Complex Numbers Questions

Complex Numbers Questions and answers are given here in an easily understandable way. Students can try solving the questions given here and verify their answers with the solutions provided. Also, they can practise the additional questions given at the end of the page. Before practising questions on complex numbers, let’s recall what complex numbers are.

What are complex numbers?

Complex numbers can be expressed as a combination of real and imaginary numbers. The standard notation of a complex number is given by z = x + iy, where x is the real part of z and iy is the imaginary part of the complex number z. Also, “i” is called the “iota” and i² = -1.

If z₁ = a + ib and z₂ = c + id are two complex numbers such that;

• (a + ib) + (c + id) = (a + c) + i(b + d)
• (a + ib) – (c + id) = (a – c) + i(b – d)
• (a + ib). (c + id) = (ac – bd) + i(ad + bc)
• (a + ib) / (c + id) = [(ac + bd)/ (c² + d²)] + i[(bc – ad) / (c² + d²)]

Learn more about complex numbers here.

Complex Numbers Questions and Answers

1. If z = 2 – 3i, then find z².

Solution:

Given,

z = 2 – 3i

z² = z.z

= (2 – 3i)(2 – 3i)

= 2(2) – 2(3i) – (3i)(2) + (3i)(3i)

= 4 – 6i – 6i + 9i² {since i² = -1}

= 4 – 12i + 9(-1)

= 4 – 12i – 9

= -5 – 12i

Therefore, z² = -5 – 12i.

2. Suppose z = (2 – i)² + [(7 – 4i)/(2 + i)] – 8, express z in the form of x + iy such that x and y are real numbers.

Solution:

Given,

z = (2 – i)² + [(7 – 4i)/(2 + i)] – 8

= (2)² – 2(2)(i) + (i²) + [(7 – 4i)(2 – i)/ (2 + i)(2 – i)] – 8

= (4 – 4i – 1) + [(14 – 7i – 8i + 4i²)/ (4 – i2)] – 8

= (3 – 4i) + [(14 – 15i – 4)/(4 + 1)] – 8 {since i² = -1}

= (3 – 4i) + [(10 – 15i)/5] – 8

= (3 – 4i) + (2 – 3i) – 8

= -3 – 7i

This is of the form x + iy such that x = -3 and y = -7.

3. Simplify:

Solution:

This can be written as:

Therefore,

4. If z₁ = 2 + 8i and z₂ = 1 – i, then find |z₁/z₂|.

Solution:

Given,

z₁= 2 + 8i and z₂ = 1 – i

z₁/z₂ = (2 + 8i)/(1 – i)

= (2 + 8i)(1 + i)/ (1 – i)(1 + i)

= [2 + 2i + 8i + 8i²]/ [1 – i²]

= (2 + 10i – 8)/ (1 + 1) {since i² = -1}

= (-6 + 10i)/2

= -3 + 5i

Now, |z₁/z₂| = √[(-3)² + (5)²]

= √(9 + 25)

= √34

5. If |z² – 1| = |z²| + 1, then show that z lies on an imaginary axis.

Solution:

Let z = x + iy be the complex number.

Now, z² = z.z = (x + iy)(x + iy)

= x² + ixy + ixy + (iy)²

= x² + 2ixy – y² {since i² = -1}

z² – 1 = x² + 2ixy – y² – 1

= (x² – y² – 1) + i(2xy)

Thus, |z² – 1| = √[(x² – y² – 1)² + (2xy)²]

= √[(x² – y² – 1)² + 4x²y²]

|z|² + 1 = [√(x² + y²)]² + 1

= x² + y² + 1

Given that,

|z² – 1| = |z²| + 1

So, √[(x² – y² – 1)² + 4x²y²] = x² + y² + 1

Squaring on both sides, we get;

(x² – y² – 1)² + 4x²y² = (x² + y² + 1)²

[x² – (y² + 1)]² + 4x²y² = [x² + (y² + 1)]²

[x² – (y² + 1)]² – [x² + (y² + 1)]² + 4x²y² = 0

As we know, (a – b)² – (a + b)² = -4ab,

-4x²(y² + 1) + 4x²y² = 0

-4x²y² – 4x² + 4x²y² = 0

4x² = 0

x = 0

Therefore, z lies on the y-axis.

6. Find the conjugate of z₁ – z₂ if z₁ = 2 + 3i and z₂ = 5 + 2i.

Solution:

Given,

z₁ = 2 + 3i

z₂ = 5 + 2i

z₁ – z₂ = (2 + 3i) – (5 + 2i)

= (2 – 5) + i(3 – 2)

= -3 + i

As we know the conjugate of z = x + iy = x – iy.

Conjugate of z₁ – z₂ = -3 – i

7. Simplify: i⁵⁹

Solution:

We know that,

i² = -1, i³ = -i, i⁴ = 1

We can write 59 as: 59 = 4 × 14 + 3

So, i⁵⁹ = i^{(4 × 14) + 3}

= i^{(4 × 14)} . i³

= 1.i³

= -i

Therefore, i⁵⁹ = -i.

8. Find real x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

(x – iy)(3 + 5i) = 3x + 5ix – 3iy – 5yi²

= 3x + i(5x – 3y) + 5y {since i² = -1}

= (3x + 5y) + i(5x – 3y)

Given that (x – iy)(3 + 5i) is the conjugate of -6 – 24i.

Here, the conjugate of -6 – 24i = -6 + 24i.

So, 3x + 5y = -6

5x – 3y = 24

Solving these two equations, we get; x = 3 and y = -3.

9. Find the relation between a and b if z = a + ib if |(z – 3)/(z + 3)| = 2.

Solution:

Given,

z = a + ib

|(z – 3)/(z + 3)| = 2

|(a + ib – 3)/(a + ib + 3)| = 2

|(a – 3) + ib|= 2|(a + 3) + ib|

√[(a – 3)² + b²] = 2√[(a + 3)² + b²]

Squaring on both sides, we get;

(a – 3)² + b² = 4[(a + 3)² + b²]

a² – 6a + 9 + b² = 4(a² + 6a + 9 + b²)

4a² + 24a + 36 + 4b² – a² + 6a – 9 – b² = 0

3a² + 30a + 27 + 3b² = 0

a² + 10a + 9 + b² = 0

(a² + 10a + 25) + (b² + 9 – 25) = 0

(a + 5)² + b² = 16

(a + 5)² + b² = 4²

10. If |z + 1| = z + 2 (1 + i), then find z.

Solution:

Let z = x + iy be the complex number.

Given,

|z + 1| = z + 2 (1 + i)

⇒ |x + iy + 1| = x + iy + 2 (1 + i)

We know,

|z| = √(x² + y²)

√[(x + 1)² + y²] = (x + 2) + i(y + 1)

Comparing real and imaginary parts,

⇒ √((x + 1)² + y²) = x + 2

And 0 = y + 2

⇒ y = -2

Substituting the value of y in √((x + 1)² + y²) = x + 2, we get;

(x + 1)² + (-2)² = (x + 2)²

x² + 2x + 1 + 4 = x² + 4x + 4

⇒ 2x = 1

⇒ x = ½

Therefore, z = x + iy = ½ – 2i.

Practice Questions on Complex Numbers

1. Find the conjugate of complex number (1 – i)/(1 + i).
2. The complex number z = a + ib, where a and b are real numbers, satisfies the equation z² + 16 – 30i = 0.
3. Calculate the modulus value of the complex number −2√3 – 2i.
4. Simplify: (1 + 6i) + (6 − 2i) − (−7 + 5i)
5. If [(1 – i)/(1 + i)]¹⁰⁰ = a + ib, then find the values of a and b.