Complex Numbers Important Questions & Answers | Problems & Solutions | Practice Questions download free pdf

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)]

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:

\(\begin{array}{l}\frac{1-2i}{3+4i}-\frac{2+i}{5i}\end{array} \)

Solution:

\(\begin{array}{l}\frac{1-2i}{3+4i}-\frac{2+i}{5i}\end{array} \)

This can be written as:

\(\begin{array}{l}=\frac{1-2i}{3+4i}.\frac{3-4i}{3-4i}-\frac{2+i}{5i}.\frac{-i}{-i}\\=\frac{(3-4i-6i+8i^2)}{9-16i^2} -\frac{(-2i-i^2)}{-5i^2}\\=\frac{(3 – 10i – 8)}{(9+16)}-\frac{(-2i+1)}{5}\\=\frac{(-5-10i)}{25}+\frac{2}{5}i – \frac{1}{5}\\=-\frac{1}{5}-\frac{2}{5}i+\frac{2}{5}i – \frac{1}{5}\\=-\frac{2}{5}\end{array} \)

Therefore,

\(\begin{array}{l}\frac{1-2i}{3+4i}-\frac{2+i}{5i}=-\frac{2}{5}\end{array} \)

Modulus and Conjugate of a Complex number If z = x + iy is a complex number then, mod of z is given by |z| = √(x2 + y2). If z = x + iy then the conjugate of z is z̄ = a – ib.

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.

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