If $$ z = \dfrac {-1}{2} + i \dfrac {\sqrt3}{2} $$, then $$ 8 + 10z + 7z^2 $$ is equal to :

$$ - \dfrac {1} {2} + i \dfrac {3\sqrt3}{2} $$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 6

Let

${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 } \le \dfrac { 1 }{ n } \right\}$ for all integers

$n\ge 1$ . Then,

$\displaystyle\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } }$ is

0%
A singleton set
0%
Not a finite set
0%
An empty set
0%
A finite set with more than one element

Explanation

Given,

${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 }\le \dfrac { 1 }{ n } \right\}$

$=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ n } \right\}$

$\therefore { X }_{ 1 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 1 \right\}$

${ X }_{ 2 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 2 } \right\}$

${ X }_{ 3 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 3 } \right\}$
.............................................
.............................................
.............................................

${ X }_{ \infty }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 0 \right\}$

$\therefore \bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } ={ X }_{ 1 }\cap { X }_{ 2 }\cap { X }_{ 3 }\cap \dots \cap { X }_{ \infty }$

$=\left\{ { x }^{ 2 }+{ y }^{ 2 }=0 \right\}$
Hence,

$\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } }$ is a singleton set.

$z_1, z_2$ and

$z_3$ are complex numbers such that

$|z_1| = |z_2| = |z_3| = \left | \displaystyle \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right | = 1,$ 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

$|z_1| = |z_2| = |z_3| = 1$ (given)
Now

$|z_1| = 1 \Rightarrow |z_r|^2 = 1$

$\Rightarrow z_1 \bar z_1 = 1$
Similarly

$z_2 \bar z_2 = 1, z_3 \bar z_3 = 1$
Now

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

$\Rightarrow |\bar z_1 + \bar z_2 + \bar z_3| = 1$

$\Rightarrow |\overline{z_1 + z_2 + z_3}| = 1 \Rightarrow |z_1 + z_2 + z_3| = 1$

The modulus of

$\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 }$ is

0%
$\sqrt { 5 }$ unit
0%
$\dfrac { \sqrt { 11 } }{ 5 }$ unit
0%
$\dfrac { \sqrt { 5 } }{ 5 }$ unit
0%
$\dfrac { \sqrt { 12 } }{ 5 }$ unit

Explanation

Let

$z=\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 }$

$=\dfrac { 5-5i+12i-4 }{ 5\left( 3+i \right) } =\dfrac { 1+7i }{ 5\left( 3+i \right) }$

$=\dfrac { \left( 1+7i \right) \left( 3-i \right) }{ 5\left( 9+1 \right) }$

$=\dfrac { 10+20i }{ 50 } =\dfrac { 1+2i }{ 5 }$

$\therefore \left| z \right| =\sqrt { { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 5 } \right) }^{ 2 } } =\dfrac { 1 }{ 5 } \sqrt { 1+4 } =\dfrac { \sqrt { 5 } }{ 5 }$

Find the product. Write the answer in standard form.

$i\left( 6-2i \right) \left( 7-5i \right)$

0%
$52+16i$
0%
$10{i}^{3}+44{i}^{2}+42i$
0%
$-44-32i$
0%
$44+32i$

Explanation

Given

$z=i(6-2i)(7-5i)$

$z=i[6(7-5i)-2i(7-5i)]$

$=i[42-30i-14i+10i^2]$

$=i[32-44i]$ ............[since,

$i^2=-1$ ]

$=-44i^2+32i$

$z=44+32i$

The number of composite number between

$101$ and

$120$ are

0%
$11$
0%
$12$
0%
$13$
0%
$None$

Explanation

A composite number is a positive integer that has atleast one positive divisor other than one or the number itself. In other words, a composite number is any integer greater than one that is not a prime number

Now between

$101$ and

$120,$

Numbers are

$102,104,105,106,108,110,111,112,114,115,116,117,118,119$

Total :

$14.$

The product of

$(3-2i)$ and

$\left(\dfrac { 5 }{ 2 } -4i\right)$ , if

$i=\sqrt { -1 }$ , is:

0%
$-\dfrac { 1 }{ 2 } -17i$
0%
$14+\dfrac{9}{2}i$
0%
$2-8i-14{i}^{2}$
0%
$i\left(8+\dfrac{9}{2}\right)$

Explanation

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

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

$= \dfrac{15}{2} - 12i - 5i - 8$

$= \dfrac{-1}{2} - 17i$

The resultant complex number when

$(4+6i)$ is divided by

$(10-5i)$ is

0%
$\dfrac {2}{25} + \dfrac {16}{25}i$
0%
$\dfrac {2}{25} - \dfrac {16}{25}i$
0%
$\dfrac {2}{5} + \dfrac {6}{5}i$
0%
$\dfrac {2}{5} - \dfrac {6}{5}i$

Explanation

The value of

$\cfrac{4 + 6i}{10 - 5i}$

$= \cfrac{4 + 6i}{10 - 5i} \times \cfrac{10 + 5i}{10 + 5i}$

$= \cfrac{(4 + 6i)(10 + 5i)}{(10 - 5i)(10 + 5i)}$

$= \cfrac{40 + 20i + 60i - 30}{100 + 25}$

$= \cfrac{10 + 80i}{125}$

$= \cfrac{2}{25} + \cfrac{16}{25}i$

$A = (3 - 4i)$ and

$B = (9 + ki)$ , where

$k$ is a constant.

$AB - 15 = 60$ , then the value of

$k$ is

0%
$6$
0%
$24$
0%
$12$
0%
$3$

Explanation

Given, $A=3-4i, B=9+ki, Ab-15=0$

$\therefore AB= (3-4i)(9+ki)$

$\therefore 27 + 3ki – 36i – 4ki^2-15=60$

$\therefore -48+3ki-36i+4k=0$

Separate the real and imaginery part equal to zero, then we get the value of $k$ ,

$\therefore -48 + 4k = 0, 3k – 36 = 0$

$\therefore -48 = -4k, 3k = 36$

$\therefore k = 12, k = 12$

So, the value of $k$ is $12$ .

The simplest form of

$\sqrt {-18} \times \sqrt {-50}$ is

0%
$-30$
0%
$-30i$
0%
$30$
0%
$30i$

Explanation

We know that

$i^2=-1$
So,

$\sqrt{18i^2}\times\sqrt{50i^2}$

$=$

$i\sqrt{18}\times i\sqrt{50}$

$=$

$i^2\sqrt{18\times50}$

$=$

$-\sqrt{900}$

$= - 30$

Express

$\dfrac {(-5-i)(-7+8i)}{(2-4i)}$ in the form of a complex number

$a+bi$ .

0%
$\dfrac{109}{10}-\dfrac{53}{10}i$
0%
$-\dfrac{109}{10}-\dfrac{53}{10}i$
0%
$\dfrac{109}{10}+\dfrac{53}{10}i$
0%
$-\dfrac{109}{10}+\dfrac{53}{10}i$

Explanation

We need to express

$\dfrac {(-5-i)(-7+8i)}{(2-4i)}$ in the form of

$a+bi$

On rationalising the given expression as follows:

$\dfrac {(-5-i)(-7+8i)(2+4i)}{(2-4i)(2+4i)}$

$=\dfrac {(35-40i+7i-8i^2)(2+4i)}{(4-16i^2)}$

$=\dfrac {(43-33i)(2+4i)}{(4+16)}$

$=\dfrac {(86+172i-66i-132i^2)}{20}$

$=\dfrac {(218+106i)}{20}$

$=\dfrac {109}{10}+\dfrac {53}{10}i$

Simplify

$(2+8i)(1-4i)-(3-2i)(6+4i)$

(Note

$:i=\sqrt{-1}$ )

0%
$8$
0%
$26$
0%
$34$
0%
$50$

Explanation

The value of

$\left ( 2+8i \right )\left ( 1-4i \right )-\left ( 3-2i \right )\left ( 6+4i \right )$

$=$

$\left ( 2+8i-8i-32i^{2} \right )-\left ( 18-12i+12i-8i^{2} \right )$

$=$

$2-32i^{2}-18+8i^{2}$

Put the given value of

$i=\sqrt{-1}$ , we get

$=$

$2-32(\sqrt{-1})^{2}-18+8(\sqrt{-1})^{2}$

$=$

$2-32(-1)-18+8(-1)$

$=$

$2+32-18-8$

$=$

$34-26=8$

$\displaystyle \left|\dfrac{\sqrt{3}+i}{(1+i)(1+\sqrt{3}i)}\right|=$

0%
$1$
0%
$\sqrt{2}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{\sqrt{2}}$

Explanation

The value of

$\displaystyle \left |\frac{\sqrt{3} + i}{(1 + i)(1 + \sqrt{3}i)} \right |$

$= \dfrac{|\sqrt{3} + i|}{|1 + i||1 + \sqrt{3}i|} = \dfrac{\sqrt{3 + 1}}{\sqrt{1 + 1} \times \sqrt{1 + 3}}$

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

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

What is the approximate magnitude of

$8 + 4i$ ?

0%
4.15
0%
8.94
0%
12.00
0%
18.64
0%
32.00

Explanation

Magnitude of

$8+4i$ is

$=\sqrt { { 8 }^{ 2 }+{ 4 }^{ 2 } }$

$=\sqrt { 64+16 }$

$=\sqrt { 80 }$

$=8.94$

The imaginary number

$i$ is defined such that

$i^2=-1$ . What is the value of

$(1 - i \sqrt {5}) ( 1 + i\sqrt {5})$ ?

0%
$\sqrt5$
0%
$5$
0%
$6$
0%
$\sqrt6$

Explanation

Imaginary number is

$i=\sqrt {-1}$

Value of

$(1-i\sqrt{5})(1+i\sqrt{5})$

It is in the form of

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

So,

$(1-i\sqrt{5})(1+i\sqrt{5})={(1)}^2-{(i\sqrt {5})}^2$

$\Rightarrow 1-(i^2\times 5)$

$\Rightarrow 1-(-1\times 5)$

$\Rightarrow 6$

$(1-i\sqrt{5})(1+i\sqrt{5})=6$

How many of the prime factors of

$30$ are greater than

$2$ ?

0%
One
0%
Two
0%
Three
0%
Four
0%
Five

Explanation

$30=2\times 3\times 5$
So, only

$3$ and

$5$ are the two prime factors that are greater than

$2$ .

$i = \sqrt {-1}$ , find the values of

$n$ such that

$i^{n} + (i)^{n}$ have a positive value.

0%
$23$
0%
$24$
0%
$25$
0%
$26$
0%
$27$

Explanation

${ i }^{ n }+{ i }^{ n }=2{ i }^{ n }$
We know that

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

${ i }^{ 4 }=1$ , so

$n$ should be multiple of

$4$ .
Among given options

$24$ will be divisible by

$4$ .
Therefore the answer is

$24$ .

The value of

$(a+2i)(b-i)$ is

0%
$a+b-i$
0%
$ab+2$
0%
$ab+(2b-a)i+2$
0%
$ab-2$
0%
$ab+(2b-a)i-2$

Explanation

The value of

$(a+2i)(b-i)$

$=a(b-i)+2i(b-i)$

$= ab-ai+2bi-2{i}^{2}$

$= ab+(2b-a)i+2$

Find

$(5 + 2i)(5 - 2i)$

0%
$25 - 4i$
0%
$25 - 20i$
0%
$21$
0%
$29$
0%
$0$

Explanation

Since

$i^2=-1$ and

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

$\Rightarrow (5+2i)(5-2i)$

$=(5)^2-(2i)^2$

$=25-4i^2$

$=25-4(-1)$

$=25+4$

$=29$

Find the number of prime numbers between 301 and 320?

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

Explanation

Prime numbers between 301 and 320 are 307, 311, 313, 317. So there are total $4$ prime numbers between 301 and 320

So correct answer will be option C

$(2 - i)\times(a - bi) = 2 + 9i$ , where i is the imaginary unit and a and b are real numbers, then a equals

0%
$3$
0%
$2$
0%
$1$
0%
$0$
0%
$-1$

Explanation

Given,

$(2-i)(a-bi) = 2a-2bi-ai-b$

$= (2a-b)+i(-a-2b)$

$= 2+9i$
By comparing, we get

$2a-b=2$ and

$a+2b=-9$
By solving, we get

$a=-1$ and

$b=-4$

Every composite number has _____________.

0%
no prime divisor
0%
one and only one prime divisor
0%
atleast one prime divisor
0%
atleast two prime divisor

Explanation

A composite number is that has at least one prime divisor other than

$1$ and itself. For example :

$10$ is composite number that has

$2, 5$ as its prime divisors.

The modulus of the complex quantity

$(2-3i)(-1+7i)$ .

0%
$5\sqrt{13}$
0%
$5\sqrt{26}$
0%
$13\sqrt{5}$
0%
$26\sqrt{5}$

Explanation

The given complex quantity is

$(2-3i)(-1+7i)$ .

Let

$z_{1}=2-3i$ and

$z_{2}=-1+7i$

Therefore,

$|z_{1}|=\sqrt{2^2+(-3)^2}=\sqrt{13}$

$|z_{2}|=\sqrt{(-1)^2+7^2}=\sqrt{50}=5\sqrt{2}$

Thus, required modulus =

$|z_{1}z_{2}|=\sqrt{13} \times 5\sqrt{2}=5\sqrt{26}$ .

The real part of

${ \left( 1-\cos { \theta } +i\sin { \theta } \right) }^{ -1 }$ is

0%
$\cfrac{1}{2}$
0%
$\cfrac { 1 }{ 1+\cos { \theta } }$
0%
$\tan { \cfrac { \theta }{ 2 } }$
0%
$\cot { \cfrac { \theta }{ 2 } }$

Explanation

Given that:

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

$=\dfrac{1}{2\sin^2\frac{\theta}{2}+i\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}$

$=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}\cdot \dfrac{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$ , rationalize denominator

$=\dfrac{1}{2\sin\frac{\theta}{2}}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})$

$=\dfrac{1}{2}-\dfrac{i}{2}\cot\frac{\theta}{2}$

The real part is

$\dfrac{1}{2}$

$z = \dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}$ , then

$|z|$ is equal to

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

Explanation

Uisng the identity

$|a+ib|=\sqrt{a^2+b^2}$
Since

$\bigg|\dfrac{z_1z_2}{z_3} \bigg|=\dfrac{|z_1||z_2|}{|z_3|}$

Therefore

$|z| = \bigg|\dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}\bigg|$

$\quad =\dfrac{|(\sqrt 3+i)^3||(3i+4)^2|}{|(8+6i)^2|}$

$\quad =\dfrac{|\sqrt 3+i|^3|3i+4|^2}{|8+6i|^2}$

$\quad =\dfrac{2^3\cdot 5^2}{10^2}=2$

Given :

$u = 1+i \sqrt{3}$ and

$v = \sqrt{3} + i$

Calculate $\dfrac{u^3 }{ v^4}$

0%
$(1/4) - i \sqrt{1/4}$
0%
$(3/4) - i \sqrt{3}/4$
0%
$(1/4) - i \sqrt{3}/4$
0%
none of these

Explanation

$u=1+i\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1}\sqrt 3)}\\ \implies 2e^{i\pi/3}$

$v=i+\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1} 1/\sqrt 3)}\\ \implies 2e^{i\pi/6}$

$\cfrac{u^3}{v^4}=\cfrac{2^3e^{i\pi}}{2^4e^{2i\pi/3}}=(1/2)e^{-i\pi/3}$

$\implies \cfrac{1}{2} [\cos(\pi/3)-i \sin(\pi/3)] \implies \cfrac{1}{2}(\cfrac{1}{2}-\cfrac{i\sqrt 3}{2})$

$\implies (\cfrac{1}{4}-\cfrac{i\sqrt 3}{4})$

The expression

$\dfrac{3-4i}{5+3i}$ is equivalent to

0%
$\dfrac{27-29i}{34}$
0%
$\dfrac{27-29i}{16}$
0%
$\dfrac{3-29i}{34}$
0%
$\dfrac{1}{8}$
0%
$15-8i$

Explanation

Given,

$\dfrac { 3-4i }{ 5+3i }$
Multiply and divide it with

$5-3i$ , we get

$\dfrac { 3-4i }{ 5+3i } \times \dfrac { 5-3i }{ 5-3i }$

$=\dfrac {(3-4i)(5-3i)}{5^2-(3i)^2}$

$=\dfrac { 3-29i }{ 34 }$

The fraction

$\dfrac{1}{1+i}$ is equivalent to

0%
$1-i$
0%
$\dfrac{1+i}{2}$
0%
$\dfrac{1-i}{2}$
0%
$i$
0%
$-i$

Explanation

Given,

$\dfrac{1}{1+i}$
Multiply and divide it with

$1-i$ , we get

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

$p + iq = (2 - 3i) (4 + 2i)$ then

$q$ is

0%
$14$
0%
$-14$
0%
$-8$
0%
$8$

Explanation

$p+iq=(2-3i)(4+2i)$

$=(8+4i-12i+6)$

$=14-8i$
Hence comparing the real parts give us

$p=14$ and comparison of the imaginary parts give us

$q=-8$ .

Perform the indicated operations:

$(5+3i)(3-2i)$

0%
$21-2i$
0%
$19-3i$
0%
$11-2i$
0%
$21-i$

Explanation

The value of

$(5+3i)(3-2i)$

$=5(3-2i)+3i(3-2i)$

$= 15-10i+9i-6{i}^{2}$

$= 21-i$

${ x }^{ 2 }+{ y }^{ 2 }=1$ then value of

$\dfrac { 1+x+iy }{ 1+x-iy }$ is

0%
$x-iy$
0%
$2x$
0%
$-2iy$
0%
$x+iy$

Explanation

Since

$x^2+y^2=1$

So let

$x=\cos\theta$ and

$y=\sin\theta$

$\therefore \dfrac{1+x+iy}{1+x-iy}=\dfrac{(1+\cos\theta)+i\sin\theta}{(1+\cos\theta)-i\sin\theta}$

$=\dfrac{2\cos^2\cfrac{\theta}{2}+2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{2\cos^2\cfrac{\theta}{2}-2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}$

$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}$

$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}\times$

$\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}$

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

$=\dfrac{\cos^2\cfrac{\theta}{2}+i^2\sin^2\cfrac{\theta}{2}+2i\cos\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{\cos^2\cfrac{\theta}{2}+\sin^2\cfrac{\theta}{2}}$

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

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

$f\left( z \right) =\dfrac { 1-{ z }^{ 3 } }{ 1-z }$ , where

$z=x+iy$ with

$z\neq 1$ , then

$Re\overline { \left\{ f\left( z \right) \right\} } =0$ reduces to

0%
${ x }^{ 2 }+{ y }^{ 2 }+x+1=0$
0%
${ x }^{ 2 }-{ y }^{ 2 }+x-1=0$
0%
${ x }^{ 2 }-{ y }^{ 2 }-x+1=0$
0%
${ x }^{ 2 }-{ y }^{ 2 }+x+1=0$
0%
${ x }^{ 2 }-{ y }^{ 2 }+x+2=0$

Explanation

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

$=1+z+{ z }^{ 2 }$
Put

$z=x+iy$ , we get

$f\left( z \right) =1+x+iy+{ \left( x+iy \right) }^{ 2 }$

$=1+x+iy+{ x }^{ 2 }+2xyi+{ i }^{ 2 }{ y }^{ 2 }$

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

$\Rightarrow f\left( \overline { z } \right) =\left( 1+x+{ x }^{ 2 }-{ y }^{ 2 } \right) -i\left( y+2xy \right)$
Now,

$Re\left\{ \overline { f\left( z \right) } \right\} =0$

$\Rightarrow 1+x+{ x }^{ 2 }-{ y }^{ 2 }=0$

$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }+x+1=0$

$x + i y = \dfrac{3}{2 + cos \theta + i sin \theta}$ , then

$x^2 + y^2$ is equal to

0%
$3x - 4$
0%
$4x - 3$
0%
$4x +3$
0%
None of these

Explanation

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

$= \dfrac{3(2+cos \theta-i sin \theta)}{(2+cos \theta)^2+sin \theta}$

$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{4 + cos^2 \theta + 4 cos \theta + sin^2 \theta}$

$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{5 + 4 cos \theta}$

$= \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos \theta} \right ) + \left( \dfrac{-3 sin \theta}{5 + 4 cos \theta} \right )$
On equating real and imaginary parts, we get

$x = \dfrac{3 (2 + cos \theta)}{5 + 4 cos \theta}$
and

$y = \dfrac{-3 sin \theta}{5 + 4 cos \theta}$

$\therefore x^2 + y^2 = \dfrac{9 [4 + cos^2 \theta + 4 cos \theta + sin^2 \theta]}{(5 + 4 cos \theta)^2}$

$= \dfrac{9}{5 + 4 cos \theta} = 4 \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos^2 \theta} \right ) - 3$

$= 4x - 3$

Hence, option B is correct.

The modulus of

$\dfrac { \left( 3+2i \right) ^{ 2 } }{ \left( 4-3i \right) }$ is:

0%
$\frac { 13 }{ 5 }$
0%
$\frac { 11 }{ 5 }$
0%
$\frac { 9 }{ 5 }$
0%
$\frac { 7 }{ 5 }$

Let

$z = x + iy$ , where

$x$ and

$y$ are real. The points

$(x, y)$ in the

$X-Y$ plane for which

$\dfrac {z + i}{z - i}$ is purely imaginary lie on

0%
A straight line
0%
An ellipse
0%
A hyperbola
0%
A circle

Explanation

Let

$z = x + iy$

$\therefore \dfrac {z + i}{(z - i)}=\dfrac{x+iy+i}{x+iy-i}$

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

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

Since,

$\dfrac{z+i}{z-i}$ should be purely imaginary

$\therefore Re\left (\dfrac {z + i}{z - i}\right ) = 0$

$\implies \dfrac {x^{2} + (y^{2} - 1)}{x^{2} + (1 - y)^{2}} = 0$

$\implies x^{2}+y^{2}-1=0$

$\implies x^{2} + y^{2} = 1$ which is the equation of a circle

Hence,

$(x,y)$ lie on a circle.

The complex number

$z$ satisfying the equation

$|z - i| = |z + 1| = 1$ is

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

Explanation

Given :

$|z-i|=|z+1|=1$

Let

$z=x+iy$

$|x+iy-i|=1$

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

$\implies x^{2}+y^{2}-2y+1=1$

$x^{2}+y^{2}-2y=0$ ....

$(i)$

Also,

$|z+1|=1$

$\implies |x+iy+1|=1$

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

$\implies x^{2}+y^{2}+2x=0$ .....

$(ii)$

Equating

$(i)$ and

$(ii)$ we get

$-y=x$

For

$y=1, x=-1$

$\therefore z=-1+i$

Also,

$y=0\implies x=0$

$\therefore z=0$

The expression

$\dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}}$ equals.

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

Explanation

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

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

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

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

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

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

Hence,

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

$\left( \dfrac{1 + i}{1 - i} \right)^m = 1$ , then the least positive integral value of m is

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

Explanation

Consider

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

Rationalize the denominator we get,

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

Here we see that denominator is in the form of

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

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

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

we know that

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

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

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

We know that

$i^2=-1$ , substituting this we get,

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

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

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

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

Given that

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

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

So,

$i^m=1$

We know that

$i^2=-1$

Squaring on both sides we get,

$i^4=(-1)^2$

$i^4=1$

Therefore the smallest value of

$m$ for which

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

$m=4$

Let Z and w be complex numbers. If

$Re(z)=|z-2|, Re(w) = |w-z|$ and

$arg(z-w)=\dfrac{\pi}{3}$ , then the value of

$Im(z+w)$ , is

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

Explanation

$Re\left( z \right) =\left| z-2 \right|$

$Re\left( w \right) =\left| w-z \right|$

$arg\left( z-w \right) =\cfrac { \pi }{ 3 }$

Now,

$=\left| w-2 \right| =2$

$arg\left( z-w \right) =\cfrac { y }{ x }$

$\tan { \cfrac { \pi }{ 3 } } =\cfrac { Im\left( z+w \right) }{ Re\left( z+w \right) }$

$\Rightarrow Im\left( z+w \right) =\cfrac { 2 }{ \sqrt { 3 } }$
Hence (B) is the correct answer

$z_1$ and

$z_2$ are complex numbers with

$|z_1|=|z_2|$ , then which of the following is/are correct?
1.

$z_1=z_2$
2. Real part of

$z_1 =$ Real part of

$z_2$
3. Imaginary part of

$z_1 =$ Imaginary part of

$z_2$
Select the correct answer using the statements given below :

0%
1 only
0%
2 only
0%
3 only
0%
None

Explanation

Solution:

We have,

$|z_1|=|z_2|$

Let,

$z_1=x_1+iy_1$ and

$z_2=x_2+iy_2$

$\because|z_1|=|z_2|$

$\therefore x_1^2+y_1^2=x_2^2+y_2^2$

or,

$(x_1^2-x_2^2)+(y_1^2-y_2^2)=0$

or,

$x_1^2-x_2^2=0$ and

$y_1^2-y_2^2=0$

or,

$x_1=\pm x_2$ and

$y_1=\pm y_2$

Let,

$z_1=1+i$ then

$z_2=-1-i$

$|z_1|=|z_2|=\sqrt2$

But,

$Re(z_1)\neq Re(z_2)$ and

$Im(z_1)\neq Im(z_2)$ and

$z_1\neq z_2$

Hence, D is the correct option.

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

$|z|$ is equal to

0%
$0$
0%
$1$
0%
$2$
0%
None of these

Explanation

Given :

$iz^{3} + z^{2} - z + i = 0$
Dividing both sides by

$i$ and using

$\dfrac {1}{i} = -i$ .
We have,

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

$\Rightarrow z^{2}(z - i) + i(z - i) = 0$ ......

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

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

$\therefore z = i$ or

$z^{2} = -i$

$\therefore |z| = |i| = 1$
and

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

$\therefore |z| =1$ .

If '

$\omega$ ' is a complex cube root of unity,then

$\omega ^{ \begin{pmatrix} \frac { 1 }{ 3 } & +\frac { 2 }{ 9 } +\frac { 4 }{ 27 } ...\infty \end{pmatrix} }+\omega^{ \begin{pmatrix} \frac { 1 }{ 2 } & +\frac { 3 }{ 8 } +\frac { 9 }{ 32 } ...\infty \end{pmatrix} }=$

0%
$1$
0%
$-1$
0%
$\omega$
0%
$i$

Explanation

$\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{27}+......+\infty$ (infinte G.P.)

$S_\omega = \dfrac{a}{1-r} =1$

$\dfrac{1}{2}+\dfrac{3}{8}+\dfrac{9}{32}+......+\infty$ (infinte G.P.)

$S_\omega = \dfrac{a}{1-r} =2$

$\therefore\, \omega^1+\omega^2=-1$

The principal argument of the complex number

$z=\cfrac { 1+\sin { \cfrac { \pi }{ 3 } } +i\cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } -i\cos { \cfrac { \pi }{ 3 } } }$ is?

0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { 2\pi }{ 3 }$
0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { \pi }{ 4 }$

Explanation

$arg(z)=arg(Numerator)-arg(Denominator)$

$=\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| } +\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| }$

$=2\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| }$

$=2\tan ^{ -1 }{ (2-\sqrt { 3 } } )=2\times {15}^{o}=\cfrac { \pi }{ 6 }$

The value of

$\sum _{ k=0 }^{ n }{ (i^k + i^{k+1} ) } ,$ where

$i^2 = -1 ,$ is equal to :

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

Explanation

The value of

$\sum _{ k=0 }^{ n }{ (i^k + i^{k+1} ) } = \sum _{ k=0 }^{ n }{ i^k ( {i +1} ) }$ is,

$= (i+1) \left[ \dfrac {1(1-i^{n+1})} {(1-i)} \right] \times \dfrac {1+i}{1+i}$

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

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

$= i - i^{n+2}$

$z_{1}$ and

$z_{2}$ be complex numbers such that

$z_{1} + i(\overline {z_{2}}) = 0$ and

$arg (\overline {z_{1}}z_{2}) = \dfrac {\pi}{3}$ . Then,

$arg (\overline {z_{1}})$ is equal to

0%
$\dfrac {\pi}{3}$
0%
$\pi$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {5\pi}{12}$
0%
$\dfrac {5\pi}{6}$

Explanation

Given,

$z_{1} + i \,\overline {(z_{2})} = 0$

$\Rightarrow z_{1} = (-i) \overline {z_{2}}$
Taking argument on both sides, we get

$arg (z_{1}) = arg \left \{(-i) \cdot \overline {z_{2}}\right \}$

$\Rightarrow arg (z_{1}) = arg (-i) + arg (\overline {z_{2}})$ (by property)

$\Rightarrow arg(z_{1}) - arg (\overline {z_{2}}) = \tan^{-1} \left (\dfrac {-1}{0}\right ) = -\dfrac {\pi}{2}$

$\Rightarrow arg (z_{1}) + arg (z_{2}) = \dfrac {-\pi}{2} ..... (i)$

$[\because arg (\overline {z}) = -arg (z)]$
and

$arg (\overline {z_{1}}z_{2}) = \dfrac {\pi}{3}$

$\Rightarrow arg (\overline {z_{1}}) + arg (z_{2}) = \dfrac {\pi}{3}$

$\Rightarrow -arg (z_{1}) + arg (z_{2}) = \dfrac {\pi}{3}... (ii)$
On adding Eqs. (i) and (ii), we get

$2 arg (z_{2}) = \dfrac {-\pi}{6}$

$\Rightarrow arg (z_{2}) = \dfrac {-\pi}{12}$
Therefore, from Eq. (i),

$arg (z_{1}) = \dfrac {-\pi}{2} + \dfrac {\pi}{12} = \dfrac {-5\pi}{12}$

$\Rightarrow -arg (z_{1}) = \dfrac {5\pi}{12}$

$\Rightarrow arg (z_{1}) = \dfrac {5\pi}{12}$ .

The inequality

$\left| z-4 \right| <\left| z-2 \right|$ represents the region given by:

0%
$Re(z)\ge 0$
0%
$Re(z)<0$
0%
$Re(z)>0$
0%
None of these

Explanation

We know that,

$\left| z-{ z }_{ 1 } \right| >\left| z-{ z }_{ 2 } \right|$ represents the region on right side of perpendicular bisector of

${z}^{1}$ and

${z}^{2}$ .
Thus, the given inequality

$\left| z-2 \right| >\left| z-4 \right|$ represents the region on the right side of perpendicular bisector of

$2$ and

$4$ .

$\Rightarrow Re(z)>3$ and

$\text{Im}(z)\in R$

Let

$z$ =

$\cos\theta + i \sin\theta$ . Then the value of

$\sum\limits_{m=1}^15Im( z^{2m-1})$ at

$\theta = 2^0$ is

0%
$\dfrac{1}{sin2^0}$
0%
$\dfrac{1}{3sin2^0}$
0%
$\dfrac{1}{2sin2^0}$
0%
$\dfrac{1}{4sin2^0}$

Explanation

using the sum of series of sine

$\sum_{m=1}^{n}sin(a+k.d)=\dfrac{sin\dfrac{n\times d}{2}}{sin\dfrac{d}{2}}\times sin(\dfrac{2a+(n-1).d}{2})$

Given,

$\sum_{m=1}^{15}Im(z^{2m-1})\rightarrow$

$sin\theta+sin3\theta+sin5\theta+.......sin29\theta=$

$\rightarrow\dfrac{sin\dfrac{15\times 2\theta}{2}}{sin\dfrac{2\theta}{2}}\times sin(\dfrac{2\theta+(14\times2\theta)}{2})$

$\rightarrow\dfrac{sin^215\theta}{sin\theta}$

put

$\theta = 2^{\circ}$

$\rightarrow\dfrac{sin^230^{\circ}}{sin2^{\circ}}$

$\rightarrow\dfrac{1}{4.sin2^{\circ}}$

If the complex numbers

$z_1, z_2$ and

$z_3$ denote the vertices of an isosceles triangle, right angled at

$z_1$ , then

$(z_1 - z_2)^2 + (z_1 - z_3)^2$ is equal to

0%
$0$
0%
$(z_2 + z_3)^2$
0%
$2$
0%
$3$
0%
$(z_2 - z_3)^2$

Explanation

Since, the triangleis isosceles and right angled, we can say that

$(z_3-z_1) = (z_2-z_1) (\cos 90^o + i \sin 90^o)$

$\Rightarrow (z_3-z_1) = i(z_2-z_1)$

Squaring bothe sides, we get

$(z_3-z_1)^2 = -(z_2-z_1)^2$

$\Rightarrow (z_3-z_1)^2 + (z_2-z_1)^2 = 0$

$z = \dfrac {-1}{2} + i \dfrac {\sqrt3}{2}$ , then

$8 + 10z + 7z^2$ is equal to :

0%
$- \dfrac {1} {2} - i \dfrac {\sqrt3}{2}$
0%
$\dfrac {1} {2} + i \dfrac {\sqrt3}{2}$
0%
$- \dfrac {1} {2} + i \dfrac {3\sqrt3}{2}$
0%
$\dfrac {\sqrt3}{2} i$
0%
$- \dfrac {\sqrt3}{2} i$

Explanation

Given,

$z = - \dfrac {1} + i \dfrac {\sqrt3}{2}$
Therefore,

$z^2 = \dfrac {1}{4} - \dfrac {3}{4} - 2i \times \dfrac {1}{2} \dfrac {\sqrt3}{2} = - \dfrac {1}{2} - i \dfrac {\sqrt3}{2}$

$\Rightarrow 8 + 10z + 7z^2 = 8 + 10 \left( - \dfrac {1}{2} + i \dfrac { \sqrt3}{2} \right) + 7 \left( - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2}i \right)$

$= 8 - 5 + i 5 \sqrt3 - \dfrac {7}{2} - i \dfrac {7 \sqrt3}{2}$

$= - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2} i$

$z$ is a complex number such that

$z + |z| = 8 + 12i$ , then the value of

$|z^{2}|$ is

0%
$228$
0%
$144$
0%
$121$
0%
$169$
0%
$189$

Explanation

Let

$z = x + iy$
Then, we have

$z + |z| = 8 + 12i$

$\Rightarrow(x + iy) + |x + iy| = 8 + 12l$

$\Rightarrow (x + \sqrt {x^{2} + y^{2}}) + iy = 8 + 12i$
On comparing the real and imaginary part, we get

$y = 12$
and

$x + \sqrt {x^{2} + y^{2}} = 8$

$\Rightarrow \sqrt {x^{2} + 144} = 8 - x$
On squaring both sides, we get

$x^{2} + 144 = 64 + x^{2} - 16x$

$\Rightarrow 16x = -80$

$\Rightarrow x = -5$

$\therefore z = x + iy = -5 + i\cdot 12$
Then,

$|z| = \sqrt {25 + 144} = \sqrt {169} = 13$

$\Rightarrow |z|^{2} = 169$

$\Rightarrow |z^{2}| = 169 (\because |z^{n}| = |z|^{n})$ .

Let

$P(e^{i\theta_1})$ ,

$Q(e^{i\theta_2})$ and

$R(e^{i\theta_3})$ be the vertices of a triangle PQR in the Argand Plane. The orthocenter of the triangle PQR is

0%
$2e^(\theta_1+\theta_2+\theta_3)$
0%
$\frac{2}{3}e^({\theta_1+\theta_2+\theta_3})$
0%
$e^{\theta_1}$ + $e^{\theta_2}$ + $e^{\theta_3}$
0%
None of these

Explanation

$P,Q,R$ lie on the unit circle and hence origin is the cirumcenter , The centrid is easily found as

$\dfrac{(p+q+r)}{3}$ ,Using euler line theorem that centroid divides the line joining the line joining orthocenter and circumcenter in the ratio of

$2:1$

Let orthocenter be

$x,y$

$Using \quad section \quad formula;$

$Re(\dfrac{(p+q+r)}{3})=\dfrac{(1.x+2.0)}{3}$

$Im(\dfrac{(p+q+r)}{3})=\dfrac{(1.y+2.0)}{3}$

we get orthocenter as

$p+q+r\rightarrow$

$e^{\theta_1}+$

$e^{\theta_2}+$

$e^{\theta_3}$

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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