MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 5

$\displaystyle\ \alpha,\beta$ be two complex numbers then

$\displaystyle\ |\alpha|^{2}+|\beta|^{2}$ is equal to

0%
$\displaystyle\ \frac{1}{2}(|\alpha+\beta|^{2}-|\alpha-\beta|^{2})$
0%
$\displaystyle\ \frac{1}{2}(|\alpha+\beta|^{2}+|\alpha-\beta|^{2})$
0%
$\displaystyle\ |\alpha+\beta|^{2}+|\alpha-\beta|^{2}$
0%
None of these

Explanation

We know,

$|(\alpha +\beta )|^{2}=\left | \alpha \right |^{2}+\left | \beta \right |^{2}+2\left | \beta \right |\left | \alpha \right |i$

$|(\alpha -\beta )|^{2}=\left | \alpha \right |^{2}+\left | \beta \right |^{2}-2\left | \beta \right |\left | \alpha \right |i$
add equations

$\therefore \left | \alpha \right |^{2}+\left | \beta \right |^{2}=\frac{1}{2}|\left( \alpha +\beta \right )|^{2}+|\left ( \alpha -\beta \right )|^{2}]$

For

${ { Z }_{ 1 }=\sqrt [ 6 ]{ \dfrac { 1-i }{ 1+i\sqrt { 3 } } } };\quad { { Z }_{ 2 }=\sqrt [ 6 ]{ \dfrac { 1-i }{ \sqrt { 3 } +i } } ;\quad { { Z }_{ 3 }=\sqrt [ 6 ]{ \dfrac { 1-i }{ \sqrt { 3 } -i } } } }$ which of the following holds good?

0%
$\sum { { \left| { Z }_{ 1 } \right| }^{ 2 } } =\dfrac { 3 }{ 2 }$
0%
${ \left| { Z }_{ 1 } \right| }^{ 4 }+{ \left| { Z }_{ 2 } \right| }^{ 4 }={ \left| { Z }_{ 3 } \right| }^{ -8 }$
0%
$\sum { { \left| { Z }_{ 1 } \right| }^{ 3 }+ } { \left| { Z }_{ 2 } \right| }^{ 3 }={ \left| { Z }_{ 3 } \right| }^{ -6 }$
0%
${ \left| { Z }_{ 1 } \right| }^{ 4 }+{ \left| { Z }_{ 2 } \right| }^{ 4 }={ \left| { Z }_{ 3 } \right| }^{ 8 }$

$\displaystyle\ z=1+i\ \tan \alpha$ , where

$\displaystyle\ \pi < \alpha < \frac{3\pi }{2}$ is

$|z|$ is equal to

0%
$\displaystyle\ \sec \alpha$
0%
$\displaystyle\ -\sec \alpha$
0%
$\displaystyle\ cosec\alpha$
0%
none of these

Explanation

$z=1+i\tan\alpha$

$\Rightarrow |z| =\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|$
Now we know that, modulus of any complex number is always non-zero
and it is given that, alpha lies in third quadrant in which

$\sec\alpha <0$
Hence

$|z|=-\sec\alpha>0$
Hence option 'B' is correct choice.

$\displaystyle u_{i}=1-\frac{1}{i}$ then

$\displaystyle u_{2}\cdot u_{3}\cdot ... \cdot u_{n}$ is equal to

0%
$\displaystyle \frac{1}{n}$
0%
$\displaystyle \frac{1}{n!}$
0%
$1$
0%
none of these

Explanation

$u_i = 1-\cfrac{1}{i}=\cfrac{i-1}{i}$ , here

$i$ is not a complex number

$\displaystyle \therefore u_2.u_3...........u_{n-1}.u_n = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{n-2}{n-1}.\frac{n-1}{n} = \frac{1}{n}$

$\displaystyle\ z=x-iy\displaystyle\$ such that

$\displaystyle\ |z+1|=|z-1|$ and amp

$\displaystyle\ \frac{z-1}{z+1}= \frac{\pi}{4}$ then

0%
$\displaystyle\ x=\sqrt{2}+1, y=0$
0%
$\displaystyle\ x=0, y=\sqrt{2}+1$
0%
$\displaystyle\ x=0, y=\sqrt{2}-1$
0%
$\displaystyle\ x=\sqrt{2}-1, y=0$

Explanation

Given

$z=x-iy$

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

$\Rightarrow$

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

$\Rightarrow 2x(2)=0$

$\Rightarrow 4x=0$

$\therefore x=0$

$\dfrac{x-1+iy}{x+1+iy}=\dfrac{\pi}{4}$

Now

$x=0$ , we get

$\Rightarrow \dfrac{-1+iy}{1+iy}$

$\Rightarrow \dfrac{1-y^{2}}{1+y^{2}}+\dfrac{-2iy}{1+y^{2}}$

Argument is

$\dfrac{\pi}{4}$

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

$\Rightarrow 1-y^{2}=-2y$

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

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

$\Rightarrow y-1=\sqrt{2}$

$\therefore y=\sqrt{2}+1$

Hence, option 'B' is correct.

The complex number which satisfies the equation

$z+\sqrt { 2 } \left| z+1 \right| +i=0$ is

0%
$2-i$
0%
$-2-i$
0%
$2+i$
0%
$-2+i$

Explanation

Since

$z+\sqrt { 2 } \left| z+1 \right| +i=0$

$\therefore x+i\left( y+1 \right) +\sqrt { 2 } \left| x+iy+1 \right| =0$

By comparing real and imaginary parts

$\Rightarrow y+1=0$

$(\because \left| x+iy+1 \right|$ is real

$)$

$\Rightarrow y=-1$

$\therefore x+\sqrt { 2 } \left| x-i+1 \right| =0$

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

$\Rightarrow { x }^{ 2 }+4x+4=0\Rightarrow { \left( x+2 \right) }^{ 2 }=0$

$\Rightarrow x=-2$

$\therefore z=-2-i$

$z$ be a complex number, then the minimum value of

$\left | z-7 \right |+\left | z \right |$ is

0%
$\displaystyle \frac{\sqrt{3}-1}{2}$
0%
$-\sqrt{7}$
0%
$\displaystyle \frac{7+\sqrt{2}}{2}$
0%
$7$

Explanation

$\left | z-7 \right |$ implies distance between

$(x,y)$ and

$(7,0)$ .

$\left | z\right |$ means distance between

$(x,y)$ and

$(0,0)$

Minimum distance would be straight line between

$(0,0)$ and

$(7,0)$ which is

$7$ .

$\displaystyle z= 1+i\sqrt{3}$ ,then

$\displaystyle z^{6}$ equals

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32
0%
-32
0%
64
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None of these

Explanation

$\displaystyle z= 1+i\sqrt{3}$

$\displaystyle z^{6}= (1+\sqrt{3})^{6}$ using Binomial Theorem.

$\displaystyle = ^{6}\:C_{0}+^{6}\:C_{1}\:(i\sqrt{3})+^{6}\:C_{2}(i\sqrt{3})^{2}+^{6}\:C_{3}(i\sqrt{3})^{3}+^{6}\:C_{4}(i\sqrt{3})^{4}+^{6}\:C_{5}(i\sqrt{3})^{5}+^{6}\:C_{6}\:(i\sqrt{3})^{6}\:$

$\displaystyle = ^{6}\:C_{0}+^{6}C_{2}(3)+^{6}C_{4}(9)-^{6}C_{6}(27)+i\sqrt{3}(^{6}C_{1}-3^{6}C_{3}+9^{6}C_{5})$

$\displaystyle = (1-15\times 3+135-27)+i\sqrt{3}(60-60)$

$\displaystyle = (136-72)+i\sqrt{3}(0)= 64$

Write all the composite numbers between

$10$ and

$35.$

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$7, 12, 14, 15, 16, 17, 20, 21, 22, 24, 25, 26, 27, 28, 30,33$ and $34$
0%
$7, 12, 14, 15, 16, 18, 20, 21, 21, 24, 25, 26, 27, 28, 32, 33$ and $34$
0%
$7, 12, 14, 15, 16, 19, 20, 21, 22, 24, 25, 26, 27, 28, 32, 33$ and $34$
0%
None of these

Explanation

All composite numbers between

$10$ and

$35$ are

$12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33$ and

$34.$

The locus of

$\displaystyle z= x+iy$ which satisfying the inequality

$\displaystyle \log_{1/2}\left | z-1 \right |> \log_{1/2}\left | z-i \right |$ is given by

0%
$\displaystyle x+y< 0$
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$\displaystyle x-y> 0$
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$\displaystyle x-y< 0$
0%
$\displaystyle x+y> 0$

Explanation

$log_{0.5}|z-1|>log_{0.5}|z-i|$

$-log_{2}|z-1|>-log_{2}|z-i|$

$log_{2}|z-1|<log_{2}|z-i|$
Hence

$|z-1|<|z-i|$
Now
Let

$z=x+iy$
Hence by squaring both sides

$(x-1)^{2}+y^{2}<x^{2}+(y-1)^{2}$

$(2y-1)<(2x-1)$

$2(y-x)<0$

$y-x<0$
Or

$x-y>0$

The complex number

$z$ satisfying the equations

0%
$\sqrt { 3 } -i$
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$2\sqrt { 3 } -2i$
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$-2\sqrt { 3 } -2i$
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$0$

Explanation

We have two equation

$\left| z \right| -4=0$ and

$\left| z-i \right| -\left| z+5i \right| =0$

Putting

$z=x+iy$ i.e

${ x }^{ 2 }+{ y }^{ 2 }=16$ ....(1)

and

$\left| x+iy-i \right| =\left| x+iy+5i \right|$

$\Rightarrow{ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ x }^{ 2 }+{ \left( y+5 \right) }^{ 2 }$

$\Rightarrow y=-2$ ...(2)

Putting

$y=-2$ in (1),

${ x }^{ 2 }+4=16\Rightarrow x=\pm 2\sqrt { 3 }$

Hence the complex numbers

$z$ satisfying the given equation are

${ z }_{ 1 }=2\sqrt { 3 } -2i$ and

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

Solve :

$(2-\sqrt{-100})(1+\sqrt{-36})$

0%
$62+2i$
0%
$52+2i$
0%
$-52+2i$
0%
$-88+2i$

Explanation

$(2-\sqrt{-100})(1+\sqrt{-36})$

$=(2-10i)(1+6i)$

$=2+12i-10i+60$

$=62+2i$

$\displaystyle z=1+i\cot\alpha,-\frac{\pi}{2}<\alpha<0,$ then

$|z|$ is equal to

0%
$cosec\alpha$
0%
$-cosec\alpha$
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$cosec\alpha$ or $-cosec\alpha$
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none of these

Explanation

$z=1+i\cot\alpha$

$\Rightarrow |z|=\sqrt{1+\cot^2\alpha}=\sqrt{co\sec^2\alpha}$

$=|co\sec\alpha|=-co\sec\alpha,$ as

$-\pi/2<\alpha<0$

$z=4+i\sqrt7$ , then value of

$z^3-4z^2-9z+91$ equals

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

Explanation

We have,

$z-4 = i\sqrt 7$

$z^2 - 8z = -23$

$z^3 -4z^2-9z+91$

$= z(z^2-8z) + 4(z^2-8z) +23z +91$

$=-23z -92 +23z+91$

$= -1$

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

0%
$\displaystyle i$
0%
$\displaystyle 1$
0%
$\displaystyle -i$
0%
$\displaystyle -1$

Explanation

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

$\displaystyle =\left ( \cos ^{2}\theta -i^{2}\sin ^{2}\theta \right )^{-1}$

$\displaystyle =\left ( \cos ^{2}\theta +\sin ^{2}\theta \right )^{-1}=1.$
Alt.

$\displaystyle \left ( e^{i\theta } .e^{-i\theta }\right )^{-1}=1$

Ans: B

$\left| z \right| \ge 5$ then the least value

$\left| {z + \frac{2}{z}} \right|$ is

0%
$\frac{{23}}{5}$
0%
$\frac{{24}}{5}$
0%
$5$
0%
none of these

For a complex number

$z$ , the minimum value of

$\left | z \right |+\left | z-\cos\alpha-i\sin\alpha \right |$ is

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

Find the modulus and amplitude of

$-2 + 2 \sqrt 3i$

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$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {\pi}{3}$
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$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {2\pi}{3}$
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$|z|=4; amp(z)=\dfrac {2\pi}{3}$
0%
$|z|=4; amp(z)=\dfrac {\pi}{3}$

Explanation

Given

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

Hence,

$|z|=2\sqrt{1+3}\ \Rightarrow 2\sqrt{4}$

$\Rightarrow 2\times2=4$
And let

$\alpha$ be the argument of the complex number.
Then

$tan\alpha=-\sqrt{3}$

Now

$Re(z)<0$ and

$Im(z)>0$

$\therefore$ the number lies in the II quadrant
Hence,

$\alpha=\pi-\dfrac{\pi}{3}$

$=\dfrac{2\pi}{3}$

The inequality

$|z-4| < |z-2|$ represents the region given by

0%
$Re(z) > 1$
0%
$Re(z) < 2$
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$Re(z) > 0$
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None of these

Explanation

Let,

$z=x+iy$

Putting in the given inequality,

$|(x-4)-iy|<|(x-2)+iy|$

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

squaring both sides,

$\Rightarrow (x-4)^2+y^2<(x-2)^2+y^2$

$\Rightarrow -8x+16<-4x+4$

$\Rightarrow 4x>12\Rightarrow x>3\Rightarrow \text{Re(z)}>3$

Solve:

$\displaystyle \left ( x+iy \right )\left ( 2-3i \right )= 4+i$

0%
$\displaystyle x= \left ( 8/13 \right ), y= -\left ( 14/13 \right ).$
0%
$\displaystyle x= \left ( 5/13 \right ), y= \left ( 14/13 \right ).$
0%
$\displaystyle x= -\left ( 14/13 \right ), y= \left ( 5/13 \right ).$
0%
$\displaystyle x= \left ( 14/13 \right ), y=- \left ( 8/13 \right ).$

Explanation

Given,

$\displaystyle \left ( x+iy \right )\left ( 2-3i \right )= 4+i$

$\displaystyle \left ( 2x+3y \right )+i\left ( -3x+2y \right )= 4+i.$

Equating real and imaginary parts,

$\displaystyle 2x+3y= 4, -3x+2y= 1.$

Solving these, we get

$\displaystyle x= \left ( \dfrac {5}{13} \right ), y= \left (\dfrac {14}{13} \right ).$

Ans: B

Find the value of

$x^3 + 7x^2 -x + 16$ , where

$x = 1 + 2i$

0%
$-17 + 24i$
0%
$24 + 17i$
0%
$17 - 24i$
0%
$24 - 17i$

Explanation

$f\left(x\right)={x}^{3}+7{x}^{2}-x+16$

$f(1+2i)=(1+2i)^3+7(1+2i)^2-(1+2i)+16$

$=\left[1+8{i}^{3}+6i\left(1+2i\right)\right]+7\left[1+4{i}^{2}+4i\right]-1-2i+16$

$=\left[1-8i+6i-12\right]+7\left[1-4+4i\right]-2i+15$

$=\left[-11-2i\right]+7\left[4i-3\right]-2i+15$

$=-11-2i+28i-21-2i+15$

$=\left(-32+15\right)+24i$

$=-17+24i$

Consider the following statements:
A. The sum of two prime numbers is a prime number.
B. The product of two prime numbers is a prime number.
Which of these statements is/are correct ?

0%
Neither A nor B
0%
A alone
0%
B alone
0%
Both A and B

Explanation

Neither the sum nor the product of two prime numbers is a prime number

Eg.

$3+5=8$ not a prime number.

$3\times 5=15$ not a prime number.

Which one of the following is the largest prime number of three digits?

0%
$997$
0%
$999$
0%
$991$
0%
$993$

Explanation

There are two numbers

$997$ and

$991$ are prime numbers out of given numbers.

$\therefore$ largest prime no. is

$997$

Option

$A$ is correct.

$z = re^{i\theta}$ , then the value of

$|e^{iz}|$ is equal to

0%
$e^{rcos\theta}$
0%
$e^{-rcos\theta}$
0%
$e^{rsin\theta}$
0%
$e^{-rsin\theta}$

Explanation

Ginen,

$z=r{ e }^{ i\theta }$

To find

$\left| { e }^{ iz } \right|$

solution,

for given,

$z=r{ e }^{ i\theta }\\ z=r(\cos\theta +i\sin\theta )\left\{ { e }^{ i\theta }=(\cos\theta +i\sin\theta ) \right\} \\ iz=ir(\cos\theta +i\sin\theta )\\ iz=r(i\cos\theta +{ i }^{ 2 }\sin\theta )\\ iz=r(-\sin\theta +i\cos\theta )\\ iz=-r\sin\theta +ir\cos\theta \\ { e }^{ iz }={ e }^{ -r\sin\theta +ir\cos\theta }\\ \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \left| { e }^{ ir\cos\theta } \right|$

We know that

$\left| { e }^{ iz } \right| =1\\ \therefore \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \longrightarrow \{ always\quad positive\} \\ { e }^{ iz }={ e }^{ -r\sin\theta }$

Find the modulus and amplitude of

$-2i$

0%
$|z|=2; amp(z)=-\dfrac {3\pi}{2}$
0%
$|z|=2i; amp(z)=\dfrac {\pi}{2}$
0%
$|z|=2; amp(z)=\dfrac {\pi}{2}$
0%
$|z|=2; amp(z)=-\dfrac {\pi}{2}$

Explanation

$z=-2i$

$|z|=2$

$=2(-i)$

$=2e^{i\frac{-\pi}{2}}$

$=|z|(e^{i\arg(z)})$

Hence

$|z|=2$ and

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

A student was asked to find the sum of all the prime numbers between

$10$ and

$40.$ He found the sum as

$180.$ Which of the following statements is true?

0%
He missed one prime number between $10$ and $20$
0%
He missed one prime number between $20$ and $30$
0%
He added one extra prime number between $10$ and $20$
0%
None of these

Explanation

Prime numbers between

$10$ and

$40$ are

$11,13,17,19,23,29,31,37$
Sum of these prime numbers

$= 180$

$\therefore$ Option

$D$ is correct.

Which of the following statement is true?

0%
1 is the smallest prime number
0%
Every prime number is an odd number
0%
The sum of two prime numbers is always a prime number
0%
None of these

Explanation

${\because}$ All the given statements are false.

The number

$10$ has four factors:

$1,2, 5$ and

$10$ . The table below lists the number of factors for some numbers

Numbers	Number of factors
$21$	$4$
$23$	$2$
$25$	$3$
$27$	$4$
$29$	$2$

From this, we can say that the number of prime numbers between

$20$ and

$30$ is:

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

Explanation

We know that prime numbers have two factors that is

$1$ and number itself.

In the given table only

$23$ and

$29$ have

$2$ factors so there will be two prime numbers that is

$23$ and

$29$ so correct answer will be option B

Find the modulus and amplitude of

$-\sqrt 3-i$

0%
$|z|=2; amp(z)=-\dfrac {5\pi}{6}$
0%
$|z|=4; amp(z)=\dfrac {5\pi}{6}$
0%
$|z|=4; amp(z)=-\dfrac {\pi}{6}$
0%
none of these

Explanation

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

$|z|=2$

$=2(\dfrac{-\sqrt{3}-i}{2})$

$=2e^{i(\frac{\pi}{6}-\pi)}$

$=2e^{i\frac{-5\pi}{6}}$

$=|z|.e^{iarg(z)}$

Hence

$|z|=2$ and

$arg(z)=\dfrac{-5\pi}{6}$ .

(i) Every prime number is odd.

(ii) The product of any two prime numbers is odd.

Which of the above statements(s) is/are correct?

0%
Statement I alone.
0%
Statement II alone.
0%
Both statements I and II.
0%
Neither statement I nor statement II.

$z_1, z_2, \varepsilon C$ are such that

$|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$ then

$\displaystyle \frac{z_1}{z_2}$ is

0%
zero
0%
purely real
0%
purely imaginary
0%
complex

Explanation

$\because |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$

$\Rightarrow |z_1|^2 + |z_2|^2 + 2 R (z_1 \bar z_2) = |z_1|^2 + |z_2|^2$

$\Rightarrow R(z_1 \bar z_2) = 0 \Rightarrow z_1 \bar z_2$ is imaginary

Now

$\displaystyle \frac{z_1}{z_2} = \frac{z_1 \bar z_2}{z_2 \bar z_2} = \frac{\text{imaginary}}{|z_2|^2} =$ imaginary

Let

$\displaystyle \left| Z_{r}-r\right| \leq r,\forall r=1,2,3,.......n.$ Then

$\displaystyle \left| \sum_{r=1}^{n} Z_{r} \right|$ is less than

0%
$n$
0%
$2n$
0%
$n(n+1)$
0%
$\displaystyle \frac{n(n+1)}{2}$

Explanation

$|Z_r - r| \le r$

$\therefore Z_r \le 2r$

$|Z_r| \le 2r$

also

$\sum_{r=1}^n |Z_r| \le 2 \sum_{r=1}^n r$

also

$|\sum_{r=1}^n Z_r| \le \sum_{r=1}^n |Z_r|$

$\{$ extending triangle inequality

$\}$

$\therefore |\sum_{r=1}^n Z_r| \le 2\times \dfrac{n(n+1)}{2}$

$\therefore |\sum_{r=1}^n Z_r| \le {n(n+1)}$

$\therefore$ option C is correct.

$(1 + i)^8 + (1 -i)^8 =$

0%
$16$
0%
$-16$
0%
$32$
0%
$-32$

Explanation

${ \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$

$\Rightarrow { \left( 1+i \right) }^{ 8 }={ \left[ { { \left( 1+i \right) }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i+{ i }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i-1 } \right] }^{ 4 }$

$={ \left[ { { \left( 2i \right) }^{ 2 } } \right] }^{ 2 }$

$={ \left( -4 \right) }^{ 2 }$

$=16$

Similarely

${ \left( 1-i \right) }^{ 8 }=16$

$\therefore { \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$

$=16+16$

$=32.$

Hence, the answer is

$32.$

If z = x + iy and |z 1 + 2i | = | z + 1 2i |,then the locus of z is

0%
circle
0%
parabola
0%
straight line
0%
None of these

Explanation

We know that

$z=x+iy$

$\Rightarrow \left| z-1+2i \right| =\left| z+1-2i \right| ,$ locus of

$x=?$

$\Rightarrow \left| x+iy-1+2i \right| =\left| x+iy+1-2i \right|$

$\Rightarrow \left| \left( x-1 \right) +i\left( y+2 \right) \right| =\left| \left( x+1 \right) +i\left( y-2 \right) \right|$

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

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

$\Rightarrow -2x+4y+5=2x-4y+5$

$\Rightarrow 4x=8y$

$\Rightarrow x=2y$

$\therefore x-2y=0$

$\therefore$ The locus of

$z$ is a straight line.

Hence, the answer straight line.

Modulus of

$\displaystyle \frac{cos \theta- isin\theta }{sin\theta - icos\theta} is$

0%
0
0%
2 $\theta$
0%
$\pi - 2\theta$
0%
none of these

Explanation

$\displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|$

$=\displaystyle \frac{|\cos \theta- i\sin\theta| }{|\sin\theta - i\cos\theta|}$

$= \displaystyle \frac{\sin^2 \theta +\cos^2 \theta}{\sin^2 \theta +\cos^2 \theta}=1$

$\therefore \displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|=1$
Hence, option D.

$z_1$ and

$z_2$ are any two complex numbers, then

$\displaystyle \frac{z_2 + z_1}{||z_2| - |z_1||}$ is

0%
$\leq 1$
0%
$\geq 1$
0%
$\geq -1$
0%
none of these

Find the value of:

$i^2 + i^4 + i^6$ +..... upto

$(2n +1)$ terms.

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

Explanation

$i^2$ +

$i^4$ +

$i^6$ + ...upto

$(2n + 1 )$ terms

$=(-1 +1)+ (-1 +1)+( -1 +$ .....upto

$(2n +1))$ terms

$\Rightarrow$ (odd terms)

$= 0+ 0 + ....-1$

$= -1$

Hence, option D is correct.

$i^n + i^{n + 1} + i^{n + 2}+ i^{n + 3} (n \in N)$ is equal to

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

Explanation

$\textbf{Step-1: Expanding the terms}$

${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}$

$= {{i}^{n}}+{{i}^{n}}.i+{{i}^{n}}.{{i}^{2}}+{{i}^{n}}.{{i}^{3}}$

$\textbf{Step-2: Taking } \mathbf{{i}^{n}} \textbf{ common}$

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

$= {{i}^{n}}(1+i-1-i)$ $\mathbf{[\because {{i}^{2}}=-1,{{i}^{3}}=-i]}$

$= {{i}^{n}}(0)=0$

$\textbf{Hence the correct option is (C) 0}$

Solve:

$\displaystyle \left|(1 + i)\frac{(2+i)}{(3 + i)}\right|$

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

Explanation

Given,

$\displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}|$ =

$\displaystyle |(1 + i)|\frac{|(2+i)|}{|(3 + i)|}$

Since,

$|Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$ and

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

$=\displaystyle \sqrt{2}\cdot \frac{\sqrt{5}}{\sqrt{10}}=1$

$\therefore \displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| = 1$

Hence, option C.

If x and y are real then which one of the following is true

0%
|x - y| = |x| - |y|
0%
|x + y| $\leq$ |x| - |y|
0%
|x - y| $\geq$ |x| - |y|
0%
|x + y| = |x| + |y|

Explanation

We know that

$|x+y|\leq |x|+|y|$

$\Rightarrow |x-y+y|\leq |x-y|+|y|$

$\Rightarrow |x|-|y|\leq |x-y|$

$\therefore |x-y|\geq |x|-|y|$
Hence, option C.

The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to

0%
$\sqrt10$
0%
$\sqrt 5$
0%
5
0%
10

Explanation

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

$|1+i||1+2i||1+3i|$ (

$\because |Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$ )

$=\sqrt {2}\cdot\sqrt {5}\cdot\sqrt {10}$

$\therefore |(1 + i) (1 + 2i) (1 + 3i)|=10$
Hence, option D.

$|z -2 + i| = |z-3-i|$ , then the locus of z is

0%
$2x -4y -5 = 0$
0%
$2x + 4y -5 = 0$
0%
$x -2y + 5 = 0$
0%
none of these

Explanation

Given:

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

To find the locus of:

$z$ .

Solution:

Let,

$z=x+iy$

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

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

$|(x-2)+i(y+1)|=|(x-3)+i(y-1)|$

On squaring both sides, we get

$\Rightarrow { \left( \sqrt { { (x-2) }^{ 2 }+{ (y+1) }^{ 2 } } \right) }^{ 2 }={ \left( \sqrt { { (x-3) }^{ 2 }+{ (y-1) }^{ 2 } } \right) }^{ 2 } \quad ...\left\{ |x+iy|=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right\}$

$\Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1+2y$

$= { x }^{ 2 }+9-6x+{ y }^{ 2 }+1-2y$

$\Rightarrow 2x+4y-5=0$

The value of

$(x - 1) \displaystyle \left ( x + \frac{1}{2} - \frac{\sqrt{3}}{2} i \right )\left ( x + \frac{1}{2} + \frac{\sqrt 3}{2} i \right )$ is

0%
$x^3 + x^2 + x 1$
0%
$x^3 -1$
0%
$x^3 + 1$
0%
$x^3 - x^2 + x + 1$

Explanation

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

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

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

$=\left( x-1 \right) \left[ { x }^{ 2 }+x+\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } \right]$

$=\left( x-1 \right) \left[ { x }^{ 2 }+x+1 \right]$

$=x^3+x^2+x-x^2-x-1$

$=x^3-1$

Hence, the answer is

$x^3-1.$

Number of complex numbers

$z$ satisfying

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

Explanation

Let the complex number

$z$ be

$x+iy$

Given that

$|2z|=|2z-1|$

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

$\Rightarrow 3{y}^{2}+4x=1$

Given that

$|2z|=|2z+1|$

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

$\Rightarrow 3{y}^{2}-4x=1$

By solving above two equation , we get

$x=0 , y=\pm\frac{1}{\sqrt3}$

But this point doesnt satisfy the equation

$|2z-1|=|2z+1|$

Therefore the number of points which satisfies the equation

$|2z|=|2z+1|=|2z-1|$ is

$0$

The locus represented by

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

0%
a circle of radius $1$ unit
0%
an ellipse with foci at $(1, 0)$ and $(0, 1)$
0%
a straight line through the origin
0%
a circle on the line joining $(1, 0)$ and $(0, 1)$ as diameter

Explanation

Given,

$|z−1|=|z+i|$

To find the locus of the equation.

Solution,

Given that,

$|z−1|=|z+i|$

Say

$x=x+iy$

$\left| x+iy-1 \right| =\left| x+iy+1 \right|$

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

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

Squaring both sides,

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

$\Rightarrow 1-2x=2y+1$

$\therefore \ x+y=0$ [line passes through origin]

Hence, C is the right answer

Fill in the blanks.
Every composite number can be expressed as a product of ____, and this factorization is unique except for the order in which the prime factors occur.

0%
Co-primes
0%
Twin primes
0%
Primes
0%
All of these

$\displaystyle z= (i)^{(i)^{(i)}}$ where

$i = \sqrt{-1}$ , then

$z$ is equal to

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

Explanation

$Z=(i)^{(i)^{(i)}}$

$i=e^{i\pi/2}$

$i^i=e^{i.(i\pi/2)}=e^{-\pi/2}$

$i^{i^i}=e^{i(-\pi/2)}\\ \implies \cos(\pi/2)-i\sin(\pi/2)\\ \implies 0-i=-i$

${z}_{1},{z}_{2}$ are two complex numbers and

$c>0$ such that

${ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right| }^{ 2 }+k{ \left| { z }_{ 2 } \right| }^{ 2 },$ then

$k=$

0%
$1-c$
0%
$c-1$
0%
$1+{c}^{-1}$
0%
$1-{c}^{-1}$

Explanation

Since Re

$\displaystyle \left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le \left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right|$

$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+Re\left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right|$

$\displaystyle \Rightarrow { \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$ ...(1)

Also, Since

$A.M.\ge G.M.$

$\displaystyle \therefore \frac { { \left( \sqrt { c } \left| { z }_{ 1 } \right| \right) }^{ 2 }+{ \left( \frac { 1 }{ \sqrt { c } } \left| { z }_{ 2 } \right| \right) }^{ 2 } }{ 2 } \ge { \left\{ c.{ \left| { z }_{ 1 } \right| }^{ 2 }.\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 } \right\} }^{ \frac { 1 }{ 2 } }\left( \because c>0 \right)$

$\displaystyle \Rightarrow c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 1 } \right| }^{ 2 }\ge 2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 }$

$\displaystyle \Rightarrow { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le \left( 1+{ c }^{ -1 } \right) \left( { \left| { z }_{ 2 } \right| }^{ 2 } \right)$ ...(2)

From (1) and (2), we get

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

$\displaystyle \therefore k=1+{ c }^{ -1 }$

Which of the following number is composite?

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

Find the product and write the answer in standard form.

$\left( 2-4i \right) \left( 3+7i \right)$

0%
$34+2i$
0%
$-5-3i$
0%
$5+3i$
0%
$5-3i$

Explanation

Given

$z=(2-4i)(3+7i)$

$z=[2(3+7i)-4i(3+7i)]$

$=[6+14i-12i-28i^2]$

$=[34+2i]$ ............[since,

$i^2=-1$ ]

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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