Properties of Complex Numbers MCQ Quiz - Objective Question with Answer for Properties of Complex Numbers - Download Free PDF

Concept:

1. The amplitude (or argument) of a complex number \( z = r(\cos\theta + i\sin\theta) \) is given by \( \text{amp}(z) = \theta \).

2. The conjugate of \( z \), denoted as \( \overline{z} \), has an amplitude \( \text{amp}(\overline{z}) = -\theta \), because conjugating a complex number reflects it across the real axis in the Argand plane.

Formula Used:

\( \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 \).

Calculation:

\( z = r(\cos\theta + i\sin\theta) \)

\( \overline{z} = r(\cos\theta - i\sin\theta) \)

\( \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 \)

Conclusion:

\( \therefore \text{amp}(z) + \text{amp}(\overline{z}) = 0 \).

Hence, the correct answer is Option 1.

Calculation:

Given,

The condition is \( \bigl|\frac{z - 2i}{z + i}\bigr| = 2,\quad z \neq -i\).

We need to find the locus of all complex numbers \(z\) satisfying this.

Step 1: Write in Cartesian form.

Let \(z = x + i\,y \) Then

\(z - 2i = x + i(y - 2),\quad z + i = x + i(y + 1).\)

Step 2: Translate the modulus equation.

\(\bigl|\tfrac{z - 2i}{z + i}\bigr| = 2 \)

\(⇒(\lvert z - 2i\rvert = 2\,\lvert z + i\rvert\)

Squaring both sides:

\(x^2 + (y-2)^2 = 4\bigl[x^2 + (y+1)^2\bigr].\)

Step 3: Expand and simplify.

Left: \(x^2 + y^2 - 4y + 4 \)
Right: \(4x^2 + 4y^2 + 8y + 4 \)

Bring all terms together and divide by 3:

\(x^2 + y^2 + 4y = 0.\)

Step 4: Complete the square.

\(x^2 + (y+2)^2 = 4.\)

This is a circle of radius \(2\) centered at \((0,-2) \)

Hence, the correct answer is Option 4. 

Concept:

Locus of Points Defined by Difference of Squares of Distances:

• In complex geometry, a point \( z \) such that \( |z - z_1|^2 - |z - z_2|^2 \) is constant represents a geometric locus.
• If the difference is constant, it can represent a straight line.
• The equation \( |z - z_1|^2 - |z - z_2|^2 = c \) simplifies to linear form in many cases.
• Here, it simplifies to a straight line equation.

Complex Number:

• Definition: A complex number is of the form \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.
• SI Unit: Dimensionless
• Modulus: \( |z| = \sqrt{x^2 + y^2} \)

Distance in Complex Plane:

• Definition: Distance between two complex numbers \( z_1 \) and \( z_2 \) is \( |z_1 - z_2| \)
• Formula: \( |z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \)

Calculation:

Given,

\( z_1 = 2 + 3i,\quad z_2 = 3 + 4i \)

\( z \in \mathbb{C} \) such that \( |z - z_1|^2 - |z - z_2|^2 = |z_1 - z_2|^2 \)

Let \( z = x + iy \)

⇒ \( |z - z_1|^2 = (x - 2)^2 + (y - 3)^2 \)

⇒ \( |z - z_2|^2 = (x - 3)^2 + (y - 4)^2 \)

⇒ \( |z_1 - z_2|^2 = (-1)^2 + (-1)^2 = 2 \)

⇒ \( (x - 2)^2 + (y - 3)^2 - (x - 3)^2 - (y - 4)^2 = 2 \)

⇒ \( [x^2 - 4x + 4 + y^2 - 6y + 9] - [x^2 - 6x + 9 + y^2 - 8y + 16] = 2 \)

⇒ \( x^2 - 4x + 4 + y^2 - 6y + 9 - x^2 + 6x - 9 - y^2 + 8y - 16 = 2 \)

⇒ \( 2x + 2y - 12 = 2 \)

⇒ \( 2x + 2y = 14 \Rightarrow x + y = 7 \)

⇒ Equation of line: \( x + y = 7 \)

⇒ x-intercept = 7 (when y = 0), y-intercept = 7 (when x = 0)

∴ The locus represents a straight line with sum of intercepts = 14.

Concept:

Power of i:

• i = \(\sqrt{-1}\)
• i2 = -1
• i3 = -i × i2 = -i
• i4 = (i2)2 = (-1)2 = 1
• i4n = 1

Calculation:

Given that,

i4n + 1, where \({\rm{i}} = \sqrt { - 1} \)

= i⁴ⁿ × i

= 1 × i

= i

Explanation:

\(\rm \frac{1}{2}+Re\left(\frac{Z_1}{Z_2}\right) = \frac{1}{2}+ Re (\frac{\omega }{\omega^2})\)

= \(\frac{1}{2}Re(\omega)^2\)

= \(\frac{1}{2}+ Re [ -\frac{1}{2} - \frac{\sqrt3}{2}i\)

= \(\frac{1}{2} + (-\frac{1}{2}) = 0\)

∴ Option (b) is correct.

Concept:

i² = -1

i³ = - i

i⁴ = 1

i⁴ⁿ = 1

Calculation:

We have to find the value of (i² + i⁴ + i⁶ +... + i²ⁿ)

(i² + i⁴ + i⁶ +... + i²ⁿ) = (i² + i⁴) + (i⁶ + i⁸) + …. + (i^2n-2 + i²ⁿ)

= (-1 + 1) + (-1 + 1) + …. (-1 + 1)

= 0 + 0 + …. + 0

= 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: (1 + i) (x + iy) = 2 + 4i

⇒ x + iy + ix + i²y = 2 + 4i

⇒ (x – y) + i(x + y) = 2 + 4i

Equating real and imaginary part,

x - y = 2         …. (1)

x + y = 4        …. (2)

Adding equation 1 and 2, we get

x = 3

Now,

5x = 5 × 3 = 15

Concept:

Modulus of z = |z| = \(\rm \sqrt {x^2+y^2} = \sqrt {Re (z)^2+Im (z)^2}\)

Calculations:

Let \(\rm z= x + iy = \dfrac{4+2i}{1-2i}\)

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

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

As we know i2 = -1 

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

\(\rm x + iy =\dfrac{10i}{5} = 0 + 2i\)

As we know that if z = x + iy be any complex number, then its modulus is given by,|z| = \(\rm \sqrt{x^2+y^2}\)

∴ |z| = \(\rm \sqrt{0^2+2^2} = 2\)

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;

Modulus of a complex number z = x + iy is given by:

|z| = \(\rm \sqrt{x^2 + y^2}\)

(a + b)(a - b) = a² - b²

Calculation: 

If z = \(\rm \frac{z_1}{z_2}\) so, modulus of |z| = \(\rm \frac{|z_1|}{|z_2|}\)

z = \(\rm \frac {1+i}{1+\sqrt3 i}\)

|z| = \(\rm \frac {\sqrt {(1)^2 + (1)^2}}{\sqrt {(1)^2 + (\sqrt 3)^2}} = \frac{\sqrt 2}{2} = \frac {1}{\sqrt 2}\) 

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:

z = x + iy

arg (z) = arg (x + iy) = tan^-1(y/x)

Therefore, the argument θ is represented as:

θ = tan^-1 (y/x)

tan( \(\rm \frac{π}{2}\)) = infinity

power of iota, i² = -1

Calculation:

let z  = \(\rm \left ( \frac{i}{5} + \frac{5}{i} \right )\)

z = \(\rm \frac{i}{5} + \frac{5i}{i^{^{2}}}\)

z = \(\rm \frac{i}{5} - 5i\) = \(\rm \frac{-24}{5}\)i

So, arg (z) = tan^-1 \(\rm \left ( \frac{-24/5}{0} \right )\) 

= - tan-1 \(\infty\) = -\(\rm \frac{π}{2}\)

Mistake PointsOption (3) is incorrect here because quadrant is also important here. In complex plane, this lies on the negative imaginary axis. Since anticlockwise movement from positive real axis is negative, the correct value of argument will be -π/2. 

Concept:

Modulus of z = |z| = \(\rm \sqrt {x^2+y^2} = \sqrt {Re (z)^2+Im (z)^2}\)

Calculations:

Let \(\rm z = x+iy =\left(\frac{1+i}{1-i} - \frac{1-i}{1+i}\right)\)

\(\rm =\frac{(1+i)^2-(1-i)^2}{1^2-i^2}\\=\frac{1+2i+i^2-1+2i-i^2}{1+1}\\=\frac{4i}{2}=2i\)

z = x + iy = 0 + 2i

As we know that if z = x + iy be any complex number, then its modulus is given by, |z| = \(\rm \sqrt{x^2 + y^2}\)

∴ |z| = \(\rm \sqrt{0^2+2^2} = 2\)

Calculation:

Let z = \(\rm {1 + i\over1- i}\)

Multiply 1 + i on both numerator and denominator

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

z = \(\rm (1+i)^2\over1^2-i^2\)

z = \(\rm 1+2i+i^2\over1+1\)

z = \(\rm 2i\over2\)= i