Complex Numbers(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

COMPLEX NUMBERS MCQs

Total Questions : 30 | Page 1 of 3 pages

For positive integers n₁,n₂ the value of the expression $(1 + i)^{n_{1}}$ + $(1 + i^{3})^{n_{1}}$ + $(1 + i^{5})^{n_{2}}$ + $(1 + i^{7})^{n_{2}}$ where i= $\sqrt[2]{- 1}$ is a real number if and only if

$n_{1}$ = $n_{2}$ +1
$n_{1}$ = $n_{2}$ -1
$n_{1}$ = $n_{2}$
$n_{1}$ >0, $n_{2}$ >0

Discuss Question

Answer: Option D. ->

n_{1}

>0,

n_{2}

>0
:
D

Using $i^{3} = - i, i^{5} = i a n d i^{7} = - i$ , we can write the given expression as
$(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}} w h e r e i = \sqrt[2]{1} = 2 [^{n_{1}} C_{0} +^{n_{1}} C_{2} (i)^{2} +^{n_{1}} C_{4} (i)^{4} +^{n_{1}} C_{6} (i)^{6} + \dots \dots] + 2 [^{n_{2}} C_{0} +^{n_{2}} C_{2} (i)^{2} +^{n_{2}} C_{4} (i)^{4} +^{n_{2}} C_{6} (i)^{6} + \dots \dots] = 2 [^{n_{1}} C_{0} -^{n_{1}} C_{2} +^{n_{1}} C_{4} -^{n_{1}} C_{6} + \dots \dots] + 2 [^{n_{2}} C_{0} -^{n_{2}} C_{2} +^{n_{2}} C_{4} -^{n_{2}} C_{6} + \dots \dots]$
This is a real number irrespective of values of $n_{1}$ and $n_{2}$ .

Question 2.

The conjugate of $\frac{(2 + i)^{2}}{3 + i}$ , in the form of a+ib, is

$\frac{13}{2} + i (\frac{15}{2})$
$\frac{13}{10} + i (\frac{- 15}{2})$
$\frac{13}{10} + i (\frac{- 9}{10})$
$\frac{13}{10} + i (\frac{9}{10})$

Discuss Question

Answer: Option C. ->

\frac{13}{10} + i (\frac{- 9}{10})

:
C

z = $\frac{(2 + i)^{2}}{3 + i} = \frac{3 + 4 i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{13}{10} + i \frac{9}{10}$

Conjugate = $\frac{13}{10} - i \frac{9}{10}$ .

Question 3.

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ $n^{2}$ ) is equal to

$a^{2}$ - $b^{2}$
$a^{2}$ + $b^{2}$
$\sqrt{a^{2} + b^{2}}$
$\sqrt{a^{2} - b^{2}}$

Discuss Question

Answer: Option B. ->

a^{2}

b^{2}

:
B
we have

(1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i)

$\Rightarrow$ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)

Multiplying (i) and (ii), we get 2.5.10.....(1+ $n^{2}$ ) = $a^{2}$ + $b^{2}$

Question 4.

$\frac{1 - i}{1 + i}$ is equal to

cos $\frac{π}{2}$ +isin $\frac{π}{2}$
cos $\frac{π}{2}$ -isin $\frac{π}{2}$
sin $\frac{π}{2}$ +icos $\frac{π}{2}$
None of these

Discuss Question

Answer: Option B. -> cos

\frac{π}{2}

-isin

\frac{π}{2}

:
B

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

Which can be written as cos $\frac{π}{2}$ -isin $\frac{π}{2}$

Question 5.

If x=-5+ $\sqrt{- 4}$ , then the value of the expression $x^{4}$ +9 $x^{3}$ +35 $x^{2}$ -x+4 is

160
-160
60
-60

Discuss Question

Answer: Option B. -> -160
:
B

x+5 =4i $\Rightarrow$ $x^{2}$ +10x+25=-16

Now, $x^{4}$ +9 $x^{3}$ +35 $x^{2}$ -x+4

=( $x^{2}$ +10x+41)( $x^{2}$ -x+4)-160=-160

Question 6.

The values of x and y for which the numbers 3+i $x^{2}$ y and $x^{2}$ +y+4i are conjugate complex are

(-2,-1) or (2,-1)
(1,-2) or (-2,1)
(1,2) or (-1,-2)
None of these

Discuss Question

Answer: Option A. -> (-2,-1) or (2,-1)
:
A

According to condition, 3-i $x^{2}$ y = $x^{2}$ + y + 4i

⇒ $x^{2}$ + y = 3 And $x^{2}$ y = -4 ⇒ x = ±2, y = -1

⇒ (x,y) = (2,-1) or (-2,-1)

Question 7.

For the complex number z, which of the following is true?

$z + ¯ z$ is real and $z ¯ z$ is imaginary
$z + ¯ z$ is imaginary and $z ¯ z$ is real
Both $z + ¯ z$ and $z ¯ z$ are real.
Both $z + ¯ z$ and $z ¯ z$ are imaginary numbers

Discuss Question

Answer: Option C. -> Both

z + ¯ z

and

z ¯ z

are real.
:
C

Let $z = x + i y$
Here, $z + ¯ z = (x + i y) + (x - i y) = 2 x$ (Real)

And $z ¯ z = (x + i y) (x - i y) = x^{2} + y^{2}$ (Real).

Question 8.

If z is a complex number such that $z^{2}$ = ( $¯ z)^{2}$ ,then

z is purely real
z is purely imaginary
Either z is purely real or purely imaginary
None of these

Discuss Question

Answer: Option C. -> Either z is purely real or purely imaginary
:
C

Let z = x+iy, then its coonjucate $¯ z$ = x-iy

Given that $z^{2} = (¯ z)^{2}$

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

if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0

Question 9.

If z is a complex number, then z. $¯ z$ = 0 if and only if

z=0
Re(z)=0
Im(z)=0
None of these

Discuss Question

Answer: Option A. -> z=0
:
A

Let z = x+iy, $¯ z$ = x-iy

∴ z $¯ z$ = 0 ⇒ (x+iy)(x-iy) = 0 ⇒ $x^{2} + y^{2}$ = 0

It is possible onle when x and y oth simultaneously zero

i.e, z = 0 +0i = 0

Question 10.

The number of solutions of the equation $z^{2}$ + $¯ z$ = 0 is

Discuss Question

Answer: Option D. -> 4
:
D

Let z = x+iy, so that $¯ z$ = x - iy, therefore

$z^{2} + ¯ z = 0 \Leftrightarrow (x^{2} - y^{2} + x) + i (2 x y - y)$ = 0

Equating real and imaginary parts , we get

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

And 2xy - y = 0 ⇒ y = 0 or x = $\frac{1}{2}$

if y = 0 , then (i) gives $x^{2}$ + x = 0 ⇒ x = 0 or

x = -1

If x = $\frac{1}{2}$ ,

Then $x^{2} - y^{2} + x = 0 \Rightarrow y^{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \Rightarrow y = \pm \frac{\sqrt{3}}{2}$

Hence, there are four solutions in all.

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