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

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 = \(5\sqrt 2\)

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

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

Calculation

Let z = r.e^iθ

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

Calculation

Given \(\bar{z}(1-i)=z^{2}(1+i)\)    ---(i)

So \(|\bar{z}||1-i|=|z|^{2}|1+i|\)

\(\Rightarrow|z|=|z|^{2} \Rightarrow|z|=0 \text { or }|z|=1\)

Let arg (z) = θ 

So from (i) we get

\(2 n \pi-\theta-\frac{\pi}{4}=2 \theta+\frac{\pi}{4}\)

\(\Rightarrow \theta=\frac{1}{3}\left(\frac{4 n-1}{2}\right) \pi=\frac{(4 n-1) \pi}{6}\)

So we will get 3 distinct values of θ.

Hence there will be total 4 possible values of complex number z.

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

Concept:

Euler's Formula on Complex Numbers:

• e^ix = 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

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

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.

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.

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 

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:

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

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 

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 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.

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

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.

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 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