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

12th Grade > Mathematics

COMPLEX NUMBERS MCQs

Complex Numbers

Total Questions : 60 | Page 2 of 6 pages

I f | z | = 1 a n d ω = \frac{z - 1}{z + 1} (w h e r e, z \neq - 1), t h e n R e (ω) i s

0
1|z+1|2
∣∣1z+1∣∣.1|z+1|2
√2|z+1|2

Discuss Question

Answer: Option A. -> 0
:
A

S i n c e, | z | = 1 a n d ω = \frac{z - 1}{z + 1} \Rightarrow z - 1 = ω z + ω \Rightarrow z = \frac{1 + ω}{1 - ω} \Rightarrow | z | = \frac{| 1 + ω |}{| 1 - ω |} \Rightarrow | 1 - ω | = | 1 + ω | [∴ | z | = 1] O n s q u a r i n g b o t h s i d e s, w e g e t 1 + | ω |^{2} - 2 R e (ω) = 1 + | ω |^{2} + 2 R e (ω) [u s i n g | z_{1} \pm z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} \pm 2 R e ({¯ z}_{1} z_{2})] \Rightarrow 4 R e (ω) = 0 \Rightarrow R e (ω) = 0

Question 12. The number of complex numbers z such that

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

equals

Discuss Question

Answer: Option A. -> 1
:
A
Let z = x + iy

| z - 1 | = | z + 1 | \Rightarrow (x - 1)^{2} + y^{2} = (x + 1)^{2} + y^{2} \Rightarrow R e (z) = 0 \Rightarrow x = 0 | z - 1 | = | z - i | \Rightarrow (x - 1)^{2} + y^{2} = x^{2} + (y - 1)^{2} \Rightarrow x = y | z + 1 | = | z - i | \Rightarrow (x + 1)^{2} + y^{2} = x^{2} + (y - 1)^{2}

Only (0, 0) satisfies all conditions.

Question 13. If -1+

\sqrt{- 3}

e^{i θ}

, then

θ

is equal to

π3
-π3
2π3
-2π3

Discuss Question

Answer: Option C. -> 2π3
:
C
Here -1+

\sqrt{- 3}

e^{i θ}

\Rightarrow

-1+

i \sqrt{3}

e^{i θ}

=rcos

θ

=-1 And rsin

θ

\sqrt{3}

Hence tan

θ

\sqrt{3}

\Rightarrow

tan

θ

=tan

\frac{2 π}{3}

.Hence

θ

\frac{2 π}{3}

Question 14. If

z_{1}, z_{2} a n d z_{3}

are complex numbers such that
|

z_{1}

|=|

z_{2}

|=|

z_{3}

∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1

,
then |

z_{1}

z_{2}

z_{3}

Equal to 1
Less than 1
Greater than 3
Equal to 3

Discuss Question

Answer: Option A. -> Equal to 1
:
A
1=

∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣

∣ ∣ ∣ \frac{z_{1} * {¯ z}_{1}}{z_{1}} + \frac{z_{2} * {¯ z}_{2}}{z_{2}} + \frac{z_{3} * {¯ z}_{3}}{z_{3}} ∣ ∣ ∣

(hence,

{| z_{1} |}^{2}

=1=

z_{1}

_{1}

, etc)

\Rightarrow

z_{1}

z_{2}

z_{3}

|=|

¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2} + z_{3}

|= |

z_{1}

z_{2}

z_{3}

|
(hence,|

_{1}

|=|

z_{1}

Question 15. 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

n1=n2+1
n1=n2-1
n1=n2
n1>0,n2>0

Discuss Question

Answer: Option D. -> n1>0,n2>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 16. 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 17. The conjugate of

\frac{(2 + i)^{2}}{3 + i}

, in the form of a+ib, is

132+i(152)
1310+i(−152)
1310+i(−910)
1310+i(910)

Discuss Question

Answer: Option C. -> 1310+i(−910)
:
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 18. 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 19. If x=-5+

\sqrt{- 4}

, then the value of the expression

x^{4}

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}

x^{3}

+35

x^{2}

-x+4
=(

x^{2}

+10x+41)(

x^{2}

-x+4)-160=-160

Question 20. 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

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