Exercise 4.1
Express (5i)(–(3/5)i) in the form a + ib.
Answer
(5i)(–3/5 · i) = –3i² = –3(–1) = 3 = 3 + 0i
Express i⁹ + i¹⁹ in the form a + ib.
Answer
i⁹ = i⁴ˣ²⁺¹ = i; i¹⁹ = i⁴ˣ⁴⁺³ = i³ = –i
→ i⁹ + i¹⁹ = i + (–i) = 0 = 0 + 0i
Express i⁻³⁹ in the form a + ib.
Answer
i⁻³⁹ = i⁻³⁹⁺⁴⁰ = i¹ = i = 0 + i
Express 3(7 + i7) + i(7 + i7) in the form a + ib.
Answer
= 21 + 21i + 7i + 7i² = 21 + 28i – 7 = 14 + 28i
Express (1 – i) – (–1 + i6) in the form a + ib.
Answer
(1 – i) – (–1 + 6i) = 1 – i + 1 – 6i = 2 – 7i
Express ((1/5) + i(2/5)) + (4 + i(5/2)) in the form a + ib.
Answer
Real part: 1/5 + 4 = 21/5; Imaginary part: 2/5 + 5/2 = 4/10 + 25/10 = 29/10
Result: 21/5 + i(29/10)
Express ((1/3 + i·7/3) + (4 + i·1/3)) – (–4/3 + i) in the form a + ib.
Answer
Real: 1/3 + 4 + 4/3 = 5/3 + 4 = 17/3
Imaginary: 7/3 + 1/3 – 1 = 8/3 – 3/3 = 5/3
Result: 17/3 + i(5/3)
Express (1 – i)⁴ in the form a + ib.
Answer
(1 – i)² = 1 – 2i + i² = 1 – 2i – 1 = –2i
⇒ (1 – i)⁴ = (–2i)² = 4i² = –4 = –4 + 0i
Express (1/3 + 3i)³ in the form a + ib.
Answer
Using (a + b)³ with a = 1/3, b = 3i:
= (1/3)³ + 3(1/9)(3i) + 3(1/3)(–9) + (27i³)
= 1/27 + i – 9 – 27i
= (1/27 – 9) + i(1 – 27)
= –242/27 – 26i
Express (–2 – i/3)³ in the form a + ib.
Answer
Let a = –2, b = –i/3:
= (–2)³ + 3(4)(–i/3) + 3(–2)(–1/9) + (–i³/27)
= –8 – 4i + 2/3 + i/27
= (–8 + 2/3) + i(–4 + 1/27)
= –22/3 + i(–107/27)
Find the multiplicative inverse of 4 – 3i.
Answer
z = 4 – 3i; z̄ = 4 + 3i; |z|² = 16 + 9 = 25
z⁻¹ = z̄/|z|² = (4 + 3i)/25 = 4/25 + 3i/25
Find the multiplicative inverse of √5 + 3i.
Answer
z = √5 + 3i; z̄ = √5 – 3i; |z|² = 5 + 9 = 14
z⁻¹ = (√5 – 3i)/14 = √5/14 – 3i/14
Find the multiplicative inverse of –i.
Answer
z = –i; z̄ = i; |z|² = 1
z⁻¹ = i/1 = i = 0 + i
Express ((3 + i√5)(3 – i√5)) / ((√3 + √2·i)(√3 – √2·i)) in the form a + ib.
Answer
Numerator: (3)² – (i√5)² = 9 – (–5) = 14
Denominator: (√3)² – (√2·i)² = 3 – (–2) = 5
Result: 14/5 + 0i
Miscellaneous Exercise
Evaluate: [i¹⁸ + (1/i)²⁵]³
Answer
i¹⁸ = i⁴ˣ⁴⁺² = i² = –1
(1/i)²⁵ = i⁻²⁵ = i⁻²⁵⁺²⁸ = i³ = –i
[–1 + (–i)]³ = (–1 – i)³ = –(1 + i)³
(1 + i)² = 2i; (1 + i)³ = 2i(1 + i) = 2i + 2i² = –2 + 2i
–(–2 + 2i) = 2 – 2i
For any two complex numbers z₁ and z₂, prove that Re(z₁z₂) = Re z₁ · Re z₂ – Im z₁ · Im z₂.
Answer
Let z₁ = a + ib, z₂ = c + id
z₁z₂ = (ac – bd) + i(ad + bc)
∴ Re(z₁z₂) = ac – bd = Re z₁ · Re z₂ – Im z₁ · Im z₂ ∎
Reduce (1/(1–4i) – 2/(1+i)) × (3–4i)/(5+i) to standard form.
Answer
1/(1–4i) = (1+4i)/17; 2/(1+i) = (2–2i)/2 = 1–i
(1+4i)/17 – (1–i) = (1+4i – 17+17i)/17 = (–16+21i)/17
(–16+21i)(3–4i) = –48+64i+63i–84i² = 36+127i
(17)(5+i) = 85+17i
(36+127i)/(85+17i) × (85–17i)/(85–17i)
Numerator: 3060–612i+10795i+2159 = 5219+10183i; Denominator: 7225+289 = 7514
= 307/442 + (599/442)i
If (a–ib)/(x–iy) = c–id, prove that (a²+b²)/(x²+y²) = c²+d².
Answer
Taking modulus: |a–ib|/|x–iy| = |c–id|
→ √(a²+b²)/√(x²+y²) = √(c²+d²)
Squaring: (a²+b²)/(x²+y²) = c²+d² ∎
If z₁ = 2 – i, z₂ = 1 + i, find |(z₁+z₂+1)/(z₁–z₂+1)|.
Answer
Numerator: (2–i)+(1+i)+1 = 4; Denominator: (2–i)–(1+i)+1 = 2–2i
|4/(2–2i)| = 4/|2–2i| = 4/√8 = 4/(2√2) = √2
If a + ib = (x+i)²/(2x²+1), prove that a² + b² = (x²+1)²/(2x²+1)².
Answer
(x+i)² = (x²–1) + 2xi
a = (x²–1)/(2x²+1), b = 2x/(2x²+1)
a² + b² = [(x²–1)² + 4x²]/(2x²+1)² = (x⁴+2x²+1)/(2x²+1)² = (x²+1)²/(2x²+1)² ∎
Let z₁ = 2 – i, z₂ = –2 + i. Find:
Answer
(i) Re(z₁z₂/z̄₁)
Ans. z₁z₂ = (2–i)(–2+i) = –4+2i+2i–i² = –3+4i; z̄₁ = 2+i
(–3+4i)/(2+i) × (2–i)/(2–i) = (–6+3i+8i–4i²)/5 = (–2+11i)/5
Re = –2/5
(ii) Im(1/z̄₁)
1/(2+i) = (2–i)/5; Im = –1/5
Find real x, y if (x–iy)(3+5i) is the conjugate of –6–24i.
Answer
Conjugate of –6–24i is –6+24i
(x–iy)(3+5i) = (3x+5y) + i(5x–3y)
→ 3x+5y = –6 and 5x–3y = 24
Solving: multiply (1) by 3 and (2) by 5: 9x+15y = –18 and 25x–15y = 120
Adding: 34x = 102 → x = 3; then y = –3
Find the modulus of (1+i)/(1–i) – (1–i)/(1+i).
Answer
(1+i)/(1–i) = (1+i)²/2 = 2i/2 = i; (1–i)/(1+i) = –i
i – (–i) = 2i; |2i| = 2
If (x+iy)³ = u+iv, show that u/x + v/y = 4(x²–y²).
Answer
(x+iy)³ = (x³–3xy²) + i(3x²y–y³)
u = x(x²–3y²) → u/x = x²–3y²
v = y(3x²–y²) → v/y = 3x²–y²
u/x + v/y = x²–3y²+3x²–y² = 4x²–4y² = 4(x²–y²) ∎
If α and β are different complex numbers with |β| = 1, find |(β–α)/(1–ᾱβ)|.
Answer
|β–α|² = |β|²–αβ̄–ᾱβ+|α|² = 1–αβ̄–ᾱβ+|α|²
|1–ᾱβ|² = 1–ᾱβ–αβ̄+|α|²|β|² = 1–ᾱβ–αβ̄+|α|²
They are equal, so |(β–α)/(1–ᾱβ)| = 1
Find the number of non-zero integral solutions of |1–i|ˣ = 2ˣ.
Answer
|1–i| = √2, so (√2)ˣ = 2ˣ
⇒ 2^(x/2) = 2ˣ
⇒ x/2 = x
⇒ x = 0
Only x = 0 satisfies, which is not non-zero. Answer: 0 non-zero integral solutions.
If (a+ib)(c+id)(e+if)(g+ih) = A+iB, show that (a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B².
Answer
|a+ib|·|c+id|·|e+if|·|g+ih| = |A+iB|
√(a²+b²)·√(c²+d²)·√(e²+f²)·√(g²+h²) = √(A²+B²)
Squaring: (a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B² ∎
If ((1+i)/(1–i))ᵐ = 1, find the least positive integral m.
Answer
(1+i)/(1–i) = (1+i)²/2 = 2i/2 = i
iᵐ = 1 when m is a multiple of 4. Least positive value: m = 4
Important Extra Questions
If z₁ = 1 + i and z₂ = 2 – 3i, find z₁/z₂ in standard form.
Answer
z₁/z₂ = (1+i)(2+3i)/[(2–3i)(2+3i)] = (2+3i+2i+3i²)/13 = (–1+5i)/13 = –1/13 + 5i/13
Find the square root of –7 – 24i.
Answer
Let √(–7–24i) = a+ib: a²–b² = –7, 2ab = –24
(a²+b²)² = 49+576 = 625 → a²+b² = 25
a² = 9, b² = 16; since ab < 0: √(–7–24i) = 3–4i or –3+4i
Solve x² + 2x + 5 = 0 over complex numbers.
Answer
x = [–2 ± √(4–20)]/2 = [–2 ± √(–16)]/2 = [–2 ± 4i]/2 = –1 ± 2i
If z = (1+i)/(1–i), find z⁴.
Answer
z = i (shown above); z⁴ = i⁴ = 1
Express the conjugate of (3–2i)(2+3i) in the form a + ib.
Answer
(3–2i)(2+3i) = 6+9i–4i–6i² = 6+5i+6 = 12+5i
Conjugate = 12 – 5i