Conjugate of Complex Number MCQ Quiz - Objective Question with Answer for Conjugate of Complex Number - Download Free PDF

Concept:

If a complex number is a ± bi, then the complex conjugate will be a ∓  bi and vice-versa.

Where a & b are real numbers and i = Imaginary number \(=\sqrt{-1}\)

Calculation:

Given the complex number z = 3i +  7 

∴ The complex conjugate of this number = -3i + 7

∴ Option 2 is correct

Given:

|z - 1 - i| = 1

Let us solve the question graphically,

|z| = 1

for |z - 1 - i|, the centre is shifted to (1, 1)

3(z - i) - 4

For (z - i), the graph is shifted down and centre is now at (1, 0)

3(z - i) (the center is shifted to (3, 0)

3(z - i) - 4,

the circle is shifted left by 4 units, 

∴ we get a circle having a radius of 3 and centre at (-1, 0).

Concept:

Conjugate of a Complex Number:

Conjugate of a complex number z = x + iy is x - iy and which is denoted as \(\overline{z}\).

For example, the conjugate of the complex number z = 3 - 4i is 3 + 4i. 

Solution:

\(\frac{1}{2-3i}\)

\(\frac{1}{2-3i}\times \frac{2+3i}{2+3i}\)

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

\(=\frac{2+3i}{4+9}\)

\(=\frac{2+3i}{13}\)

\(=\frac{2}{13}+\frac{3}{13}i\)

Conjugate of this complex number is \(\frac{2}{13}-\frac{3}{13}i\)

∴ The correct option is (2)

Concept:

Consider a complex number, z = a + ib

Conjugate of complex number = z̅ = a - ib

Modulus of complex number  = |z| = \(\rm \sqrt{(a^2 + b^2) }\)

Calculation: 

Let, z = a + ib,

zz̅ = (a + ib)(a - ib)

= \(\rm a^2-(ib)^2\)

= \(\rm a^2-i^2(b)^2\)

=\(\rm a^2-(ib)^2\)

=\(\rm a^2+b^2\cdots (\because i^2=-1)\)

And, |z|² = \(\rm (\sqrt{a^2+b^2})^2\)

= \(\rm {a^2+b^2}\)

∴ zz̅  = |z|²

Now, 

 \(\rm z^{-1}=\frac1 z=\frac{1}{a+ib}\)

=\(\rm \frac{1}{a+ib}\times \frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}\)

= \(\rm \frac{\bar z}{|z|^2}\) ≠ \(\rm \frac {z}{|z|^2}\)

Hence, option (1) is correct.

Concept:

Conjugate of a complex number:

For any complex number z = x + iy  the conjugate z̅ is given by  z̅ = x - iy

Calculation:

Let z = \(\rm {3+2i}\over{2-i}\)

z = \(\rm {3+2i\over2-i} \times{2+i\over2+i}\)

z = \(\rm {6+7i+2i^2\over2^2-(i)^2}\)

z = \(\rm {4 + 7i\over5}\)

z = \(\rm {4\over 5}+{7\over 5}i\)

Conjugate of z = z̅ = \(\rm {4\over 5}- {7\over 5}i\)

1Concept:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• For calculating the conjugate, replace i with -i.
• Conjugate of z = x – iy

Calculation:

Let z = (i - i²)³

⇒ z = i³ (1 - i) 3  = - i (1 - i)³

For calculating the conjugate, replace i with -i.

⇒ z̅  =  -(- i) (1 - (- i))³

⇒ z̅  =  i(1 + i)³

Using (a + b) 3 = a3 + b3 + 3a2b + 3ab2

⇒ z̅  =  i(1 + i³ +3 ×1² × i + 3 × i² × 1 ) 

⇒ z̅  =  i(1 - i + 3i - 3) 

⇒ z̅  =  i(-2 + 2i)

⇒ z̅  = -2i + 2i²

⇒ z̅  = -2 - 2 i

So, the conjugate of  (i - i2)3 is -2 - 2i

Concept: 

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z = z̅ = x – iy

Calculation:

Given complex number is z = \(\rm 3i+4\over2-3i\)

z = \(\rm {3i+4\over2-3i}\times{2+3i\over2+3i}\)

z = \(\rm 6i+8-9+12i\over2^2-(3i)^2\)

z = \(\rm 18i-1\over13\)

z = \(\rm {-1\over13}+{18\over13}i\)

Conjugate of z = (z̅) = \(\rm {-1\over13}-{18\over13}i\)

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:

Let z = \(\rm \frac{(3-i)(1+2i)}{(2+i)(1-3i)}\)

⇒ z = \(\rm \frac{3+6i-i-2i^{2}}{2-6i+i-3i^{2}}\)  = \(\rm \frac{3+6i-i+2}{2-6i+i+3}\)

⇒ z = \(\rm \frac{5+5i}{5-5i}\)  = \(\rm\frac{1+i}{1-i}\) 

⇒ z = \(\rm\frac{1+i}{1-i}\) × \(\rm\frac{1+i}{1+i}\) = \(\rm \frac{1+i^{2}+2i}{1-i^{2}}\) = i 

⇒ z = i  

We know that, Conjugate of z = z̅ = x - iy   

⇒ \(\rm\overline{z}\) = - i  . 

The correct option is 1

Concept:

Conjugate of a complex number:

For any complex number z = x + iy  the conjugate z̅ is given by  z̅ = x - iy

Calculation:

z = \(\rm i + 3 \over 2i + 1\)

z = \(\rm { i + 3 \over 2i + 1}\times {-2i+1\over-2i +1}\)

z = \(\rm -2i^2-6i+i+3\over1 - (2i)^2\)

z = \(\rm {5-5i\over1 + 4}\)

z = \(\rm {5(1 - i)\over5}\)

z = 1 - i

Conjugate of z = z̅  = 1 + i

Concept: 

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z = z̅ = x – iy

Calculation:

Given complex number is z = \(\rm i+2\over1-3i\)

z = \(\rm {i+2\over1-3i}\times{1+3i\over1+3i}\)

z = \(\rm i+2+6i+3i^2\over1^2-(3i)^2\)

z = \(\rm 7i-1\over10\)

z = \(\rm {-1\over10}+{7\over10}i\)

Conjugate of z = (z̅) = \(\rm {-1\over10}-{7\over10}i\)

NOTE:

The complex conjugate has the same real part as z and the same imaginary part but with the opposite sign.

Concept:

The circle is defined as the locus of a point that moves in a plane such that its distance from a fixed point in that plane is constant.

Equation of circle having a center (h, k) and radius a is  given as,

(x - h)² + (y - k)² = a²

But when a circle passes through the origin, then the equation of the circle is

x² + y² - 2 × h × x - 2 × k × y = 0 ----(1)

a² = h² + k²  -----(2)

Calculation:

Given:

z z̅  + (1 + i) z +(1 - i) z̅  = 0

As we know,

z = x + iy, z̅ = x - iy using these value in the above given equation we get:

(x + iy)(x - iy) + (1 + i)(x + iy) + (1 - i)(x - iy) = 0

x² + y² + x + iy + ix - y + x - iy - ix - y = 0

x² + y² + 2x - 2y = 0 -----(3)

Comparing (3) with (1) we get:

h = -1 and k = 1 which are centre of circle.

By using h, k values in (2) we get the radius of the circle as √2.

The general equation of a non-degenerate conic section is: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B and C are all not zero

The above given equation represents a non-degenerate conics whose nature is given below in the table:

Concept:

Conjugate of a complex number:

For any complex number z = x + iy  the conjugate z̅ is given by  z̅ = x - iy

Calculation:

Let z = \(\rm {3+2i}\over{2-i}\)

z = \(\rm {3+2i\over2-i} \times{2+i\over2+i}\)

z = \(\rm {6+7i+2i^2\over2^2-(i)^2}\)

z = \(\rm {4 + 7i\over5}\)

z = \(\rm {4\over 5}+{7\over 5}i\)

Conjugate of z = z̅ = \(\rm {4\over 5}- {7\over 5}i\)

Concept:

Consider a complex number, z = a + ib

Conjugate of complex number = z̅ = a - ib

Modulus of complex number  = |z| = \(\rm \sqrt{(a^2 + b^2) }\)

Calculation: 

Let, z = a + ib,

zz̅ = (a + ib)(a - ib)

= \(\rm a^2-(ib)^2\)

= \(\rm a^2-i^2(b)^2\)

=\(\rm a^2-(ib)^2\)

=\(\rm a^2+b^2\cdots (\because i^2=-1)\)

And, |z|² = \(\rm (\sqrt{a^2+b^2})^2\)

= \(\rm {a^2+b^2}\)

∴ zz̅  = |z|²

Now, 

 \(\rm z^{-1}=\frac1 z=\frac{1}{a+ib}\)

=\(\rm \frac{1}{a+ib}\times \frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}\)

= \(\rm \frac{\bar z}{|z|^2}\) ≠ \(\rm \frac {z}{|z|^2}\)

Hence, option (1) is correct.

Concept:

• The conjugate of the complex function is given by changing the sign of i.
• If z = x + iy , then  \( \overline{z} \) = x - iy

Explanation:

Given that conjugate of cos x - i sin 2x is sin x + i cos 2x .

and conjugate of cos x - i sin 2x  is cos x + i sin 2x .

According to the question,

cos x + i sin 2x = sin x + i cos 2x

Comparing the real and imaginary parts of the above equation, we have

cos x = sin x and sin 2x = cos 2x 

⇒ tan x = 1 and  tan 2x = 1

⇒ x = π/ 4 and x = π/ 8

Which is not possible for at the same time. 

Therefore, no solution exists.

Concept: 

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z = z̅ = x – iy

Calculation:

Given that:

\(\rm \frac{1 - i√3}{1 + i√3}\)

Multiply numerator and denominator both by (1 - i√3), we get

\(\rm (\frac{1 - i√3}{1 + i√3}) \times (\frac {1 - i\sqrt3}{1 - i \sqrt3})\) = \(\rm \frac{(1 - i√3)^2}{(1)^2 - (i√3)^2}\) = \(\rm \frac{[1 + (i\sqrt3)^2 - 2i\sqrt3]}{1 + 3}\) = \(\rm \frac{(- 2 - 2i\sqrt3]}{4}\) = \(\rm \frac{-1}{2} - \frac{\sqrt3}{2}i\)

Conjugate of z = z̅ = x – iy

Here z =  \(\rm \frac{-1}{2} - \frac{\sqrt3}{2}i\) 

 z̅ =  \(\rm \frac{-1}{2} + \frac{\sqrt3}{2}i\)