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 \(i=\sqrt{-1}\) ?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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,
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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|?
Answer (Detailed Solution Below)
Real and Imaginary parts Question 5 Detailed Solution
Calculation
Let z = r.eiθ
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
Answer (Detailed Solution Below)
Real and Imaginary parts Question 6 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Real and Imaginary parts Question 7 Detailed Solution
Download Solution PDFConcept:
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.
Answer (Detailed Solution Below)
Real and Imaginary parts Question 8 Detailed Solution
Download Solution PDFConcept:
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 \(\rm \frac{10}{1-i}\)
Answer (Detailed Solution Below)
Real and Imaginary parts Question 9 Detailed Solution
Download Solution PDFConcept:
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} = ?
Answer (Detailed Solution Below)
Real and Imaginary parts Question 10 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Real and Imaginary parts Question 11 Detailed Solution
Download Solution PDFConcept:
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 \(z=\frac{1-i}{i}\)
Answer (Detailed Solution Below)
Real and Imaginary parts Question 12 Detailed Solution
Download Solution PDFConcept:
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:
\(\Rightarrow z=\frac{1-i}{i}\)
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?
Answer (Detailed Solution Below)
Real and Imaginary parts Question 13 Detailed Solution
Download Solution PDFConcept:
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 i2 = -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.
Answer (Detailed Solution Below)
Real and Imaginary parts Question 14 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Real and Imaginary parts Question 15 Detailed Solution
Download Solution PDFConcept:
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 i2 = -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