If $$z_1$$ and $$z_2$$ are two complex numbers, then $$Re(z_1z_2)$$ is:

$$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 4

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

0%
$\sqrt{5}$
0%
$5\sqrt{2}$
0%
$\sqrt{10}$
0%
$5$

Explanation

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

$=|\dfrac{6-3i+2i+1}{1+i}|$

$=|\dfrac{7-i}{1+i}|$

$=|\dfrac{(7-i)(1-i)}{2}|$

$=|\dfrac{7-7i-i-1}{2}|$

$=|\dfrac{6-8i}{2}|$

$=|3-4i|$

$=\sqrt{3^{2}+4^{2}}$

$=\sqrt{25}$

$=5$

$z_1$ and

$z_2$ are two complex numbers, then

$Re(z_1z_2)$ is:

0%
$Re(z_1)Re(z_2)$
0%
$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$
0%
$Im(z_1).Re(z_2)$
0%
$Re(z_1).Im(z_2)$

Explanation

$z_{1}=a_{1}+ib_{1}; z_{2}=a_{2}+ib_{2}$

$z_{1}z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})$

$=a_{1}a_{2}+(a_{1}b_{2})i+(a_{2}b_{1})i-b_{1}b_{2}$

$=a_{1}a_{2}-b_{1}b_{2}+(a_{1}b_{2}+a_{2}b_{1})i$

$\therefore Re(z_{1}z_{2})=a_{1}a_{2}-b_{1}b_{2}$

$=Re(z_{1})\cdot Re(z_{2})-Im(z_{1})\cdot Im(z_{2})$

The real part of

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

0%
$1$
0%
$16$
0%
$16\omega ^2$
0%
$\displaystyle \frac{-3}{25}$

Explanation

$\displaystyle \frac{1+i}{3-i}=\frac{(1+i)(3+i)}{(3-i)(3+i)}=\frac{3-1+4i}{3^{2}+(1)^{2}}=\frac{2+4i}{10}=\frac{1+2i}{5}$

$\displaystyle \therefore \left ( \frac{1+2i}{5} \right )^{2}$

$\displaystyle =\frac{1^{2}-2^{2}+4i}{25}$

$\displaystyle =\frac{-3+4i}{25}$

$\displaystyle \therefore$ Real part

$=\dfrac{-3}{25}$

$z=-1+3i$ then

$z^2+2z+10=$

0%
$0$
0%
$1$
0%
$–1$
0%
$2$

Explanation

$z^{2}+2z+10$

$=(z+1)^{2}+10-1$

$=(z+1)^{2}+9$

$=(-1+3i+1)^{2}+9$

$=(3i)^{2}+9$

$=-9+9$

$=0$

Hence,

$z^{2}+2z+10=0$ where

$z=-1+3i$ .

The value of

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

0%
$0$
0%
$5i$
0%
$4i$
0%
$3i$

Explanation

The given series is a G.P with a common ratio of

$(1+i)$ .
Hence

$1+(1+i)+(1+i)^{2}+(1+i)^{3}$

$=\dfrac{1-(1+i)^{4}}{1-(1+i)}$

Now

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

$=\sqrt{2}(e^{i\frac{\pi}{4}})$
Hence

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

$=\dfrac{1-(\sqrt{2}(e^{i\frac{\pi}{4}}))^{4}}{-i}$

$=\dfrac{1-4(e^{i\pi})}{-i}$

$=\dfrac{1-4(-1)}{-i}$

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

$=5i$

$\left ( 5+3i \right )(x+iy)=3-4i$ then

$34x =$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$(5+3i)(x+iy)=3-4i$
L.H.S.

$=(5x-3y)+(5y+3x)i=3-4i$

$5x-3y=3$

$5y+3x=-4$
Solving the equations simultaneously,

$x=\dfrac{3}{34} ; y=\dfrac{-29}{34}$

$\therefore 34x=34\times \dfrac{3}{34}$

$=3$

The simplified value of

$\displaystyle \frac{1-i}{1+i}$ is:

0%
$i$
0%
$-i$
0%
$1$
0%
$-2i$

Explanation

Let

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

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

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

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

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

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

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

$=-i$

The minimum value of

$|\mathrm{z}|+|\mathrm{z}-1|+|\mathrm{z}-2|$ is

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

Let

$z = x+iy$

$f(z) = |z| + |z-1| +|z-2|$

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

For minimum value,

$y= 0$

$f(x) = |x| + |x-1| +|x-2|$ for all

$x$

The function can be expressed as follows

$f(x) = 3x -3$ ,

$x \geq 2$

$f(x) = x+1,$

$1\leq x<2$

$f(x) = 3- x$ ,

$0\leq x\leq 1$

$f(x) = -3x +3$ ,

$x <0$

Hence,

$f(min) = f(1) = 2$

$z_{1},\ z_{2}$ are roots of equation

$z^{2}-az+a^{2}=0$ , then

$|\displaystyle \frac{z_{1}}{z_{2}}|=$

0%
$a^{2}$
0%
a
0%
2a
0%
1

Explanation

$z_{1}$ and $z_{2}$ are roots of $z^{2}-az+a^{2}=0$
$\therefore z_{1},z_{2}=\dfrac{a\pm \sqrt{a^{2}-4a^{2}}}{2}$
$=a(\dfrac{1\pm i\sqrt{3}}{2})$
$\therefore z_{1} , z_{2}$ are multiple of complex cube roots of unity.
$\therefore |\dfrac{z_{1}}{z_{2}}|=|\dfrac{\omega ^{2}}{\omega }|=|\omega |=1$

$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$ then

0%
$x>1$
0%
$x<1$
0%
$x<2$
0%
$x>2$

Explanation

$Let \quad z= x+iy$

$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$

$\Rightarrow |\mathrm{z}-2| < |z|$

$\Rightarrow |x - 2 + iy| < |x + iy|$

$\Rightarrow \sqrt{x^2 - 4x + 4 + y^2} < \sqrt{x^2 + y^2}$

$\Rightarrow -4x + 4 < 0$

$\Rightarrow x > 1.$

The modulus of

$(1 + i) (3 + 4i) =$

0%
$\sqrt{50}$
0%
$\sqrt{25}$
0%
$10$
0%
$10 \sqrt{2}$

Explanation

$z_{1},z_{2}$ being two complex numbers,

$|z_{1}.z_{2}|=|z_{1}|.|z_{2}|$

$\therefore |(1+i)(3+4i)|$

$=|1+i|\cdot|3+4i|$

$=\sqrt{1^{2}+1^{2}}\times \sqrt{3^{2}+4^{2}}$

$=\sqrt{2}\sqrt{25}$

$=\sqrt{50}$

The region represented by z such that

$\left | \dfrac{\mathrm{z}-a}{z+a} \right |=1({\rm Im} (a) = 0)$ is

0%
$y=0$
0%
$x=0$
0%
$x+y=0$
0%
$x-y=0$

Explanation

Let

$z=x+iy$

$\Rightarrow \left | (x-a)+iy \right |=\left | (x+a)+iy \right |$

$\Rightarrow (x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}$

$\Rightarrow \not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax$

$\Rightarrow 0$

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

0%
4
0%
3
0%
2
0%
1

Explanation

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

$=\left | \dfrac{\not{1}+2i-\not{1}-(\not{1}-2i)-\not{1} }{(1^{2}+1^{2})^{2}}\right |$

$=\left | \dfrac{4i}{4} \right |=1$

$z_{1},\ z_{2}$ are any two complex numbers then

$\left |z_{1^{+}}\sqrt{\mathrm{z}_{1}^{2}-\mathrm{z}_{2}^{2}} \right |+\left|\mathrm{z}_{1}-\sqrt{\mathrm{z}_{1}^{2}-\mathrm{z}_{2}^{2}} \right|$ is equal to

0%
$|\mathrm{z}_{1}|$
0%
$|\mathrm{z}_{2}|$
0%
$|z_{1}+\mathrm{z}_{2}|+|\mathrm{z}_{1}-\mathrm{z}_{2}|$
0%
$|\mathrm{z}_{1}+\mathrm{z}_{2}|-|\mathrm{z}_{1}-\mathrm{z}_{2}|$

Explanation

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

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

$=4{ z }_{ 1 }^{ 2 }\longrightarrow \left( 1 \right)$ (On simplifying )

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

$=4{ z }_{ 1 }^{ 2 }\longrightarrow \left( 2 \right)$ (On Simplifying )

$\Rightarrow (1)=(2)$

Hence, the answer is

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| +\left| { z }_{ 1 }-{ z }_{ 2 } \right|.$

$z_1$ ,

$z_2$ ,

$z_3$ are complex numbers such that

$\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =\left| \dfrac { 1 }{ { z }_{ 1 } } +\dfrac { 1 }{ { z }_{ 2 } } +\dfrac { 1 }{ { z }_{ 3 } } \right| =1$ , then

$|z_1+z_2+z_3|$ is:

0%
Equal to 1
0%
Less than 1
0%
Greater than 3
0%
Equal to 3

Explanation

Let

$z_1 = cis \theta_1$ ,

$z_2 = cis\theta_2$ and

$z_3 =cis\theta_3$

We have,

$\left | \dfrac{1}{z_1} + \dfrac{1}{z_2} +\dfrac{1}{z_3} \right| = 1$

$| cis(-\theta_1) + cis(-\theta_2) + cis (-\theta_3) | = 1$

Hence,

$| (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) - i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | = 1$

Hence,

$| (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) + i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | =1$

Hence,

$|z_1+ z_2+ z_3 | =1$

$\left| \frac { { z }_{ 1 }+{ z }_{ 2 } }{ { z }_{ 1 }-{ z }_{ 2 } } \right| =1$ if

$\frac{z_1}{z_2}$ is purely imaginary
II. If z is purely real then

$z=\overline{z}$

0%
Only II is true
0%
Only I is true
0%
Both I and II are true
0%
I is false but II is true

Explanation

$\displaystyle \left | \frac{z_{1}-z_{2}}{z_{1}-z_{2}} \right |=1$
Dividing by

$z_2$

$\displaystyle \left | \frac{z_{1}/z_{2}-1}{z_{1}/z_{2}+1} \right |=1$
Let

$z_1/z_2$ be a complex number

$u=x+iy$

$\displaystyle \Rightarrow \left | \frac{u-1}{u+1} \right |=1$

$\displaystyle \Rightarrow \left | u-1 \right |^{2}=\left | u+1 \right |^{2}$

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

$\therefore (x-1)^{2}=(x+1)^{2}$
Which is possible only if x=0

$\therefore u=iy$ ( purely imaginary)
ii) If

$z$ is purely real, then

$z=x$ (imaginary part zero)
Hence

$\bar{z}$ is also equal to

$x$ .

$\Rightarrow z = \bar{z}$

$(x+iy)(2+cos\theta+isin\theta)=3$ then

$x^{2}+y^{2}-4x+3$ is

0%
$0$
0%
$1$
0%
$3$
0%
$4$

Explanation

$(x+iy)(2+cos\theta+isin\theta)=3$

$2x+x(cos\theta+isin\theta)+i2y+iy(cos\theta+isin\theta)=3$

$2x+xcos\theta-ysin\theta+i(xsin\theta+2y+ycos\theta)=3$

Therefore, equating the real and imaginary parts of LHS and RHS gives us

$2x+xcos\theta-ysin\theta=3$

$xsin\theta+ycos\theta+2y=0$

$\Rightarrow xcos\theta-ysin\theta=3-2x$

Squaring both sides give us

$x^{2}cos^{2}\theta+y^{2}sin^{2}\theta-2xycos\theta.sin\theta=(2x-3)^{2}$ ...(i)

And

$xsin\theta+ycos\theta=-2y$

$x^{2}sin^{2}\theta+y^{2}cos^{2}\theta+2xysin\theta.cos\theta=4y^{2}$ ...(ii)

Adding both i and ii gives us

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

$x^{2}+y^{2}=4y^{2}+4x^{2}-12x+9$

$3x^{2}+3y^{2}-12x+9=0$

$x^{2}+y^{2}-4x+3=0$

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

$z^{4}-4z^{2}+8z+35$ is :

0%
$6$
0%
$0$
0%
$1$
0%
$2$

Explanation

Given

$z = 2 - i \sqrt3,$

$\therefore$

$z - 2 = - i \sqrt3$
Squaring both sides,

$\therefore$

$z^2 - 4z + 4 = -3$

$\therefore$

$z^2 - 4z = -7$
Now

$z^4 - 4z^2 + 8z + 35 = (z^2 - 4z + 7)(z^2 + 4z + 5)$
Thus

$z^4 - 4z^2 + 8z + 35 = 0.$

$Z_{1},Z_{2}$ are two unimodular Complex numbers then

$\left |\displaystyle \frac{1}{Z_{1}}+\frac{1}{Z_{2}} \right|=$

0%
$1$
0%
$|Z_{1}+Z_{2}|$
0%
$|Z_{1}|+|Z_{2}|$
0%
$2$

Explanation

$\left |\dfrac{1}{z_1}+\dfrac{1}{z_2}\right|$

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

$=\dfrac{\left|z_1+z_2\right|}{\left|z_1\right|\left|z_2\right|}$

$= \dfrac{|z_1+z_2| }{1\times 1}$

$=|z_1+z_2|$

The real value of

$\theta$ for which the expression,

$\displaystyle \frac{1+i\cos\theta}{1-2i\cos\theta}$ is real number is

0%
$n\displaystyle \pi\pm\frac{\pi}{2}$
0%
$n\displaystyle \pi-\frac{\pi}{2}$
0%
$n\displaystyle \pi+\frac{\pi}{2}$
0%
$2n\displaystyle \pi\pm \cfrac {\pi}{2}$

Explanation

Let

$z=\cfrac{1+i\cos\theta}{1-2i\cos\theta}$

$=\cfrac{(1+i\cos\theta)(1+2i\cos\theta)}{(1-2i\cos\theta)(1+2i\cos\theta)}$

$=\cfrac{1-2\cos^{2}\theta+i3\cos\theta}{1+4\cos^{2}\theta}$

$=\cfrac{1-2\cos^{2}\theta}{1+4\cos^{2}\theta}+i\left (\dfrac{3\cos\theta}{1+4\cos^{2}\theta}\right)$
If the above complex number is purely real, then

$Im(z)=0$

$3\cos\theta=0$

$\theta=\cfrac{(2n\pm1)\pi}{2}$ , where

$n\epsilon N$

$\alpha,\ \beta,\ \gamma$ are modulus of the complex number

$3+4i, -5+12i,\ 1-i$ , then the increasing order for

$\alpha, \beta$ and

$\gamma$ is

0%
$\alpha,\ \gamma,\ \beta$
0%
$\alpha,\ \beta,\ \gamma$
0%
$\gamma,\ \alpha,\ \beta$
0%
can't be determined

Explanation

$\alpha =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5$

$\beta =\sqrt{(5)^{2}+(12)^{2}}=13$

$\gamma =\sqrt{1+1^{2}}=\sqrt{2}$

$\Rightarrow \gamma < \alpha < \beta$

$\mathrm{l}\mathrm{n}$ a G.

$\mathrm{P}$ first term is

$\sqrt{3}+i$ and common ratio is

$\sqrt{3}-i$ then the modulus of the

$n^{th}$ term of the G.

$\mathrm{P}$ . is

0%
4
0%
$2^{n-1}$
0%
$2^{n}$
0%
$2^{n+1}$

Explanation

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

$\downarrow$ $\downarrow$

first ratio

$|a_{n}|=|\sqrt{3}+i|(|\sqrt{3-i}|)^{n-1}$

$[as |z^{n-1}|=(|z|)^{n-1}]$

$=(\sqrt{3+1})\times(\sqrt{3+1})^{n-1}$

$=2\times2^{n-1} = 2^n$

$\displaystyle \log_{(\frac{1}{3})}|z+1|>\log_{(\frac{1}{3})}|z-1|$ , then

0%
${\rm Re}(z)\geq 0$
0%
${\rm Re}(z)<0$
0%
${\rm Im}(z)>0$
0%
${\rm Im}(z)\leq 0$

Explanation

$\displaystyle \log(\frac{1}{3})|z+1|>\log(\frac{1}{3})|z-1|$

$\Rightarrow |z + 1| < |z - 1|$

$\Rightarrow (x^2 + 2x + 1 + y^2) < (x^2 - 2x + 1 + y^2)$

$\Rightarrow | 4x | < 0$
Thus,

$Re(z) < 0.$

$z_{1},z_{2}$ are complex numbers and a.b are real numbers then

$|az_{1}-bz_{2}|^{2}+|bz_{1}+az_{2}|^{2}=$

0%
$(a^{2}+b^{2})|z_{1}^{2}+z_{2}^{2}|$
0%
$(a^{2}+b^{2})(|z_{1}|^{2}+|z_{2}|^{2})$
0%
$\displaystyle \frac{1}{a^{2}+b^{2}}(|z_{1}|^{2}+|z_{2}|^{2})$
0%
$(a^{2}+b^{2})|z_{1}|+|z_{2}|$

Explanation

$|az_1- bz_2|^2 = |az_1|^2 + |bz_2|^2- 2Re (az_1bz_2)$

$=|a|^2|z_1|^2 +|b|^2|z_2|^2 -2ab Re(z_1z_2)$
Similarly,

$|bz_1+ bz_2|^2 = |bz_1|^2 + |bz_2|^2 +2Re (bz_1az_2)$

$=|b|^2|z_1|^2 +|a|^2|z_2|^2 +2ab Re(z_1z_2)-(2)$
Adding,

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

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

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

$\dfrac{x+3i}{2+iy}=1-i$ , then the value of

$\left ( 5x-7y \right )^2$ is

0%
$1$
0%
$0$
0%
$2$
0%
$4$

Explanation

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

$x+3i=(1-i)(2+iy)$

$x+3i=2+iy-2i+y$

$x-y-2=i(y-5)$
Or

$(x-y-2)+i(5-y)=0$

Hence

$5-y=0$ or

$y=5$ and

$x-y-2=0$ or

$x-y=2$

Since

$y=5$

Hence

$x=y+2$ or

$x=5+2$

$x=7$

Therefore

$x=7$ and

$y=5$ .

Therefore

$(5x-7y)^{2}$

$=(5(7)-7(5))^{2}$

$=(35-35)^{2}$

$=0$

Number of solutions of the equation

$|z|^{2}+7{z}=0$ is

0%
$1$
0%
$2$
0%
$4$
0%
$6$

Explanation

Let

$\ z=x+iy$
Then,

$\left | z \right |^{2}+7z=x^{2}+y^{2}+7(x+iy)=0$

$\Rightarrow x^{2}+y^{2}+7x=0$ &

$7y=0$

$\Rightarrow x^{2}+ 7x=0$ and

$y=0$

$\Rightarrow x=-7\ or\ x=0$ and

$y=0$
So, solutions are

$z=-7$ or

$z=0$
Hence, two solutions.

The given figure represents a multiplication operation, where each alphabet represents a different number, then what is the value of A.

0%
0
0%
3
0%
2
0%
4

Explanation

$A+7=9$

$A=9-7$

$A=2$

Hence option-

$C$

Solve the equation

$\left|z\right| = z + 1 + 2i$ .

0%
$3 - 2i$
0%
$2 - \displaystyle \frac{3}{2}i$
0%
$2 +\displaystyle \frac{3}{2}i$
0%
$\displaystyle \frac{3}{2} - 2i$

Explanation

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

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

$\Longrightarrow \sqrt{x^2 +y^2} = x + 1$ and

$0 = 2 + y$ or

$y = -2$

$\Longrightarrow \sqrt{x^2 + 4} = x + 1$

$\Rightarrow x^2 + 4 = x^2 + 2x + 1$

$\Rightarrow 2x = 3$

$\therefore x = \dfrac{3}{2}$

$\Longrightarrow x + iy = \dfrac{3}{2} - 2i$

Ans: D

$z = x + iy$ and

$w = \displaystyle \frac{1 - zi}{z - i}, |w| = 1$ , then find the locus of z.

0%
z lies on the imaginary axis.
0%
z lies only on positive real axis.
0%
z lies only on negative real axis.
0%
z lies on the real axis.

Explanation

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

$\left| - i (x + iy) \right| = \left|x + iy - i\right|$ , 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: D

$z = x + iy$ and

$x^2 + y^2 = 16$ , then the range of

0%
[0, 4]
0%
[0,2]
0%
[2, 4]
0%
none of these

Explanation

$z=x+iy$
And

$|z|^{2}=16$

$|z|=4$
If z is purely real,

$x=\pm4$ and

$||x|-|y||=4$ ...(i)
If z is purely imaginary

$y=\pm4i$ and

$||x|-|y||=4$ ...(ii)
Now if

$|x|=|y|$
Thus

$2x^{2}=2y^{2}=16$

$x=y=\pm2\sqrt{2}$
Under such case

$||x|-|y||=0$
Hence the range of values of

$||x|-|y||$ is

$[0,4]$

$\left|z\right |^2 - 3 = 3\left|z\right|$ , then the value of

$\left|z\right|$ is

0%
$1$
0%
$\displaystyle \frac{3 + \sqrt{21}}{2}$
0%
$\displaystyle \frac{\sqrt{21}- 3}{2}$
0%
none of these

Explanation

$|z|^{2}-3=3|z|$

$|z|^{2}-3|z|-3=0$

$|z|= \displaystyle \frac{3\pm\sqrt{21}}{2}$

$|z|=\displaystyle \frac{3+\sqrt{21}}{2}$

Find the complex numbers z which simultaneously satisfy the equation

$\displaystyle \left | \frac{z - 12}{z - 8 i} \right | = \frac{5}{3}$ and

$\displaystyle \left | \frac{z - 4}{z - 8} \right | = 1$ .

0%
6 + 8 i or 6 + 17 i
0%
6 + 8 i or 6 - 17 i
0%
6 - 8 i or 6 + 17 i
0%
6 - 8 i or 6 - 17 i

Explanation

Put

$z=x+iy$

$\displaystyle \left| \frac { z-4 }{ z-8 } \right| =1\Rightarrow \left| \frac { x-4+iy }{ x-8+iy } \right| =1\\ \Rightarrow { \left( x-4 \right) }^{ 3 }+{ y }^{ 2 }={ \left( x-8 \right) }^{ 2 }+{ y }^{ 2 }\\ \Rightarrow x=6$

With

$\displaystyle x=6,\left| \frac { z-12 }{ z-8i } \right| =\frac { 5 }{ 3 }$

$\displaystyle \Rightarrow \left| \frac { -6+iy }{ 6+i\left( y-8 \right) } \right| =\frac { 5 }{ 3 } \\ \Rightarrow 9\left( 36+{ y }^{ 2 } \right) =25\left[ 36+\left( y-8 \right) 2 \right] \\ \Rightarrow { y }^{ 2 }-25y+136=0\\ \Rightarrow y=17,8$
Hence the required numbers are

$z=6+17i,6+8i$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect and Reason is correct

Explanation

$\displaystyle \left| { z }_{ 1 }{ +z }_{ 2 } \right| =\left| \frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } \right| \quad$

$\displaystyle \quad \Longrightarrow \left| { z }_{ 1 }{ +z }_{ 2 } \right| =\left| \frac { { z }_{ 1 }+{ z }_{ 2 } }{ { z }_{ 1 }{ z }_{ 2 } } \right|$

$\quad \because \quad { z }_{ 1 }\neq { -z }_{ 2 }$

$\quad \therefore \quad \left| { z }_{ 1 }{ z }_{ 2 } \right| =1$
Therefore

${ z }_{ 1 }{ z }_{ 2 }$ is unimodular.
But

${ z }_{ 1 }$ &

${ z }_{ 2 }$ are not unimodular.

Ans: c

Locate the complex number

$z = x + iy$ for which

$log_{1/3} \{ log_{1/2} (|z|^2 + 4 |z| + 3) \} < 0$

0%
Empty set
0%
circle of radius 6 and center at origin
0%
circle of radius 3 and center at origin
0%
circle of radius 2 and center at origin

Explanation

$\log_{ 1/3 }\{ \log_{ 1/2 }(|z|^{ 2 }+4|z|+3)\} <0$

$\Rightarrow \log_{ 1/2 }(|z|^{ 2 }+4|z|+3)>0$

$\Rightarrow |z|^{ 2 }+4|z|+3<0$

$\Rightarrow \left( \left| z \right| +1 \right) \left( \left| z \right| +3 \right) <0$
Since,

$\left| z \right| +1>0$ and

$\left| z \right| +3>0$
Therefore,

$z$ belongs to empty set.

Ans: A

$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0$ , then

0%
$\left| z \right| <1$
0%
$\left| z \right| >1$
0%
$\left| z \right| =1$
0%
$\left| z \right| =0$

Explanation

Given

$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0\Rightarrow i{ z }^{ 2 }\left( z-i \right) -\left( z-i \right) =0$

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

$\displaystyle { z }^{ 2 }=\frac { 1 }{ i } =-i$

Now

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

and

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

Thus, in both cases

$\left| z \right| =1$

Number of roots of the equation

$z^{10} - z^5 - 992 = 0$ where real parts are negative is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

We have,

$z^{10} - z^5 - 992 = 0$
Put

$z^5=u$

$\Rightarrow u^2-u-992=0$

$\Rightarrow u = \dfrac{1\pm\sqrt{1+ 4\times 992}}{2}$

$\quad =\dfrac{1\pm 63}{2}=32, -31$

$\Rightarrow z^5 =32, (-1)^531$

$\Rightarrow z=(32)^{1/5}, -(31)^5$
Hence number of roots of

$z$ having negative real parts is

$5$
corresponding to

$z^5 =-31$

Find the greatest and the least value of

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

$z_1 = 24 + 7i$ and

$\left|z_2\right| = 6.$

0%
least value is 25, greatest value is 31
0%
least value is 19, greatest value is 31
0%
least value is 19, greatest value is 25
0%
least value is 13, greatest value is 25

Explanation

$\left|z_1 + z_2\right|$ is 19 and the greatest value is 31.

Ans: C

$\displaystyle\ z_{1}\neq-z_{2}$ and

$\displaystyle\ |z_{1}+z_{2}|=\left | \frac{1}{z_{1}}+\frac{1}{z_{2}} \right |$ then

0%
at least one of $\displaystyle\ z_{1}, z_{2}$ is unimodular
0%
both $\displaystyle\ z_{1}, z_{2}$ are unimodular
0%
$\displaystyle\ z_{1}. z_{2} = 1$
0%
None of these

Explanation

Let

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

$z_{2}=x_{2}+iy_{2}$
Now

$z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})$

$\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}$

$=\bar{z_{1}}+\bar{z_{2}}$

$=(x_{1}+x_{2})-i(y_{1}+y_{2})$ .
Now

$|z_{1}+z_{2}|$

$=\sqrt{(x_{2}+x_{1})^2+(y_{2}+y_1)^{2}}$
And

$|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}|$

$=|(x_{1}+x_{2})-i(y_{1}+y_{2})|$

$=\sqrt{(x_{2}+x_{1})^2+(y_{2}+y_1)^{2}}$
Hence

$|z_{1}+z_{2}|=|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}|$ for all values of

$z_{1}$ and

$z_{2}$ .

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

0%
$1$
0%
$-1$
0%
$2$
0%
$\frac{1}{2}$

Explanation

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

$(w - \overline{w}z)/(1-z)$ is purely real where

$w = \alpha + i\beta, \beta \neq 0$ and

$z \neq 1$ , then set of the values of

$z$ is

0%
${z : \left|z\right| = 1}$
0%
${z : z = \overline{z}}$
0%
${z : z \neq 1}$
0%
${z : \left|z\right| = 1, z \neq 1}$

Explanation

$\dfrac{(\omega - \overline{\omega} z)}{1-z}$ is purely real and

$z\ne 1$

$\dfrac{(\omega - \overline{\omega} z)}{1-z} = \dfrac{\overline{\omega -\overline{\omega} z}}{(\overline{1-z})}$

$\dfrac{(\omega - \overline{\omega} z)}{1-z} = \dfrac{\overline {\omega} - \omega\overline{z}}{1-\overline z }$

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

$\omega -\omega\overline z-\overline{\omega}z + \overline{\omega} |z|^2 = \overline{\omega}+\omega|z|^2 -\overline{\omega}z-\omega\overline z$

$\omega - \overline{\omega} = |z|^2 (\omega - \overline{\omega})$

$\therefore |z|^2 = 1$ (as

$Im(\omega) = \beta \ne 0$ )

$\Rightarrow |z| =1$ But

$z\ne 1$

$\therefore$ Set of values of

$z$ is

$z:|z| =1,z\ne1$

$(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$ , then find the value of

$a^2 + b^2$ .

0%
$(a^2 + b^2) = 9$
0%
$(a^2 + b^2) = 27$
0%
$(a^2 + b^2) = 3$
0%
$(a^2 + b^2) = 1$

Explanation

Given that

$(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$
Taking modulus and squaring both sides, we get

$(8 + 1)^{50} = 3^{98}(a^2 + b^2)$
or

$9^{50}=3^{98}(a^2 + b^2)$
or

$3^{100} = 3^{98} (a^2 + b^2)$
or

$(a^2 + b^2) = 9$

Ans: A

Find the minimum value of

$|z-1|$ if

0%
$\left|z - 1\right| \ge 0$
0%
$\left|z - 1\right| \ge 1$
0%
$\left|z - 1\right| \ge 2$
0%
$\left|z - 1\right| \ge 3$

Explanation

We know that,

$|z_1+z_2|⩽||z_1|+|z_2||$

Therefore,

$\Rightarrow \left|z - 1\right| \ge 1$

Ans: B

If z is a complex number, then find the minimum value of

0%
$E = 1$
0%
$E = 2$
0%
$E = 3$
0%
$E = 4$

Explanation

Let z be a complex number such that the imaginary part of z is nonzero and a =

$z^2 + z + 1$ is real. Then a cannot take the value

0%
$-1$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$

Explanation

Let

$z=\alpha+i\beta$ . Then

$Im(z^2+z+1)=2\alpha\beta+\beta=0$

$\Rightarrow \alpha=-\dfrac{1}{2}$ since

$\beta\neq0$ . Then

$a=\alpha^2-\beta^2+\alpha+1$

$a=\dfrac{3}{4}-\beta^{2}$

$a\neq\dfrac{3}{4}$ .

$[\because \beta\neq0]$

$\left|z_1 + z_2 + z_3\right|$ .

0%
the greatest value is 6.
0%
the greatest value is 7.
0%
the greatest value is 9.
0%
the greatest value is 12.

Explanation

Given,

$\left|z_1 + z_2 + z_3\right|$

For all complex numbers

$z_1, z_2$ satisfying

$\left|z_1\right| = 12$ and

$\left|z_2 - 3 - 4i\right| = 5$ , then minimum value of

$\left|z_1 - z_2\right|$ is

0%
$0$
0%
$2$
0%
$7$
0%
$17$

Explanation

Given:-

${ z }_{ 1 }$ and

${ z }_{ 2 }$ are the complex number.

$\left| { z }_{ 1 } \right| =12 \left| { z }_{ 2 }-3-4i \right| =5$

To find minimum value of

$\left| { z }_{ 1 }-{ z }_{ 2 } \right|$

Solution:-

We know that

$\left| { z }_{ 1 }-{ z }_{ 2 } \right| \ge \left| { z }_{ 1 } \right| -\left| { z }_{ 2 } \right|$

and

$\left| { z }_{ 2 }-3-4i \right| \ge \left| { z }_{ 2 } \right| -\left| 3+4i \right|$

$\left| { z }_{ 2 }-3-4i \right| \ge \left| { z }_{ 2 } \right| -\sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } \left\{ \left| a+bi \right| \right\} =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \\ 5\ge \left| { z }_{ 2 } \right| -5\\ \left| { z }_{ 2 } \right| \ge 10$

for minimum value.

${ z }_{ 2 }=10$

$\therefore$ minimum value of

$\left| { z }_{ 1 }-{ z }_{ 2 } \right| \ge \left| { z }_{ 1 } \right| -\left| { z }_{ 2 } \right| \\ \left| { z }_{ 1 }-{ z }_{ 2 } \right| \ge 12-10\\ \left| { z }_{ 1 }-{ z }_{ 2 } \right| \ge 2$

Hence the minimum value of

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

The value of the sum

$\displaystyle\ \sum _{n=1}^{13}\left ( i^{n}+i^{n+1} \right )$ , where

$\displaystyle\ i=\sqrt{-1}$

0%
$i$
0%
$i-1$
0%
$-i$
0%
$0$

Explanation

$\sum (i^n+i^{n+1})=i+2(i^2+i^3+i^4.........i^{13})+i^{14}$

$=i+2(0)-1$ .........................(

$i^2=-1,i^3=-i,i^4=1$ )

$=i-1$

$z_{1}$ and

$z_{2}$ are two nonzero complex number such that

$\displaystyle |z_1|+|z_{2}|=|z_{1}+z_{2}|$ then

$arg\: z_{1}-arg\: z_{2}$ is equal to

0%
$-\pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle -\frac{\pi}{2}$
0%
$0$

Explanation

$|z_{1}|+|z_{2}|=|z_{1}+z_{2}|$ can only occur when

$z_{1}$ is parallel to

$z_{2}$ on argand diagram.
Therefore

$arg(z_{1})-arg(z_{2})$ will be

$0$
Hence, option 'D' is correct..

$\displaystyle \displaystyle\ |z_{1}-1|<1, |z_{2}-2|<2, |z_{3}-3|<3$ then

$\displaystyle\ |z_{1}+z_{2}+z_{3}|$

0%
$\displaystyle\$ is less than 6
0%
$\displaystyle\$ is more than 3
0%
$\displaystyle\$ is less than 12
0%
$\displaystyle\$ lies between 6 and 12

Explanation

Given the conditions in the question, the maximum values of

$z_1, z_2,z_3$ are 2, 4 and 6 respectively.
Thus, the maximum value of

$|z_1 + z_2 + z_3|$ becomes 12 when all the values are maximum simultaneously.
So,

$|z_1 + z_2 + z_3|$ is less than 12.

$\displaystyle\ z_{1}, z_{2}$ are two nonzero complex numbers such that

$\displaystyle\ |z_{1}+z_{2}|=|z_{1}|+\left| z_{ 2 } \right|$ then amp

$\displaystyle\ \frac{z_{1}}{z_{2}}$ is equal to

0%
$\displaystyle\ \pi$
0%
$\displaystyle\ -\pi$
0%
$0$
0%
$\displaystyle\ \frac{\pi}{2}$

Explanation

This is true if the the complex numbers are parallel.
i.e.

$arg(z_{1})=arg(z_{2})$
Hence,

$arg(\frac{z_{1}}{z_{2}})=0$

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.