COMPLEX CONJUGATE ROOTS & DE MOIVRE’S THEOREM
This question bank contains 16 questions covering De Moivre’s theorem, nth roots of complex numbers, conjugate root theorem, and advanced applications, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (3 Questions)
MCQ 1. Using De Moivre’s theorem, \((\text{cis } 30°)^6\) equals:
A) \(\text{cis } 180°\) B) \(\text{cis } 36°\) C) \(-1\) D) Both A and C
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Solution:
Using De Moivre’s theorem: \((\text{cis } \theta)^n = \text{cis } (n\theta)\)
\((\text{cis } 30°)^6 = \text{cis } (6 \times 30°) = \text{cis } 180°\)
Since \(\text{cis } 180° = \cos 180° + i \sin 180° = -1 + 0i = -1\)
Both expressions represent the same value.
✅ Answer: D) Both A and C
MCQ 2. If \(3 + 2i\) is a root of a polynomial with real coefficients, which other number must also be a root?
A) \(-3 + 2i\) B) \(3 – 2i\) C) \(-3 – 2i\) D) \(2 + 3i\)
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Solution:
By the complex conjugate root theorem:
If a polynomial has real coefficients and \(a + bi\) is a root, then \(a – bi\) is also a root
For \(z = 3 + 2i\), the conjugate is \(\overline{z} = 3 – 2i\)
✅ Answer: B) \(3 – 2i\)
MCQ 3. How many distinct fourth roots does the complex number \(-16\) have?
A) 1 B) 2 C) 4 D) 8
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Solution:
Every non-zero complex number has exactly \(n\) distinct \(n\)th roots
4
For fourth roots, \(n = 4\), so there are 4 distinct fourth roots
These form vertices of a square on a circle centered at the origin
✅ Answer: C) 4
📌 Paper 1 Questions (No Calculator) – 6 Questions
Paper 1 – Q1. Use De Moivre’s theorem to evaluate \((\sqrt{3} – i)^4\).
[5 marks]
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Solution:
Step 1: Convert to polar form
\(|\sqrt{3} – i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\)
\(\arg(\sqrt{3} – i) = -\arctan(1/\sqrt{3}) = -\pi/6\) (Quadrant IV)
So \(\sqrt{3} – i = 2(\cos(-\pi/6) + i\sin(-\pi/6))\)
Step 2: Apply De Moivre’s theorem
\((\sqrt{3} – i)^4 = 2^4(\cos(-4\pi/6) + i\sin(-4\pi/6))\)
\(= 16(\cos(-2\pi/3) + i\sin(-2\pi/3))\)
Step 3: Evaluate trigonometric functions
\(\cos(-2\pi/3) = \cos(2\pi/3) = -1/2\)
\(\sin(-2\pi/3) = -\sin(2\pi/3) = -\sqrt{3}/2\)
Step 4: Final result
\((\sqrt{3} – i)^4 = 16(-1/2 – i\sqrt{3}/2) = -8 – 8i\sqrt{3}\)
✅ Answer: \((\sqrt{3} – i)^4 = -8 – 8i\sqrt{3}\)
Paper 1 – Q2. Find all cube roots of \(8i\).
[5 marks]
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Solution:
Step 1: Express \(8i\) in polar form
\(|8i| = 8\), \(\arg(8i) = \pi/2\)
So \(8i = 8e^{i\pi/2}\)
Step 2: Apply cube root formula
\((8i)^{1/3} = 8^{1/3} e^{i(\pi/2 + 2\pi k)/3}\) for \(k = 0, 1, 2\)
\(= 2e^{i(\pi/6 + 2\pi k/3)}\)
Step 3: Calculate each root
\(k = 0\): \(z_1 = 2e^{i\pi/6} = 2(\sqrt{3}/2 + i/2) = \sqrt{3} + i\)
\(k = 1\): \(z_2 = 2e^{i5\pi/6} = 2(-\sqrt{3}/2 + i/2) = -\sqrt{3} + i\)
\(k = 2\): \(z_3 = 2e^{i3\pi/2} = 2(0 – i) = -2i\)
✅ Answer: \(z = \sqrt{3} + i\), \(z = -\sqrt{3} + i\), \(z = -2i\)
Paper 1 – Q3. Use De Moivre’s theorem to derive the formula for \(\sin(2\theta)\).
[4 marks]
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Solution:
Step 1: Apply De Moivre’s theorem
\((\cos\theta + i\sin\theta)^2 = \cos(2\theta) + i\sin(2\theta)\)
Step 2: Expand left side
\((\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\cos\theta\sin\theta + (i\sin\theta)^2\)
\(= \cos^2\theta + 2i\cos\theta\sin\theta + i^2\sin^2\theta\)
\(= \cos^2\theta – \sin^2\theta + 2i\cos\theta\sin\theta\)
Step 3: Equate imaginary parts
From \(\cos(2\theta) + i\sin(2\theta) = (\cos^2\theta – \sin^2\theta) + 2i\cos\theta\sin\theta\)
Imaginary parts: \(\sin(2\theta) = 2\cos\theta\sin\theta\)
✅ Answer: \(\sin(2\theta) = 2\sin\theta\cos\theta\)
Paper 1 – Q4. If \(2 – 3i\) is a root of the polynomial \(P(z) = z^2 + bz + c\) where \(b\) and \(c\) are real, find \(b\) and \(c\).
[4 marks]
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Solution:
Step 1: Apply conjugate root theorem
Since \(P(z)\) has real coefficients and \(2 – 3i\) is a root, then \(2 + 3i\) is also a root
Step 2: Form the polynomial
\(P(z) = (z – (2 – 3i))(z – (2 + 3i))\)
\(= (z – 2 + 3i)(z – 2 – 3i)\)
\(= ((z – 2) + 3i)((z – 2) – 3i)\)
Step 3: Expand using difference of squares
\(P(z) = (z – 2)^2 – (3i)^2 = (z – 2)^2 – 9i^2 = (z – 2)^2 + 9\)
\(= z^2 – 4z + 4 + 9 = z^2 – 4z + 13\)
✅ Answer: \(b = -4\), \(c = 13\)
Paper 1 – Q5. Find the square roots of \(-3 + 4i\).
[5 marks]
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Solution:
Step 1: Convert to polar form
\(|-3 + 4i| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5\)
Quadrant II: \(\theta = \pi – \arctan(4/3)\)
Let \(\alpha = \arctan(4/3)\), so \(\theta = \pi – \alpha\)
Step 2: Apply square root formula
\((-3 + 4i)^{1/2} = 5^{1/2} e^{i(\pi – \alpha + 2\pi k)/2}\) for \(k = 0, 1\)
\(= \sqrt{5} e^{i(\pi – \alpha)/2 + i\pi k}\)
Step 3: Calculate both roots
\(k = 0\): \(z_1 = \sqrt{5} e^{i(\pi – \alpha)/2}\)
\(k = 1\): \(z_2 = \sqrt{5} e^{i(\pi – \alpha)/2 + i\pi} = -z_1\)
Step 4: Convert to Cartesian form
Using \(\cos((\pi – \alpha)/2) = \sin(\alpha/2)\) and \(\sin((\pi – \alpha)/2) = \cos(\alpha/2)\)
The roots are \(\pm(1 + 2i)\)
✅ Answer: \(z = 1 + 2i\) and \(z = -1 – 2i\)
Paper 1 – Q6. Show that \(1 + \omega + \omega^2 = 0\) where \(\omega = e^{2\pi i/3}\).
[3 marks]
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Solution:
Method 1: Direct calculation
\(\omega = e^{2\pi i/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\(\omega^2 = e^{4\pi i/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} – i\frac{\sqrt{3}}{2}\)
Step 2: Sum the terms
\(1 + \omega + \omega^2 = 1 + (-\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (-\frac{1}{2} – i\frac{\sqrt{3}}{2})\)
\(= 1 – 1 + 0i = 0\)
Method 2: Using \(\omega^3 = 1\)
Since \(\omega^3 – 1 = 0\), we have \((\omega – 1)(\omega^2 + \omega + 1) = 0\)
Since \(\omega \neq 1\), we must have \(\omega^2 + \omega + 1 = 0\)
✅ Proven: \(1 + \omega + \omega^2 = 0\)
📌 Paper 2 Questions (Calculator Allowed) – 4 Questions
Paper 2 – Q1. Use De Moivre’s theorem to find \((1 + i\sqrt{3})^{15}\).
[6 marks]
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Solution:
Step 1: Convert to polar form
\(|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\)
\(\arg(1 + i\sqrt{3}) = \arctan(\sqrt{3}/1) = \pi/3\)
So \(1 + i\sqrt{3} = 2e^{i\pi/3}\)
Step 2: Apply De Moivre’s theorem
\((1 + i\sqrt{3})^{15} = (2e^{i\pi/3})^{15} = 2^{15} e^{i15\pi/3} = 32768 e^{i5\pi}\)
Step 3: Simplify using periodicity
\(e^{i5\pi} = e^{i\pi} \cdot e^{i4\pi} = e^{i\pi} \cdot (e^{i2\pi})^2 = e^{i\pi} \cdot 1^2 = e^{i\pi} = -1\)
Step 4: Final result
\((1 + i\sqrt{3})^{15} = 32768 \times (-1) = -32768\)
✅ Answer: \((1 + i\sqrt{3})^{15} = -32768\)
Paper 2 – Q2. Find all fifth roots of \(32\) and represent them in the complex plane.
[7 marks]
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Solution:
Step 1: Express \(32\) in polar form
\(32 = 32e^{i \cdot 0}\)
Step 2: Apply fifth root formula
\(32^{1/5} = (32)^{1/5} e^{i(0 + 2\pi k)/5}\) for \(k = 0, 1, 2, 3, 4\)
\(= 2 e^{i2\pi k/5}\)
Step 3: Calculate each root
\(k = 0\): \(z_1 = 2e^{0} = 2\)
\(k = 1\): \(z_2 = 2e^{i2\pi/5} = 2(\cos(72°) + i\sin(72°))\)
\(k = 2\): \(z_3 = 2e^{i4\pi/5} = 2(\cos(144°) + i\sin(144°))\)
\(k = 3\): \(z_4 = 2e^{i6\pi/5} = 2(\cos(216°) + i\sin(216°))\)
\(k = 4\): \(z_5 = 2e^{i8\pi/5} = 2(\cos(288°) + i\sin(288°))\)
Step 4: Geometric representation
The five roots form vertices of a regular pentagon on a circle of radius 2, centered at origin, with angles \(0°, 72°, 144°, 216°, 288°\)
✅ Answer: Five roots as calculated, forming regular pentagon with radius 2
Paper 2 – Q3. A polynomial \(P(x) = x^3 – 6x^2 + 13x – 10\) has real coefficients. If \(2 + i\) is a root, find all roots and factorize \(P(x)\) completely.
[8 marks]
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Solution:
Step 1: Apply conjugate root theorem
Since \(P(x)\) has real coefficients and \(2 + i\) is a root, then \(2 – i\) is also a root
Step 2: Form quadratic factor
\((x – (2 + i))(x – (2 – i)) = (x – 2)^2 – i^2 = x^2 – 4x + 4 + 1 = x^2 – 4x + 5\)
Step 3: Perform polynomial division
Divide \(P(x) = x^3 – 6x^2 + 13x – 10\) by \(x^2 – 4x + 5\)
Using long division: \(P(x) = (x^2 – 4x + 5)(x – 2)\)
Step 4: Find third root
From \(x – 2 = 0\), the third root is \(x = 2\)
Step 5: Complete factorization
\(P(x) = (x – 2)(x – (2 + i))(x – (2 – i))\)
✅ Answer: Roots are \(2, 2+i, 2-i\); \(P(x) = (x-2)(x^2-4x+5)\)
Paper 2 – Q4. Solve the equation \(z^6 + 64 = 0\) and express all solutions in both exponential and Cartesian forms.
[8 marks]
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Solution:
Step 1: Rearrange equation
\(z^6 = -64 = 64e^{i\pi}\)
Step 2: Find sixth roots
\(z = (64)^{1/6} e^{i(\pi + 2\pi k)/6}\) for \(k = 0, 1, 2, 3, 4, 5\)
\(z = 2 e^{i\pi(1 + 2k)/6}\)
Step 3: Calculate all roots in exponential form
\(k = 0\): \(z_1 = 2e^{i\pi/6}\)
\(k = 1\): \(z_2 = 2e^{i\pi/2}\)
\(k = 2\): \(z_3 = 2e^{i5\pi/6}\)
\(k = 3\): \(z_4 = 2e^{i7\pi/6}\)
\(k = 4\): \(z_5 = 2e^{i3\pi/2}\)
\(k = 5\): \(z_6 = 2e^{i11\pi/6}\)
Step 4: Convert to Cartesian form
\(z_1 = 2(\sqrt{3}/2 + i/2) = \sqrt{3} + i\)
\(z_2 = 2(0 + i) = 2i\)
\(z_3 = 2(-\sqrt{3}/2 + i/2) = -\sqrt{3} + i\)
\(z_4 = 2(-\sqrt{3}/2 – i/2) = -\sqrt{3} – i\)
\(z_5 = 2(0 – i) = -2i\)
\(z_6 = 2(\sqrt{3}/2 – i/2) = \sqrt{3} – i\)
✅ Answer: Six solutions as calculated in both exponential and Cartesian forms
📌 Paper 3 Questions (Extended Response) – 3 Questions
Paper 3 – Q1. Investigation of De Moivre’s theorem and trigonometric identities.
(a) Use De Moivre’s theorem to derive formulas for \(\cos(4\theta)\) and \(\sin(4\theta)\) in terms of \(\cos\theta\) and \(\sin\theta\). [8 marks]
(b) Hence, find exact values of \(\cos(22.5°)\) and \(\sin(22.5°)\). [6 marks]
(c) Verify your results by using the half-angle formulas. [4 marks]
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Complete solution:
(a) Deriving formulas:
Using \((\cos\theta + i\sin\theta)^4 = \cos(4\theta) + i\sin(4\theta)\)
Expand using binomial theorem and separate real/imaginary parts
Result: \(\cos(4\theta) = \cos^4\theta – 6\cos^2\theta\sin^2\theta + \sin^4\theta\)
\(\sin(4\theta) = 4\cos^3\theta\sin\theta – 4\cos\theta\sin^3\theta\)
(b) Finding exact values:
Using \(4 \times 22.5° = 90°\) and solving the equations above
Results: \(\cos(22.5°) = \frac{\sqrt{2+\sqrt{2}}}{2}\), \(\sin(22.5°) = \frac{\sqrt{2-\sqrt{2}}}{2}\)
(c) Verification:
Using half-angle formulas with \(\theta = 45°\) confirms the results
✅ Complete derivation with exact trigonometric values
Paper 3 – Q2. Complex polynomial analysis and root relationships.
(a) A quartic polynomial \(P(z)\) with real coefficients has roots \(1+2i\) and \(3-i\). Find the other two roots and express \(P(z)\) in factored form. [6 marks]
(b) If the leading coefficient of \(P(z)\) is 2, find the complete polynomial. [4 marks]
(c) Find all roots of \(P(z) = 0\) and verify by substitution. [5 marks]
(d) Analyze the geometric pattern formed by these roots in the complex plane. [3 marks]
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✅ Complete polynomial analysis with geometric interpretation of complex roots
Paper 3 – Q3. Advanced applications of roots of unity.
(a) Find all 8th roots of unity and show they form a group under multiplication. [6 marks]
(b) Investigate the relationship between primitive 8th roots and cyclotomic polynomials. [7 marks]
(c) Apply these concepts to solve \(z^8 – z^4 + 1 = 0\). [5 marks]
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✅ Advanced investigation connecting roots of unity to group theory and cyclotomic polynomials