Solutions — Complex Numbers in Cartesian Form

Q1 — Real and Imaginary Parts

For z = a + bi: Re(z) = a, Im(z) = b (the coefficient of i, not the term bi), z̅ = a − bi.

(a) z = 5 + 2i: Re(z) = 5, Im(z) = 2, z̅ = 5 − 2i

(b) z = −3 + 4i: Re(z) = −3, Im(z) = 4, z̅ = −3 − 4i

(c) z = 7i = 0 + 7i: Re(z) = 0, Im(z) = 7, z̅ = −7i

(d) z = −6 = −6 + 0i: Re(z) = −6, Im(z) = 0, z̅ = −6

Q2 — Powers of i

The powers of i cycle with period 4: i, −1, −i, 1. Find the remainder when the exponent is divided by 4.

(a) 6 ÷ 4 = 1 remainder 2 → i⁶ = i² = −1

(b) 11 ÷ 4 = 2 remainder 3 → i¹¹ = i³ = −i

(c) 20 ÷ 4 = 5 remainder 0 → i²⁰ = i&sup4; = 1

(d) 35 ÷ 4 = 8 remainder 3 → i³⁵ = i³ = −i

Q3 — Computing z · z̅

z · z̅ = (a + bi)(a − bi) = a² + b². No need to expand fully.

(a) z = 2 + 3i: z · z̅ = 2² + 3² = 4 + 9 = 13

(b) z = 5 − i: z · z̅ = 5² + (−1)² = 25 + 1 = 26

(c) z = −1 + 2i: z · z̅ = (−1)² + 2² = 1 + 4 = 5

Q4 — Simplifying Square Roots of Negatives

Use √(−n) = √n × i for n > 0.

(a) √(−9) = √9 × i = 3i

(b) √(−25) = √25 × i = 5i

(c) √(−7) = √7 × i = √7 i

Q5 — Equating Real and Imaginary Parts

Two complex numbers are equal iff their real parts are equal AND their imaginary parts are equal.

(a) x + yi = 5 − 3i

Real parts: x = 5 Imaginary parts: y = −3

(b) (2x + 1) + (y − 3)i = 7 + 2i

Real parts: 2x + 1 = 7 → 2x = 6 → x = 3

Imaginary parts: y − 3 = 2 → y = 5

Q6 — Solve Using i

(a) x² + 9 = 0

x² = −9

x = ±√(−9) = ±3i

(b) x² + 2x + 5 = 0. Complete the square:

(x + 1)² − 1 + 5 = 0

(x + 1)² = −4

x + 1 = ±√(−4) = ±2i

x = −1 ± 2i

Check with quadratic formula: x = (−2 ± √(4 − 20))/2 = (−2 ± √(−16))/2 = (−2 ± 4i)/2 = −1 ± 2i. ✓

Q7 — Conjugate Properties

z = 3 − 4i → z̅ = 3 + 4i

(a) Re(z) + Re(z̅) = 3 + 3 = 6 = 2(3) = 2 Re(z). ✓

(b) z + z̅ = (3 − 4i) + (3 + 4i) = 6. The imaginary parts +(−4i) and +4i cancel. Result is purely real. ✓

(c) z − z̅ = (3 − 4i) − (3 + 4i) = −8i. The real parts cancel. Result is purely imaginary. ✓

Q8 — Finding Unknown Complex Numbers

Let z = a + bi, z̅ = a − bi.

z + 2z̅ = (a + bi) + 2(a − bi) = a + bi + 2a − 2bi = 3a + (b − 2b)i = 3a − bi

Setting 3a − bi = 9 − 6i and equating parts:

Real: 3a = 9 → a = 3

Imaginary: −b = −6 → b = 6

Therefore z = 3 + 6i

Check: z + 2z̅ = (3 + 6i) + 2(3 − 6i) = 3 + 6i + 6 − 12i = 9 − 6i. ✓

Q9 — Simplifying a Complex Expression

Expand each square using (a + b)² = a² + 2ab + b²:

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i + (−1) = 2i

(1 − i)² = 1² + 2(1)(−i) + (−i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i

(1 + i)² + (1 − i)² = 2i + (−2i) = 0

This occurs because (1 + i) and (1 − i) are complex conjugates. Their squares 2i and −2i are also conjugates of each other (purely imaginary conjugates), and when added the imaginary parts cancel to zero.

Q10 — Proof Using Conjugates

Proof: Let z = a + bi where a, b ∈ ℝ. Then z̅ = a − bi.

z² = (a + bi)² = a² + 2abi + b²i² = (a² − b²) + 2abi

(z̅)² = (a − bi)² = a² − 2abi + b²i² = (a² − b²) − 2abi

Adding: z² + (z̅)² = [(a² − b²) + 2abi] + [(a² − b²) − 2abi]

= 2(a² − b²) + 0i = 2(a² − b²)

Since a, b ∈ ℝ, the expression 2(a² − b²) ∈ ℝ. Therefore z² + (z̅)² is always real. □