Real and Imaginary parts MCQ Quiz in বাংলা - Objective Question with Answer for Real and Imaginary parts - বিনামূল্যে ডাউনলোড করুন [PDF]

Calculation:

x + iy = \(\rm 2+3i \over 1-i\)

x + iy = \(\rm {2+3i \over 1-i}\times{1+i\over1+i}\)

x + iy = \(\rm 2+5i-3\over 1^2-(i)^2\)

x + iy = \(\rm -{1\over2}+{5\over2}i\)

∴ x = \(-{1\over2}\) and y = \(5\over2\)

Concept:

1. A complex number (Z):  Complex number is the combination of a real number and an imaginary number. It is given by

Z = x + iy, where 'x' and 'y' are the real and imaginary parts of Z and

i = √-1 or i² = -1

Re(Z) = x and Img(Z) = y

2. Two complex numbers will be equal if their real and imaginary part are equal.

Calculation:

Given that,

\(\frac{(1 + i)x - 2i}{3 + i}+\frac{(2-3i)y+i}{3 - i}= i\)

\(\Rightarrow\ \frac{[(1 + i)x - 2i](3 \ -\ i)\ +\ [(2-3i)y+i](3+i)}{9\ -\ i^2}\ =\ i\)

⇒(1 + i)(3x - ix) - 2i(3 - i) + (2 - 3i)(3y + iy) + i(3 + i) = 10i

⇒ 3x + 3xi - ix - i²x - 6i + 2i² + 6y - 9iy + 2iy - 3yi² + 3i + i² = 10i

We know that,  i² = -1

⇒ 3x + 2xi + x - 6i - 2 + 6y - 7yi + 3y + 3i -1 = 10i  

⇒ 4x + 9y − 3 + 2xi − 7yi − 13i = 0

⇒ 4x + 9y − 3 + (2x − 7y − 13)i = 0

On comparing the real part and imaginary part, we get

4x + 9y − 3 = 0     …… (1)

2x − 7y − 13 = 0  …… (2)

On solving both equations, we get

x = 3 and y = −1

Hence, the value of x, y is 3, −1.

Concept:

Let z = x + iy be any complex number. 

Conjugate of z: \(\rm \bar{z} = x - iy\) 

Calculations:

Let z = x + iy be any complex number. 

⇒\(\rm \bar{z} = x - iy\)

Given, z = -z̅

⇒ (x + iy) = - (x - iy) 

⇒ x + iy = - x + iy 

⇒ 2x = 0

i.e The real part of z is zero.

Hence, If z = -z̅, then the real part of z is zero.

Calculation

\( z \neq 1 \) and \( \dfrac{z^{2}}{z-1} \) is real so imaginary Part is 0.

Let's say \( z = x+iy \)

⇒ \( \dfrac{z^{2}}{z-1} \) \( = \dfrac{(x+iy)^{2}}{x+iy-1} = \dfrac{x^{2}-y^{2}+2ixy}{(x-1)+iy} \)

On rationalizing

\( \Rightarrow \dfrac{x^{2}-y^{2}+2ixy}{(x-1)+iy} \times \dfrac{(x-1)-iy}{(x-1)-iy} \)

\( \Rightarrow \dfrac{(x^{2}-y^{2}+2ixy)((x-1)-iy)}{(x-1)^{2}+y^{2}} \)

(Imaginary Part \( = 0 \))\( \Rightarrow 2xy(x-1)-x^{2}y+y^{3} = 0 \)

\( \Rightarrow 2x^{2}y-2xy-x^{2}y+y^{3} = 0 \)

\( \Rightarrow x^{2}y+y^{3}-2xy = 0 \)

\( \Rightarrow y(x^{2}+y^{2}-2x) = 0 \)

\( \Rightarrow y = 0 \) or \( x^{2}+y^{2}-2x = 0 \)

\( \Rightarrow \text{Real axis Line}: y=0 \quad \text{or} \quad \text{Circle}: x^2+y^2-2x=0 \Rightarrow (x-1)^2+(y-0)^2=1^2 \)

Hence option 1 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: 4x + i(2x - y) = 8 - 4i

Equating real part of the complex number 

⇒ 4x = 8

⇒ x = 2

Equating imaginary part of the complex number  

⇒ 2x - y = -4

\(⇒ 2\times2-y=-4\)

\(⇒ y=8\)

\(\therefore x=2 , y=8\)

Hence , option 2 is correct 

Concept:

A complex number z = x + iy

Real part of z = x, Imaginary part of z = y

If the Complex number is Purely imaginary so we can say that the Real part of a complex number is Zero.

If the Complex number is Purely real so we can say that the Imaginary part of a complex number is Zero.

Calculation:

Given: (x2 - 5x + 6) + i √17

For purely imaginary complex number real part is 0

So, x2 - 5x + 6 = 0

x² - 3x - 2x + 6 = 0

x(x - 3) - 2(x - 3) = 0

(x - 3)(x - 2) = 0

Here, x = 2, 3

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 complex number 

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

Multiplying the numerator and denominator with 1 - i

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

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

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

\(=\frac{1-1-2i}{2}\)

= -i

So, z = 0 - i

∴ Re (z) = 0 and Im (z) = -1

Hence,option 4 is correct  

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 = \(5\sqrt 2\)

Concept:

|z₁.z₂| = |z₁|.|z₂|

If z = x + iy, then |z| = √(x² + y²)

Calculation:

Given, 

(a + ib)(c + id)(e + if)(g + ih) = A + iB

Now taking modulus both sides we get,

|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

⇒ |(a + ib)|.|(c + id)|.|(e + if)|.|(g + ih)| = |A + iB|

⇒ √(a² + b²) .√(c2 + d2​) .√(e2 + f2​) .√(g2 + h2​) = √(A2 + B2​) 

Squaring both sides,

(a2 + b2) (c2 + d2)(e2 + f2)(g2 + h2) = A² + B²

∴ The correct answer is option (2).

Concept:

cos θ + i sin θ = e^iθ

cos(π/6) = √3/2, sin(π/6) = 1/2

cos(5π/6) = -√3/2, sin(5π/6) = 1/2

Calculation:

Given, \(z = \left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\)

⇒ \(z = \left(cos {π \over 6}+i sin {π \over 6}\right)^5 + \left(\cos {π \over 6}-i sin {π \over 6}\right)^5\)

⇒ \(z = \left(e^{{iπ \over 6}} \right)^5 + \left(e^{-{iπ \over 6}}\right)^5\)

⇒ \(z = \left(e^{{5π i\over 6}} \right) + \left(e^{-{5πi \over 6}}\right)\)

⇒ \(z = \left(cos {5π \over 6}+i sin {5π \over 6}\right) + \left(\cos {5π \over 6}-i sin {5π \over 6}\right)\)

⇒ \(z = 2 \cos(\frac{5\pi}{6}) + 0i\)

⇒ \(z = 2 \cos(\pi - \frac{\pi}{6}) + 0i\)

⇒ \(z = - 2 \cos(\frac{\pi}{6}) + 0i\)

⇒ \(z = - 2\times \frac{\sqrt3}{2} + 0i\)

⇒ \(z = -\sqrt{3} +0i\)

⇒ Im(z) = 0 and Re(z) < 0

∴ The correct answer is option (2).