate the points representing the complex number for which ${ \log }_{ 1/2 }\dfrac { \left| z-1 \right| +4 }{ \left| z-1 \right| -2 } >1$

compute.
$[ (cos \theta \, + \, i \, sin \, \theta ) (\, cos \theta \, - \, i sin \theta ) ] = ?$

Prove the identity, $|1+z_1\bar{z_2}|^2+|z_1-z_2|^2=(1+|z_1|^2)(1+|z_2|^2)$

Solve $\left(\dfrac{1+i}{1-i}\right)^{24}$

Prove the identity $|1-z_1{z_2}|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

If $\left( {a + ib} \right)\left( {c + id} \right) = A + iB$ , then show that $\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {A^2} + {B^2}$ .

If $(a+ib)(c+id)=A+iB$ , then show that $({a^2} + {b^2})({c^2} + {d^2}) = {A^2} + {B^2}$

Define and Give one example of each :
a) Parallel lines
b) Intersecting lines
c) Perpendicular line
d) Prime Number
e) Twin Prime
f) Even Number

Find the values of the following:
$(1+i\sqrt {3})^{3}$

Find the value of ${(-1+\sqrt{-3})}^{2}+{(-1-\sqrt{-3})}^{2}$ .

Find the modules and the argument of the complex no $Z=1+i$

If $a = \frac{{ - 1 + \sqrt {3i} }}{2}$ , $b = \frac{{ - 1 - \sqrt {3i} }}{2}$ then show that $\,\,{a^2} = b$ and $\,\,{b^2} = a$ .

$z=2+3i$ find $|z|$

Find the principal arguments of the following Complex Number
(i) $5 + 5i$

If ${z_1},{z_2} \in C,$ then $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right|$ .

Let $Z = x + iy$ and $\omega = \dfrac {1 - iZ}{Z - i}$ . If $|\omega| = 1$ , show that $Z$ is purely real.

Find the modulus of $\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}$

Solve: $x^2 - (3\sqrt{2} - 2i)x - \sqrt{2}i = 0$

State whether the given statement is true or false $\overline{(z^{-1})} \, = \, (\bar{z})^{-1}$

Express $\dfrac{{5 + i\sqrt 2 }}{{2i}}$ in the form of $x + iy$ .

If $\dfrac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x - iy,$ then find the value of $x + y$ .

Find the real and imaginary parts of the complex number z= $\dfrac{3i^{20}-i^{19}}{2i-1}$

If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\frac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$ , then find $(x,\,y)$ ,

For the complex number $\sqrt { 37 } + \sqrt { - 19 }$ .
Real part is $\sqrt 37$ and imaginary is $\sqrt 19$
Enter 1 if true or 0 for false

Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(1 + 2i)(-2 + i)$

Give three pairs of prime numbers whose difference is $2$ . [Remark: Two prime numbers whose difference is $2$ are called twin primes].

If $z \neq 0$ is a complex number,then prove that $Re(z) = 0 \Longrightarrow Im(z^2)=0$ .

If $(a + ib) (c + id) (e + if)(g +ih) = A + iB$ , then prove that $(a^2 + b^2) (c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$ .

If $z = x + iy$ and $w = \dfrac{(1 - iz)}{(z - i)}$ and $\left|w\right| = 1$ , then prove that $z$ is purely real.

If $(x+iy)^3=u+iv$ , then prove that $\displaystyle \frac{u}{x}+\frac{v}{y} =4(x^2 - y^2)$ .

If $z_1$ and $z_2$ are complex numbers and $u = \sqrt{z_1z_2}$ , then prove that
$\left|z_1\right| + \left|z_2\right| = \left|\displaystyle \frac{z_1 + z_2}{2}+u\right| + \left|\displaystyle \frac{z_1 + z_2}{2}-u\right|$ .

If $\left|z_1 + z_2\right| = \left|z_1\right| + \left|z_2\right|$ , then show that $arg(z_1) = arg(z_2)$ .

If $\left|z_1 - z_2\right| = \left|z_1\right| + \left|z_2\right|$ , then prove that $arg(z_1) - arg (z_2) = \pi$ . I

Find the minimum value of $\left | 1+z \right |+\left | 1-z \right |.$

$2^{i}=e^{i(lnx)}$ .

Compute the modulus of complex number $z=\sqrt5-2i$

If $\displaystyle a+ib=\dfrac { { \left( x+i \right) }^{ 2 } }{ { 2x }^{ 2 }+1 }$ , prove that $\displaystyle { a }^{ 2 }+{ b }^{ 2 }=\dfrac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ { \left( 2x^2+1 \right) }^{ 2 } }$ .

Express $\cfrac{2+i}{(1+i)(1-2i)}$ in the form of $a+ib$ . Find its modulus and argument.

Find the real and imaginary parts of the complex number $\dfrac {a + ib}{a - ib}$

Find the number of integral solution of $(1-i)^x=2^x$ .

If $z$ and $\alpha$ complex numbers such that $|z| = |\alpha| = r, r > 0$ and $\omega = \dfrac {z - \overline {\alpha}}{r^{2} + z\overline {\alpha}}$ , find $Re(\omega)$ .

If the ratio $\left (\dfrac {1 - z}{1 + z}\right )$ is purely imaginary, then find value of $|z|$ .

If $a=\cfrac { 1+i }{ \sqrt { 2 } }$ , find the value of ${ a }^{ 6 }+{ a }^{ 4 }+{ a }^{ 2 }+1$

Find the points of the plane which satisfy the following equations.
$\displaystyle \left | \, \frac{z \, - \, 2}{z \, + \, 3} \, \right | \, = \, 1.$

Find the points of the plane which satisfy the following equations.
$\displaystyle \left | \, \frac{z \, + \, i}{z \, - \, 3i} \, \right | \, = \, 1.$

Simplify the following :
$\dfrac{(1 \, - \, i)^3}{1 \, - \, i^3}$

Express the following in the form A + iB :
$\dfrac{1}{1 \, - \, \cos \theta \, + \, 2i \, \sin \theta}$

Put the following in the form A + iB :
$\dfrac{(a \, + \, ib)^2}{(a \, - \, ib)} \, - \, \dfrac{(a \, - \, ib)^2}{(a \, + \, ib)}$

If $Z_r \, = \, \left ( \cos\dfrac{r\pi }{10} \, + \, i \, \sin \dfrac{r\pi }{10}\right ).$ Then, find the value of $Z_1 \cdot Z_2 \cdot Z_3 \cdot Z_4$

If $i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0$ , then show that $\left| z \right| =1$ .

$The\quad complex\quad number\quad \cfrac { { 2 }^{ n } }{ { (1+i) }^{ 2n } } +\cfrac { { (1+i) }^{ 2n } }{ { 2 }^{ n } } ,\quad n\quad \in Z,\quad is\quad equal\quad to\quad$

Find all the values of $z$ which satisfy the equation
exp $\left| \dfrac { { \left| z \right| }^{ 2 }-\left| z \right| +4 }{ { \left| z \right| }^{ 2 }+1 } \log2 \right| ={ \log }_{ \sqrt { 2 } }\left| 3\sqrt { 15 } +11i \right|$

Find the modulus and argument of the complex number:
$\dfrac{1}{1+i}$

For any two complex number $z_1$ and $z_2$ prove that $Re(z_1z_2)= Rez_1Rez_2-I mz_1Imz_1$

let $z_1$ and $z_2$ be two complex number such that $|1-{z_1}z_2|^2-|z_1-z_2|^2=k\left(1-|z_1|^2\right)\left(1-|z_2|^2\right)$ find the value of $k$

If $\left( {a + ib} \right) = \frac{{1 + i}}{{1 - i}}$ , then prove that $\left( {{a^2} + {b^2}} \right) = 1$

How solve and what is its meaning
$z_{1}=2+3i$
$z_{2}=3+4i$
$\left|z_{1}+z_{2}\right|=?$

Let $\dfrac{1}{a+ib}=c+id$ for non-zero $a,b$ , then find $c,d$ .

Find the modulus and the arguments of each of the complex numbers is $z=-\sqrt{3}+i$

Find real values of $\theta$ for which $\left( {\dfrac{{4 + 3i\;\sin \theta }}{{1 - 2i\;\sin \theta }}} \right)$ is purely real.

Prove that ${\left( {1 + i} \right)^4}{\left( {1 + \dfrac{1}{i}} \right)^4} = 16$

$\dfrac{4+3i}{(2+3i)(4-3i)}$ .

For any two complex numbers $z_1,z_2$ and any two real numbers a and b,
$|az_1-bz_2|^2+|bz_1+az_2|^2=.....$ .

Among the complex numbers $Z$ satisfying the condition $|z+3-\sqrt{3}i| = \sqrt{3}$ , find the number having the least positive argument.

Show that:
$\left(\dfrac{1+i}{\sqrt{2}}\right)^{8}+\left(\dfrac{1-i}{\sqrt{2}}\right)^{8}=2$

Find the value of ${ x }^{ 3 }+2{ x }^{ 2 }-3x+21$ if $x=1+2i$

Let $z_{1}=2-i,z_{2}=-2+i$ . Find $Re\left(\dfrac {z_{1} z_{2}}{ \vec {z_{1}}}\right)$

Find the expanded form when $z$ is divided by $z-0.5$

Show that if $\left| \dfrac { z - 3 i } { z + 3 i } \right| = 1 ,$ then $z$ is a real number.

If $|z-1|+|z-3| \le 8$ , then find the range of $|z-4|$ .

Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(1+i)(1-i)^{-1}$

Show that $(-1 + \sqrt 3 i)^3$ is a real number.

Express $(5-3i)^{3}$ in the form $a+ib$ .

Simplify: $\left( -\sqrt { 3 } +\sqrt { -2 } \right) \left( 2\sqrt { 3 } -i \right)$ =(a + ib) $\left( -\sqrt { 3 } +\sqrt { -2 } \right)$ . Find value of a and b.

Simplify and express the result in the form of $a+ib$
$(\sqrt{3}+i)^6$

Find the possible missing twins for the following number so that they become twin primes.
$89$ .