Real and Imaginary parts MCQ Quiz - Objective Question with Answer for Real and Imaginary parts - Download Free PDF

Last updated on May 20, 2025

Latest Real and Imaginary parts MCQ Objective Questions

Real and Imaginary parts Question 1:

What is the value of \(\sqrt{12+5 i}+\sqrt{12-5 i}\) where \(i=\sqrt{-1}\) ?

  1. 24
  2. 25
  3. 5√2
  4. 5(√2 - 1)
  5. 5

Answer (Detailed Solution Below)

Option 3 : 5√2

Real and Imaginary parts Question 1 Detailed Solution

Formula Used: 

(a + b) (a - b) = a2 - b2

i2 = -1

Calculation:

Let x =\(\sqrt{12+5 i}+\sqrt{12-5 i}\), where \(i=\sqrt{-1}\)

Squaring both sides,

x2\({12+5 i}+{12-5 i} +2\sqrt{12+5 i}\sqrt{12-5 i}\)

x2\({24} +2\sqrt{144+25}\)

x2 = \({24} +2\sqrt{169}\)

x2 = \(24 +26\)

x2 = \(50\)

x = \(5\sqrt 2\)

Real and Imaginary parts Question 2:

If α and β are the roots of the equation 2z2 – 3z – 2i = 0, where \(\mathrm{i}=\sqrt{-1}\), then \(\text { 16. } \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\) is equal to 

  1. 398 
  2. 312 
  3. 409 
  4. 441

Answer (Detailed Solution Below)

Option 4 : 441

Real and Imaginary parts Question 2 Detailed Solution

Calculation

2z2 - 32 - 2i = 0

⇒ \(2\left(z-\frac{i}{z}\right)=3\)

\(\alpha-\frac{i}{\alpha}=\frac{3}{2}\)

⇒ \(\alpha^{2}-\frac{1}{\alpha^{2}}-2 i=\frac{9}{4}\)

⇒ \(\alpha^{2}-\frac{1}{\alpha^{2}}-2 i=\frac{9}{4}\)

⇒ \(\frac{9}{4}+2 i=\alpha^{2}-\frac{1}{\alpha^{2}}\)

⇒ \(\frac{81}{16}-4+9 i=\alpha^{4}+\frac{1}{\alpha^{4}}-2\)

⇒ \(\frac{49}{16}+9 i=\alpha^{4}+\frac{1}{\alpha^{4}}\)

Similarly

⇒ \(\frac{49}{16}+9 i=\beta^{4}+\frac{1}{\beta^{4}}\)

\(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right)+\beta^{15}\left(\beta^{4}+\frac{1}{\beta^{4}}\right)}{\alpha^{15}+\beta^{15}}\)

⇒ \(\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 i\right)}{\left(\alpha^{15}+\beta^{15}\right)}\)

Real = \(\frac{49}{16}\)

Im = 9

\(\text { 16. } \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) = 441\)

Hence option 4 is correct

Real and Imaginary parts Question 3:

If |z1| = |z2| = ....|zn| = 1, then the value of |z1 + z2 + .... zn| - \(\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots \ldots . .+\frac{1}{z_{n}}\right|\) is,

  1. 0
  2. 1
  3. -1
  4. None

Answer (Detailed Solution Below)

Option 1 : 0

Real and Imaginary parts Question 3 Detailed Solution

Answer (1)

\(\mathrm{z}_{1} \overline{\mathrm{z}}_{1}=\mathrm{z}_{2} \overline{\mathrm{z}}_{2}=\ldots .=\mathrm{z}_{\mathrm{n}} \overline{\mathrm{z}}_{\mathrm{n}}=1\)

⇒ \(\bar{z}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}\)

∴ \(\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|\)

\(\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0\)

Real and Imaginary parts Question 4:

If a complex number \(z\) is such that \( \frac{z - 2i}{z - 2} \) is a purely imaginary number and the locus of \(z\) is a closed curve, then the area of the region bounded by that closed curve and lying in the first quadrant is:

  1. \(2\pi\)
  2. \(\frac{\pi}{2}\)
  3. \(\pi\)
  4. \(\frac{\pi}{4}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{\pi}{2}\)

Real and Imaginary parts Question 4 Detailed Solution

Calculation

Given:

A complex number \(z\) such that \(\frac{z-2i}{z-2}\) is purely imaginary.

Let \(z = x + iy\), where \(x\) and \(y\) are real numbers. Then

\(\frac{z-2i}{z-2} = \frac{x+iy-2i}{x+iy-2} = \frac{x + i(y-2)}{(x-2) + iy}\)

\(= \frac{(x+i(y-2))((x-2)-iy)}{(x-2)^2 + y^2}\)

\(= \frac{x(x-2) + y(y-2) + i[(y-2)(x-2)-xy]}{(x-2)^2 + y^2}\)

Since \(\frac{z-2i}{z-2}\) is purely imaginary, its real part must be zero:

⇒ \(x(x-2) + y(y-2) = 0\)

⇒ \(x^2 - 2x + y^2 - 2y = 0\)

⇒ \((x-1)^2 + (y-1)^2 = 2\)

This is the equation of a circle with center \((1, 1)\) and radius \(\sqrt{2}\).

The area of the region bounded by this circle and lying in the first quadrant is \(\frac{1}{4}\) of the total area of the circle.

Area of the circle is \(\pi r^2 = \pi (\sqrt{2})^2 = 2\pi\).

The area in the first quadrant is \(\frac{1}{4}(2\pi) = \frac{\pi}{2}\).

Hence option 2 is correct

Real and Imaginary parts Question 5:

Let z̅ denote the complex conjugate of a complex number z. If z is a non-zero complex number for which both real and imaginary parts of \((\bar{z})^{2}+\frac{1}{z^{2}}\) are integers, then which of the following is/are possible value(s) of |z|?

  1. \(\left(\frac{43+3 \sqrt{205}}{2}\right)^{1 / 4}\)
  2. \(\left(\frac{7+\sqrt{33}}{4}\right)^{1 / 4}\)
  3. \(\left(\frac{9+\sqrt{65}}{4}\right)^{1 / 4}\)
  4. \(\left(\frac{7+\sqrt{13}}{6}\right)^{1 / 4}\)

Answer (Detailed Solution Below)

Option :

Real and Imaginary parts Question 5 Detailed Solution

Calculation

Let z = r.e

So, \((\bar{z})^{2}+\frac{1}{z^{2}}=\left(r^{2}+\frac{1}{r^{2}}\right) e^{-2 i \theta}=a+i b(\text { say) }\), where a, b ∈ Z

So, \(\left(r^{2}+\frac{1}{r^{2}}\right)^{2}=a^{2}+b^{2}\)

\(\Rightarrow r^{8}-\left(a^{2}+b^{2}-2\right) r^{4}+1=0\)

\(\Rightarrow \quad r^{4}=\frac{\left(a^{2}+b^{2}-2\right) \pm \sqrt{\left(a^{2}+b^{2}-2\right)^{2}-4}}{2}\)

For \(a^{2}+b^{2}=45 \text { (i.e }(a, b)=( \pm 6, \pm 3) \text { or }( \pm 3, \pm 6)\)

We get \(r=\left(\frac{43+3 \sqrt{205}}{2}\right)^{1 / 4}\)

Hence option 1 is correct

Top Real and Imaginary parts MCQ Objective Questions

If (2 - i) (x - iy) = 3 + 4i then 5x is

  1. 2
  2. 3
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 1 : 2

Real and Imaginary parts Question 6 Detailed Solution

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Concept:

Equality of complex numbers.

Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2

"OR"

Re (z1) = Re (z2) and Im (z1) = Im (z2).

Calculation:

Given:

(2 - i) (x - iy) = 3 + 4i

⇒ 2x - 2iy - ix + i2y = 3 + 4i

⇒ 2x - 2iy - ix - y = 3 + 4i                 (∵ i2 = -1)

⇒ (2x – y) + i(-x - 2y) = 3 + 4i

Equating real and imaginary parts,

2x - y = 3      ----(1)

-x - 2y = 4       ----(2)

Solving equation 1 and 2, we get

x = \(\frac 2 5\) and y = \(\frac {-11}{5}\)

Now, the value of 5x can be calculated as:

5x = 5 × \(\frac 2 5\) = 2

What is the real part of (sin x + icos x)3

  1. –cos 3x
  2. –sin 3x
  3. sin 3x
  4. cos 3x

Answer (Detailed Solution Below)

Option 2 : –sin 3x

Real and Imaginary parts Question 7 Detailed Solution

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Concept:

Euler's Formula on Complex Numbers:

  • eix = cos x + i sin x
  • e-ix = cos x - i sin x


Calculation:

(sin x + icos x)3

Take i common, we get

(sin x + icos x)3

\({{\rm{i}}^3}{\left( {\frac{{\sin {\rm{x}}}}{{\rm{i}}} + {\rm{\;}}\cos {\rm{x}}} \right)^3}\)

= -i  × (-i sin x + cos x)3                

(∵ i3 = -i and 1/i = -i)

= -i × (cos x - i sin x) 3

\(= {\rm{}} - {\rm{i\;}} \times {\rm{}}{\left( {{{\rm{e}}^{ - {\rm{ix}}}}} \right)^3} \)      

\(= {\rm{}} - {\rm{i\;}} \times {\rm{\;}}{{\rm{e}}^{ - {\rm{i}}3{\rm{x}}}}{\rm{\;}}\)

(∵e-ix = cos x - i sin x)

= -i (cos 3x – i sin 3x)

= (-i cos 3x + i2 sin 3x)

= -sin3x – i cos 3x

∴ Real part = -sin 3x

Find the value of θ for which \(\rm z=\frac {3-2i\sinθ}{2+i\sinθ}\) is purely real.

  1. θ = nπ, where n belongs to an integer
  2. No value of θ exists
  3. 0
  4. π 

Answer (Detailed Solution Below)

Option 1 : θ = nπ, where n belongs to an integer

Real and Imaginary parts Question 8 Detailed Solution

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Concept:

Let z = x + iy be a complex number, Where x is called the real part of a complex number or Re (z) and y is called the imaginary part of the complex number or Im (z)

Condition for purely real: Imaginary part equals zero.

Condition for purely imaginary: Real part equals zero.

Calculation:

\(\rm z=\frac {3-2i\sinθ}{2+i\sinθ}\)

Multiplying the numerator and denominator by 2 + i sin θ

\(\rm⇒ z=\frac{3-2isinθ}{2+isinθ}\times \frac{2-isinθ}{2-isinθ}\)

\(\rm =\frac{6-3isinθ-4isinθ+2i^2sin^2θ}{4-i^2sin^2θ}\)

\(\rm =\frac{6-7isinθ-2sin^2θ}{4+sin^2θ}\)       (∵ i2 = -1)

\(\rm⇒ z=\frac{6+2sin^2θ}{4+sin^2θ}+\frac{-7isinθ}{4+sin^2θ}\)

The imaginary part of z 

\(\rm ⇒ Im(z)=\frac{-7\sin θ}{4+\sin^2θ}\)

For z to be purely real Im(z) = 0

\(\therefore \frac{-7\sinθ}{4+\sin^2θ}=0\)

⇒ sin θ = 0

So, θ = nπ, where n belongs to an integer

Find the argument of the complex no \(\rm \frac{10}{1-i}\)

  1. \(\frac {\pi}{6}\)
  2. \(\frac{\pi}{3}\)
  3. 0
  4. \(\frac{\pi}{4}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{\pi}{4}\)

Real and Imaginary parts Question 9 Detailed Solution

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Concept:

Let z = x + iy be a complex number.

  • Modulus of z =  \(\rm \left | z \right |= \sqrt{x^{2}+y^{2}}\)
  • Arg (z) = Arg (x + iy) =  \(\rm tan^{-1}\left ( \frac{y}{x} \right )\) 
 

Calculation: 

Let \(\rm z=\frac{10}{1-i}\)

Multiplying numerator and denominator by 1 + i

\(\rm \Rightarrow z =\frac{10}{1-i}\times \frac{1+i}{1+i}\)

\(\rm =\frac{10(1+i)}{1-i^2}\)

\(\rm =\frac{10(1+i)}{2}\)

= 5 + 5i

\(\rm \Rightarrow arg(z)=tan^{-1}(5/5)\)

\(\rm \therefore arg(z)=\frac{\pi}{4}\)

Hence, option 4 is correct.

If a complex number z = (2x - 3y) + i(x2 - y2) = 0, then Re{z} = ?

  1. 3y - x
  2. 2x + x2
  3. y2
  4. None of these.

Answer (Detailed Solution Below)

Option 4 : None of these.

Real and Imaginary parts Question 10 Detailed Solution

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Concept:

Complex Numbers:

  • A complex number is a number of the form a + ib, where a and b are real numbers and i is the complex unit defined by i = \(\rm \sqrt{-1}\).
  • 'a' is called the real part Re{z} and b is called the imaginary part Im{z}.
  • If z1 = z2, then Re{z1} = Re{z2} and Im{z1} = Im{z2}.
  • The number 0 can be written as: 0 + i0.

 

Calculation:

Since, z = (2x - 3y) + i(x2 - y2) = 0, it means that both the real and the imaginary parts of the z are equal to 0.

i.e. Re{z} = 2x - 3y = 0.

If 2x2 + (x2 - y)i = (8 - 3i), find the values of x and y

  1. x = -1, 1 and y = 4
  2. x = 0, y = 3
  3. x = -2, 2 and y = 7
  4. x = 2, 3 and y = 7, 12 

Answer (Detailed Solution Below)

Option 3 : x = -2, 2 and y = 7

Real and Imaginary parts Question 11 Detailed Solution

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Concept:

Equality of complex numbers:

Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2

Or Re (z1) = Re (z2) and Im (z1) = Im (z2)

 

Calculation:

Given that, 2x2 + (x2 - y)i = (8 - 3i)

Comparing the real and imaginary parts,

⇒ 2x2 = 8 

⇒ x2 = 4

⇒ x = -2, 2

And, x2 - y = -3

⇒ 4 - y = -3               [∵ x2 = 4]

⇒ y = 7 

Hence, option (3) is correct.

Find the real and imaginary part of the complex number \(z=\frac{1-i}{i}\)

  1. 1 ,1
  2. -1 ,1
  3. 1 , -1
  4. -1 , -1

Answer (Detailed Solution Below)

Option 4 : -1 , -1

Real and Imaginary parts Question 12 Detailed Solution

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Concept:

Equality of complex numbers.

Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2

Or Re (z1) = Re (z2and Im (z1) = Im (z2).

 

Calculations:

\(\Rightarrow z=\frac{1-i}{i}\)

Multiplying numerator and denominator by i

\(\Rightarrow z=\frac{1-i}{i}\times \frac{i}{i}\)

\(=\frac{i-i^2}{i^2}\)

\(=\frac{i+1}{-1}\)

\(=-1-i\)

Re(z) = -1 

Im(z) = -1

Hence , option 4 is correct 

If \(\rm x + iy = \dfrac{3+4i}{2-i}\) where \(\rm i = \sqrt{-1}\), then what is the value of y?

  1. \(\dfrac {9}{5}\)
  2. \(\dfrac {11}{5}\)
  3. \(\dfrac {2}{5}\)
  4. \(\dfrac {13}{5}\)

Answer (Detailed Solution Below)

Option 2 : \(\dfrac {11}{5}\)

Real and Imaginary parts Question 13 Detailed Solution

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Concept:

Equality of complex numbers.

Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2

Or Re (z1) = Re (z2) and Im (z1) = Im (z2).

Calculations:

Given:  \(\rm x + iy = \dfrac{3+4i}{2-i}\)

\(⇒ \rm x + iy = \dfrac{3+4i}{2-i}\times\dfrac{2+i}{2+i}\)

\(⇒ \rm x + iy = \dfrac{6+11i+4i^2}{4-i^2}\)   

As we know i= -1 

\(⇒ \rm x + iy = \dfrac{6+11i-4}{4+1}\)

\(⇒ \rm x + iy = \dfrac{2+11i}{5} = \dfrac 25+i \dfrac{11}{5}\)

Comparing real and imaginary parts, we get.

\(\rm x= \dfrac 2 5 \; and \;y = \dfrac {11}{5}\)

1. The difference of Z and its conjugate is an imaginary number.

2. The sum of Z and its conjugate is a real number.

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 3 : Both 1 and 2

Real and Imaginary parts Question 14 Detailed Solution

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Concept:

Let z = x + iy be a complex number,

Where x is called the real part of the complex number or Re (z) and y is called the Imaginary part of the complex number or Im (z)

Conjugate of z = z̅ = x - iy  

Calculation:

1. The difference of Z and its conjugate is an imaginary number.

Consider z = a + ib           ....(i)

conjugate of z = z̅ = a - ib      ....(ii)

eq(i) - eq (ii)

z - z̅ = a + ib - a + ib

⇒ 2ib

Thus it is clear that the difference of z and its conjugate is an imaginary number.

2. The sum of Z and its conjugate is a real number.

eq (i) + eq(ii)

z + z̅ = a + ib + a - ib

2a

Thus it is clear that the sum of Z and its conjugate is a real number.

So, Both 1 and 2 are correct.

If \(\rm A + iB = \dfrac{4+2i}{1-2i}\) where \(\rm i = \sqrt{-1}\), then what is the value of A?

  1. -8
  2. 0
  3. 4
  4. 8

Answer (Detailed Solution Below)

Option 2 : 0

Real and Imaginary parts Question 15 Detailed Solution

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Concept:

Let A = x1 + iy1 and B = x2 + iy2 

If A = B then x1 = x2 and y1 = y2

 

Calculations:

Given \(\rm A + iB = \dfrac{4+2i}{1-2i}\)

\(\rm A + iB = \dfrac{4+2i}{1-2i}\times\dfrac{1+2i}{1+2i}\)

⇒  \(\rm A + iB = \dfrac{4+10i+4i^2}{1-4i^2}\)   

We know i= -1 

\(\rm A + iB = \dfrac{4+10i-4}{1+4}\)

\(\rm A + iB = \dfrac{10i}{5}\)

⇒A + iB = 2i

⇒ A + iB = 0 + 2i

Comparing real and imaginary parts, we get.

⇒A = 0

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