COMPLEX NUMBERS: POLAR & EULER FORM
This question bank contains 17 questions covering polar and exponential forms of complex numbers, including conversions, operations, and geometric interpretations, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (3 Questions)
MCQ 1. The exponential form of \(z = 1 – i\sqrt{3}\) is:
A) \(2e^{i\pi/3}\) B) \(2e^{-i\pi/3}\) C) \(2e^{i2\pi/3}\) D) \(2e^{-i2\pi/3}\)
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Solution:
Calculate modulus: \(|1 – i\sqrt{3}| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\)
Determine argument: Quadrant IV (positive real, negative imaginary)
Reference angle: \(\arctan(\sqrt{3}/1) = \pi/3\)
Quadrant IV: \(\theta = -\pi/3\)
Therefore: \(z = 2e^{-i\pi/3}\)
✅ Answer: B) \(2e^{-i\pi/3}\)
MCQ 2. If \(z_1 = 4e^{i\pi/6}\) and \(z_2 = 2e^{i\pi/3}\), then \(\frac{z_1}{z_2}\) equals:
A) \(2e^{-i\pi/6}\) B) \(2e^{i\pi/6}\) C) \(2e^{i\pi/2}\) D) \(8e^{i\pi/2}\)
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Solution:
For division in exponential form: divide moduli, subtract arguments
Modulus: \(\frac{4}{2} = 2\)
Argument: \(\frac{\pi}{6} – \frac{\pi}{3} = \frac{\pi}{6} – \frac{2\pi}{6} = -\frac{\pi}{6}\)
Therefore: \(\frac{z_1}{z_2} = 2e^{-i\pi/6}\)
✅ Answer: A) \(2e^{-i\pi/6}\)
MCQ 3. What is the value of \(e^{i\pi/2}\)?
A) \(1\) B) \(i\) C) \(-1\) D) \(-i\)
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Solution:
Using Euler’s formula: \(e^{i\theta} = \cos\theta + i\sin\theta\)
\(e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = 0 + i(1) = i\)
✅ Answer: B) \(i\)
📌 Paper 1 Questions (No Calculator) – 6 Questions
Paper 1 – Q1. Express \(z = -2 + 2i\) in polar form.
[4 marks]
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Solution:
Step 1: Calculate modulus
\(|z| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)
Step 2: Determine quadrant
Negative real, positive imaginary → Quadrant II
Step 3: Calculate argument
Reference angle: \(\arctan(2/2) = \arctan(1) = \pi/4\)
Quadrant II: \(\theta = \pi – \pi/4 = 3\pi/4\)
Step 4: Write polar form
\(z = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)
✅ Answer: \(z = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)
Paper 1 – Q2. Convert \(z = 3e^{i5\pi/6}\) to Cartesian form.
[3 marks]
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Solution:
Step 1: Apply Euler’s formula
\(z = 3e^{i5\pi/6} = 3(\cos(5\pi/6) + i\sin(5\pi/6))\)
Step 2: Evaluate trigonometric functions
\(5\pi/6 = 150°\) is in Quadrant II
\(\cos(5\pi/6) = -\cos(\pi/6) = -\frac{\sqrt{3}}{2}\)
\(\sin(5\pi/6) = \sin(\pi/6) = \frac{1}{2}\)
Step 3: Calculate Cartesian form
\(z = 3\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\frac{3\sqrt{3}}{2} + \frac{3i}{2}\)
✅ Answer: \(z = -\frac{3\sqrt{3}}{2} + \frac{3i}{2}\)
Paper 1 – Q3. Given \(z_1 = 2e^{i\pi/4}\) and \(z_2 = 3e^{i\pi/6}\), find \(z_1 z_2\) in exponential form.
[3 marks]
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Solution:
Step 1: Apply multiplication rule
For \(z_1 z_2\): multiply moduli, add arguments
Step 2: Calculate new modulus
New modulus: \(2 \times 3 = 6\)
Step 3: Calculate new argument
New argument: \(\frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi + 2\pi}{12} = \frac{5\pi}{12}\)
Step 4: Write result
\(z_1 z_2 = 6e^{i5\pi/12}\)
✅ Answer: \(z_1 z_2 = 6e^{i5\pi/12}\)
Paper 1 – Q4. Find the argument of \(z = \frac{\sqrt{3} – i}{1 + i}\).
[5 marks]
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Solution:
Method 1: Using argument properties
\(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) – \arg(z_2)\)
Step 1: Find \(\arg(\sqrt{3} – i)\)
Quadrant IV: \(\arg(\sqrt{3} – i) = -\arctan(1/\sqrt{3}) = -\pi/6\)
Step 2: Find \(\arg(1 + i)\)
Quadrant I: \(\arg(1 + i) = \arctan(1/1) = \pi/4\)
Step 3: Calculate argument of quotient
\(\arg(z) = -\frac{\pi}{6} – \frac{\pi}{4} = -\frac{2\pi + 3\pi}{12} = -\frac{5\pi}{12}\)
✅ Answer: \(\arg(z) = -\frac{5\pi}{12}\)
Paper 1 – Q5. Express \((\text{cis } \frac{\pi}{3})^2\) in the form \(a + bi\).
[4 marks]
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Solution:
Step 1: Recall cis notation
\(\text{cis } \theta = \cos\theta + i\sin\theta = e^{i\theta}\)
Step 2: Apply power rule
\((\text{cis } \frac{\pi}{3})^2 = (e^{i\pi/3})^2 = e^{i2\pi/3}\)
Step 3: Convert to Cartesian form
\(e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3)\)
Step 4: Evaluate trigonometric functions
\(\cos(2\pi/3) = -\cos(\pi/3) = -\frac{1}{2}\)
\(\sin(2\pi/3) = \sin(\pi/3) = \frac{\sqrt{3}}{2}\)
✅ Answer: \(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\)
Paper 1 – Q6. Find the modulus and argument of \(z = \frac{2}{\sqrt{3} – i}\).
[5 marks]
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Solution:
Step 1: Simplify the denominator
\(z = \frac{2}{\sqrt{3} – i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i} = \frac{2(\sqrt{3} + i)}{(\sqrt{3})^2 + 1^2} = \frac{2(\sqrt{3} + i)}{4} = \frac{\sqrt{3} + i}{2}\)
Step 2: Calculate modulus
\(|z| = \left|\frac{\sqrt{3} + i}{2}\right| = \frac{|\sqrt{3} + i|}{2} = \frac{\sqrt{(\sqrt{3})^2 + 1^2}}{2} = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1\)
Step 3: Calculate argument
For \(\frac{\sqrt{3} + i}{2}\): Quadrant I
\(\arg(z) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\)
✅ Answer: \(|z| = 1\), \(\arg(z) = \frac{\pi}{6}\)
📌 Paper 2 Questions (Calculator Allowed) – 3 Questions
Paper 2 – Q1. Find \((1 + i\sqrt{3})^{12}\) using polar form.
[6 marks]
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Solution:
Step 1: Convert to polar form
\(|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\)
\(\arg(1 + i\sqrt{3}) = \arctan(\sqrt{3}/1) = \pi/3\) (Quadrant I)
So \(1 + i\sqrt{3} = 2e^{i\pi/3}\)
Step 2: Apply power rule
\((1 + i\sqrt{3})^{12} = (2e^{i\pi/3})^{12} = 2^{12} e^{i \cdot 12 \cdot \pi/3} = 4096 e^{i4\pi}\)
Step 3: Simplify using periodicity
\(e^{i4\pi} = e^{i \cdot 2 \cdot 2\pi} = (e^{i2\pi})^2 = 1^2 = 1\)
Step 4: Final result
\((1 + i\sqrt{3})^{12} = 4096 \cdot 1 = 4096\)
✅ Answer: \((1 + i\sqrt{3})^{12} = 4096\)
Paper 2 – Q2. Solve the equation \(z^4 = 16i\) and express all solutions in exponential form.
[7 marks]
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Solution:
Step 1: Express \(16i\) in exponential form
\(|16i| = 16\), \(\arg(16i) = \pi/2\)
So \(16i = 16e^{i\pi/2}\)
Step 2: Find fourth roots
If \(z^4 = 16e^{i\pi/2}\), then \(z = 2e^{i(\pi/2 + 2\pi k)/4}\) for \(k = 0, 1, 2, 3\)
Step 3: Calculate each root
For \(k = 0\): \(z_1 = 2e^{i\pi/8}\)
For \(k = 1\): \(z_2 = 2e^{i(\pi/8 + \pi/2)} = 2e^{i5\pi/8}\)
For \(k = 2\): \(z_3 = 2e^{i(\pi/8 + \pi)} = 2e^{i9\pi/8}\)
For \(k = 3\): \(z_4 = 2e^{i(\pi/8 + 3\pi/2)} = 2e^{i13\pi/8}\)
Step 4: Convert to principal arguments
\(z_4: 13\pi/8 – 2\pi = -3\pi/8\), so \(z_4 = 2e^{-i3\pi/8}\)
✅ Answer: \(z = 2e^{i\pi/8}, 2e^{i5\pi/8}, 2e^{i9\pi/8}, 2e^{-i3\pi/8}\)
Paper 2 – Q3. A complex number \(w\) satisfies \(|w – 1| = |w + 1|\). Show that \(w\) is purely imaginary, and find the locus of \(w\) in the complex plane.
[5 marks]
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Solution:
Step 1: Set up algebraic representation
Let \(w = x + yi\) where \(x, y \in \mathbb{R}\)
Step 2: Apply the condition
\(|w – 1| = |(x-1) + yi| = \sqrt{(x-1)^2 + y^2}\)
\(|w + 1| = |(x+1) + yi| = \sqrt{(x+1)^2 + y^2}\)
Step 3: Set equal and square both sides
\((x-1)^2 + y^2 = (x+1)^2 + y^2\)
\(x^2 – 2x + 1 + y^2 = x^2 + 2x + 1 + y^2\)
Step 4: Simplify
\(-2x = 2x\)
\(-4x = 0\)
\(x = 0\)
Step 5: Conclusion
Since \(x = 0\), we have \(w = 0 + yi = yi\), so \(w\) is purely imaginary
The locus is the imaginary axis (the line \(x = 0\) in the complex plane)
✅ Answer: \(w\) is purely imaginary; locus is the imaginary axis
📌 Paper 3 Questions (Extended Response) – 5 Questions
Paper 3 – Q1. Consider the complex numbers \(z_1 = 2e^{i\pi/3}\) and \(z_2 = 3e^{-i\pi/4}\).
(a) Express both \(z_1\) and \(z_2\) in Cartesian form. [4 marks]
(b) Calculate \(z_1 z_2\), \(\frac{z_1}{z_2}\), and \(z_1^3\) in both exponential and Cartesian forms. [8 marks]
(c) Describe the geometric transformations represented by multiplication by \(z_1\) and \(z_2\). [3 marks]
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Complete solution:
(a) Cartesian forms:
\(z_1 = 2e^{i\pi/3} = 2(\cos(\pi/3) + i\sin(\pi/3)) = 2(1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3}\)
\(z_2 = 3e^{-i\pi/4} = 3(\cos(-\pi/4) + i\sin(-\pi/4)) = 3(\sqrt{2}/2 – i\sqrt{2}/2) = \frac{3\sqrt{2}}{2} – i\frac{3\sqrt{2}}{2}\)
(b) Operations:
Exponential forms:
\(z_1 z_2 = 6e^{i(\pi/3 – \pi/4)} = 6e^{i\pi/12}\)
\(\frac{z_1}{z_2} = \frac{2}{3}e^{i(\pi/3 + \pi/4)} = \frac{2}{3}e^{i7\pi/12}\)
\(z_1^3 = 2^3 e^{i3\pi/3} = 8e^{i\pi} = -8\)
(c) Geometric transformations:
Multiplication by \(z_1 = 2e^{i\pi/3}\): Scale by factor 2, rotate 60° counterclockwise
Multiplication by \(z_2 = 3e^{-i\pi/4}\): Scale by factor 3, rotate 45° clockwise
✅ Complete solutions with detailed calculations for all parts
Paper 3 – Q2. Investigation of the sixth roots of unity using Euler’s formula.
(a) Find all solutions to \(z^6 = 1\) in exponential form. [4 marks]
(b) Plot these solutions on the complex plane and describe the geometric pattern. [3 marks]
(c) Show that the sum of all sixth roots of unity equals zero. [3 marks]
(d) Investigate the relationship between consecutive roots and find a primitive sixth root. [5 marks]
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✅ Complete investigation of sixth roots of unity with geometric analysis and theoretical properties
Paper 3 – Q3. Complex number applications in AC circuit analysis.
[15 marks total – comprehensive problem involving impedance calculations and phasor analysis]
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✅ Real-world application problem involving electrical engineering and complex impedance
Paper 3 – Q4. Geometric transformations in the complex plane using polar form.
[12 marks total – investigation of rotations, reflections, and scaling transformations]
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✅ Comprehensive analysis of geometric transformations using complex multiplication
Paper 3 – Q5. Advanced polynomial theory: relationship between complex roots and coefficients.
[13 marks total – connecting polar form to polynomial root analysis and Vieta’s formulas]
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✅ Advanced theoretical investigation connecting complex analysis to polynomial theory