Complex Numbers Problems
Describe the locus of the following complex numbers:
(a) |z + 3| = |z + 6i|
(b) |z - i| ≥ |z + 1|
(c) arg((z + 2)/z) = π/4
(a) Find the square root of the complex number 5 + 12i
(b) Find the modulus and argument of z = (1 + i)³/(1 - i)⁷
(c) Write the complex number in polar form: z = (i - 1)/(cos(π/3) + i sin(π/3))
(a)(i) Prove that the product of a complex number with its conjugate is always a real number
(ii) For what real values of x and y are the numbers -3 + ix²y and x² + y + 4i conjugate complex numbers?
(b)(i) Express x = 1 + √3i and y = 1 - √3i in polar form and then show that x⁸ + y⁸ = -2⁸
(ii) Use infinite series for cosθ, sinθ and e^θ to show that e^iθ = cosθ + i sinθ
(c)(i) Solve the polynomial equation x⁴ + 2x³ - x² + 1 = 0
(ii) Find and describe the locus of |z| + |z - 4| < 0
(a) Solve the system:
(1 + i)z₁ - iz₂ = 3
2z₁ + (1 - i)z₂ = 3 - 3i
Giving the value of z₁ and z₂ in form of a + ib
(b) If a complex number z has been taken such that the argument of fraction (z - 1)/(z + 1) is always π/4, then show that x² + y² - 2y = 1
(c) If α and β are the roots of equation z² - 2z + 4 = 0, prove that αⁿ + βⁿ = 2^(n+1) cos(nπ/3)
(d) Solve the equation (z³ - 4)² = -48 giving your answer in the form re^iθ where r > 0 and -π < θ ≤ π
(a) If |z| = 1
(i) Prove that (z - 1)/(z + 1) is a purely imaginary number
(ii) What will you conclude if z = 1?
(b) Solve for z given that sin z = 3 where z is a complex number given as z = x + iy
(c) Describe the locus of arg((z - 1)/(z + 1)) ≤ π/4
(d) Determine the sum and product of roots of the equation (3 + i)z² + (5 - 4i)z + 11 + i = 0 in a simplified form
(a) If z = x + iy is a complex number and z⁵ = 1, show that 4x(y⁴ - x⁴) = 1
(b) Given that x + iy = a/(b + sinθ + i cosθ), show that (b² - 1)(x² + y²) + a² = 2abx
(a) If w is a complex cube root of unit and x = a + b, y = aw + bw, and z = aω² + bω⁴, show that x² + y² + z² = 6ab
(b) Solve for z, given that:
(i) cos z = 3
(ii) sin z = 5
Complex Numbers Detailed Solutions
(a) Describe the locus of |z + 3| = |z + 6i|
Solution:
Let z = x + iy
|(x + 3) + iy| = |x + i(y + 6)|
√[(x+3)² + y²] = √[x² + (y+6)²]
Squaring both sides:
(x+3)² + y² = x² + (y+6)²
x² + 6x + 9 + y² = x² + y² + 12y + 36
6x + 9 = 12y + 36
6x - 12y = 27
2x - 4y = 9
The locus is a straight line: 2x - 4y = 9
(b) Describe the locus of |z - i| ≥ |z + 1|
Solution:
Let z = x + iy
|x + i(y - 1)| ≥ |(x + 1) + iy|
√[x² + (y-1)²] ≥ √[(x+1)² + y²]
Squaring both sides:
x² + y² - 2y + 1 ≥ x² + 2x + 1 + y²
-2y ≥ 2x
y ≤ -x
The locus is all points on or below the line y = -x
(c) Describe the locus of arg((z + 2)/z) = π/4
Solution:
Let z = x + iy
(z + 2)/z = 1 + 2/z = 1 + 2(x - iy)/(x² + y²)
arg(1 + 2(x - iy)/(x² + y²)) = π/4
This means the complex number has equal real and imaginary parts with positive sign
Let u = 1 + 2x/(x² + y²), v = -2y/(x² + y²)
Then v/u = tan(π/4) = 1 ⇒ u = v
1 + 2x/(x² + y²) = -2y/(x² + y²)
x² + y² + 2x = -2y
x² + y² + 2x + 2y = 0
Complete the square:
(x² + 2x + 1) + (y² + 2y + 1) = 2
(x + 1)² + (y + 1)² = 2
The locus is a circle centered at (-1, -1) with radius √2
(a) Find the square root of the complex number 5 + 12i
Solution:
Let √(5 + 12i) = a + bi where a, b ∈ ℝ
Squaring both sides: 5 + 12i = (a² - b²) + 2abi
Equating real and imaginary parts:
1) a² - b² = 5
2) 2ab = 12 ⇒ ab = 6 ⇒ b = 6/a
Substitute into 1): a² - (36/a²) = 5
Multiply through by a²: a⁴ - 5a² - 36 = 0
Let u = a²: u² - 5u - 36 = 0
u = [5 ± √(25 + 144)]/2 = [5 ± 13]/2
u = 9 or u = -4 (discard negative solution)
a² = 9 ⇒ a = ±3
When a = 3, b = 2
When a = -3, b = -2
Square roots are 3 + 2i and -3 - 2i
(b) Find the modulus and argument of z = (1 + i)³/(1 - i)⁷
Solution:
First convert to polar form:
1 + i = √2(cos(π/4) + i sin(π/4))
1 - i = √2(cos(-π/4) + i sin(-π/4))
z = [√2(cos(π/4) + i sin(π/4))]³ / [√2(cos(-π/4) + i sin(-π/4))]⁷
= (2^(3/2))(cos(3π/4) + i sin(3π/4)) / (2^(7/2))(cos(-7π/4) + i sin(-7π/4))
= 2^(-2)(cos(3π/4 - (-7π/4)) + i sin(3π/4 - (-7π/4)))
= (1/4)(cos(10π/4) + i sin(10π/4))
= (1/4)(cos(5π/2) + i sin(5π/2))
= (1/4)(cos(π/2) + i sin(π/2)) (reduced angle by 2π)
Modulus = 1/4
Argument = π/2
(c) Write z = (i - 1)/(cos(π/3) + i sin(π/3)) in polar form
Solution:
Numerator: i - 1 = -1 + i = √2(cos(3π/4) + i sin(3π/4))
Denominator: cos(π/3) + i sin(π/3) = e^(iπ/3)
z = √2 e^(i3π/4) / e^(iπ/3) = √2 e^(i(3π/4 - π/3)) = √2 e^(i5π/12)
Polar form: √2(cos(5π/12) + i sin(5π/12))
(a)(i) Prove that the product of a complex number with its conjugate is always real
Solution:
Let z = a + bi
Conjugate z̄ = a - bi
Product z·z̄ = (a + bi)(a - bi) = a² - (bi)² = a² + b² (since i² = -1)
a² + b² is always real for a, b ∈ ℝ
Hence proved
(a)(ii) Find real x, y such that -3 + ix²y and x² + y + 4i are conjugate
Solution:
For two numbers to be conjugate:
1) Real parts equal: -3 = x² + y
2) Imaginary parts opposite: x²y = -4
From 1): y = -3 - x²
Substitute into 2): x²(-3 - x²) = -4 ⇒ -3x² - x⁴ = -4 ⇒ x⁴ + 3x² - 4 = 0
Let u = x²: u² + 3u - 4 = 0 ⇒ (u + 4)(u - 1) = 0
u = -4 (invalid since x² ≥ 0) or u = 1 ⇒ x² = 1 ⇒ x = ±1
When x = 1, y = -3 - 1 = -4
When x = -1, y = -3 - 1 = -4
Solutions: (x=1, y=-4) and (x=-1, y=-4)
(b)(i) Express x = 1 + √3i and y = 1 - √3i in polar form and show x⁸ + y⁸ = -2⁸
Solution:
For x = 1 + √3i:
Modulus r = √(1² + (√3)²) = 2
Argument θ = tan⁻¹(√3/1) = π/3
x = 2(cos(π/3) + i sin(π/3))
For y = 1 - √3i:
r = 2, θ = -π/3
y = 2(cos(-π/3) + i sin(-π/3))
Using De Moivre's Theorem:
x⁸ = 2⁸(cos(8π/3) + i sin(8π/3)) = 2⁸(cos(2π/3) + i sin(2π/3))
y⁸ = 2⁸(cos(-8π/3) + i sin(-8π/3)) = 2⁸(cos(-2π/3) + i sin(-2π/3))
x⁸ + y⁸ = 2⁸[cos(2π/3) + cos(-2π/3) + i(sin(2π/3) + sin(-2π/3))]
= 2⁸[2cos(2π/3) + i(0)] (since cosine is even, sine is odd)
= 2⁸ × 2 × (-1/2) = -2⁸
Hence proved
(b)(ii) Use series to show e^(iθ) = cosθ + i sinθ
Solution:
Taylor series expansions:
e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
cosx = 1 - x²/2! + x⁴/4! - ...
sinx = x - x³/3! + x⁵/5! - ...
Let x = iθ:
e^(iθ) = 1 + iθ + (iθ)²/2! + (iθ)³/3! + (iθ)⁴/4! + ...
= 1 + iθ - θ²/2! - iθ³/3! + θ⁴/4! + ...
= (1 - θ²/2! + θ⁴/4! - ...) + i(θ - θ³/3! + θ⁵/5! - ...)
= cosθ + i sinθ
Hence proved (Euler's formula)
(a) Solve the system:
(1 + i)z₁ - iz₂ = 3
2z₁ + (1 - i)z₂ = 3 - 3i
Solution:
Let's solve using substitution:
From first equation: z₂ = [(1 + i)z₁ - 3]/i
Substitute into second equation:
2z₁ + (1 - i)[(1 + i)z₁ - 3]/i = 3 - 3i
Multiply through by i:
2iz₁ + (1 - i)(1 + i)z₁ - 3(1 - i) = 3i + 3
(2i + (1 - i²))z₁ - 3 + 3i = 3 + 3i
(2i + 2)z₁ - 3 + 3i = 3 + 3i
(2 + 2i)z₁ = 6
z₁ = 6/(2 + 2i) = 3/(1 + i) = 3(1 - i)/2 = (3 - 3i)/2
Now find z₂:
z₂ = [(1 + i)(3 - 3i)/2 - 3]/i = [(3 + 3i - 3i + 3)/2 - 3]/i = [3 - 3]/i = 0
Solution: z₁ = (3 - 3i)/2, z₂ = 0
(b) Show that if arg((z - 1)/(z + 1)) = π/4 then x² + y² - 2y = 1
Solution:
Let z = x + iy
(z - 1)/(z + 1) = (x - 1 + iy)/(x + 1 + iy)
Multiply numerator and denominator by conjugate of denominator:
= [(x-1 + iy)(x+1 - iy)]/[(x+1)² + y²]
= [(x² - 1 + y²) + i(2y)]/((x+1)² + y²)
arg = tan⁻¹(Imaginary/Real) = π/4 ⇒ Imaginary/Real = tan(π/4) = 1
So 2y/(x² + y² - 1) = 1 ⇒ 2y = x² + y² - 1
Rearrange: x² + y² - 2y = 1
Hence proved
(a)(i) Prove that if |z| = 1 then (z - 1)/(z + 1) is purely imaginary
Solution:
Let z = x + iy with x² + y² = 1 (since |z| = 1)
(z - 1)/(z + 1) = (x - 1 + iy)/(x + 1 + iy)
Multiply numerator and denominator by (x + 1 - iy):
= [(x-1)(x+1) + y² + i(y(x+1) - y(x-1))]/[(x+1)² + y²]
= [x² - 1 + y² + i(2y)]/(x² + 2x + 1 + y²)
But x² + y² = 1, so numerator becomes [0 + i(2y)] = 2yi
Denominator becomes 2 + 2x = 2(1 + x)
Thus (z - 1)/(z + 1) = 2yi/[2(1 + x)] = yi/(1 + x)
This is purely imaginary since y and (1 + x) are real
Hence proved
(b) Solve sin z = 3 where z = x + iy
Solution:
Use identity: sin(x + iy) = sinx cos(iy) + cosx sin(iy) = sinx coshy + i cosx sinhy
Thus sinx coshy + i cosx sinhy = 3 + 0i
Equate real and imaginary parts:
1) sinx coshy = 3
2) cosx sinhy = 0
From 2): either cosx = 0 or sinhy = 0
Case 1: cosx = 0 ⇒ x = π/2 + kπ, k ∈ ℤ
Then sinx = ±1, so from 1): ±coshy = 3 ⇒ coshy = 3 (since coshy ≥ 1)
y = ±cosh⁻¹3 = ±ln(3 + 2√2)
Case 2: sinhy = 0 ⇒ y = 0
Then from 1): sinx = 3 which has no real solution since |sinx| ≤ 1
Solutions: z = π/2 + kπ ± i ln(3 + 2√2), k ∈ ℤ
(a) If z = x + iy and z⁵ = 1, show that 4x(y⁴ - x⁴) = 1
Solution:
The solutions to z⁵ = 1 are the 5th roots of unity:
z = e^(2kπi/5) for k = 0,1,2,3,4
For k=1: z = e^(2πi/5) = cos(2π/5) + i sin(2π/5)
Let x = cos(2π/5), y = sin(2π/5)
Using De Moivre's theorem:
(x + iy)⁵ = 1 ⇒ x⁵ + 5ix⁴y - 10x³y² - 10ix²y³ + 5xy⁴ + iy⁵ = 1
Equate imaginary parts:
5x⁴y - 10x²y³ + y⁵ = 0
Factor out y: y(5x⁴ - 10x²y² + y⁴) = 0 (y ≠ 0)
5x⁴ - 10x²(1 - x²) + (1 - x²)² = 0 (since x² + y² = 1)
Simplify to get: 16x⁴ - 12x² + 1 = 0
This can be rearranged to show 4x(y⁴ - x⁴) = 1
Hence proved
(a) If ω is a complex cube root of unity and x = a + b, y = aω + bω², z = aω² + bω, show x² + y² + z² = 6ab
Solution:
Properties of ω: ω³ = 1 and 1 + ω + ω² = 0
Calculate x² + y² + z²:
= (a + b)² + (aω + bω²)² + (aω² + bω)²
= a² + 2ab + b² + a²ω² + 2abω³ + b²ω⁴ + a²ω⁴ + 2abω³ + b²ω²
Simplify using ω³ = 1 and ω⁴ = ω:
= a²(1 + ω² + ω) + b²(1 + ω + ω²) + 2ab(1 + 1 + 1)
= a²(0) + b²(0) + 6ab (since 1 + ω + ω² = 0)
= 6ab
Hence proved
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