12 exam-style questions on exponential form, De Moivre’s theorem, \(n\)th roots and trigonometric identities. Tap Show answer for full working.
Express \( z = 2 + 2i \) in the form \( re^{i\theta} \), where \( r > 0 \) and \( -\pi < \theta \le \pi \).
\( r = |z| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \).
\( \theta = \arg(z) = \tan^{-1}(1) = \pi/4 \) (first quadrant).
So \( z = 2\sqrt{2}\, e^{i\pi/4} \).
Given \( z = 3e^{i\pi/6} \) and \( w = 2e^{-i\pi/3} \), find \( zw \) in the form \( re^{i\theta} \).
\( zw = 3\cdot 2 \cdot e^{i(\pi/6 - \pi/3)} = 6e^{i(-\pi/6)} = 6e^{-i\pi/6} \).
Use De Moivre’s theorem to express \( \cos 3\theta \) in terms of \( \cos\theta \).
\( (\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta \).
Expand: \( \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta \).
Real part: \( \cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta \).
Find the cube roots of \( 8 \) in the form \( a + bi \).
\( 8 = 8e^{i0} = 8e^{i2k\pi} \). Roots: \( 2e^{i(2k\pi)/3} \) for \( k = 0,1,2 \).
\( k=0 \): \( 2 \).
\( k=1 \): \( 2e^{i2\pi/3} = 2(-\frac12 + i\frac{\sqrt{3}}{2}) = -1 + i\sqrt{3} \).
\( k=2 \): \( 2e^{i4\pi/3} = 2(-\frac12 - i\frac{\sqrt{3}}{2}) = -1 - i\sqrt{3} \).
Express \( z = -1 + i\sqrt{3} \) in exponential form and hence find \( z^6 \).
\( r = \sqrt{1+3} = 2 \), \( \arg = 2\pi/3 \) (second quadrant).
\( z = 2e^{i2\pi/3} \). Then \( z^6 = 2^6 e^{i4\pi} = 64e^{i0} = 64 \).
Use De Moivre’s theorem to find \( \sin 5\theta \) in terms of \( \sin\theta \).
\( (\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta \).
Imaginary part of expansion: \( \sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \).
In terms of \( \sin\theta \) only (using \( \cos^2 = 1-\sin^2 \)): \( \sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \).
Find all solutions of \( z^4 = -16 \) in the form \( re^{i\theta} \) with \( r > 0 \) and \( -\pi < \theta \le \pi \).
\( -16 = 16e^{i(\pi + 2k\pi)} \). Roots: \( 2e^{i(\pi + 2k\pi)/4} = 2e^{i(\pi/4 + k\pi/2)} \), \( k=0,1,2,3 \).
\( k=0 \): \( 2e^{i\pi/4} \)
\( k=1 \): \( 2e^{i3\pi/4} \)
\( k=2 \): \( 2e^{i5\pi/4} = 2e^{-i3\pi/4} \)
\( k=3 \): \( 2e^{i7\pi/4} = 2e^{-i\pi/4} \)
Show that \( \cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1 \).
\( \cos 4\theta = \operatorname{Re}((\cos\theta + i\sin\theta)^4) \).
\( (\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4i\cos^3\sin - 6\cos^2\sin^2 - 4i\cos\sin^3 + \sin^4 \).
Real: \( \cos^4 - 6\cos^2\sin^2 + \sin^4 = \cos^4 - 6\cos^2(1-\cos^2) + (1-\cos^2)^2 \)
\( = \cos^4 - 6\cos^2 + 6\cos^4 + 1 - 2\cos^2 + \cos^4 = 8\cos^4 - 8\cos^2 + 1 \).
Find the square roots of \( 3 + 4i \) in the form \( a + bi \), where \( a,b \in \mathbb{R} \).
Let \( \sqrt{3+4i} = x + yi \). Then \( (x+yi)^2 = 3+4i \Rightarrow x^2 - y^2 = 3 \), \( 2xy = 4 \).
So \( xy = 2 \). Also \( x^2 + y^2 = |3+4i| = 5 \).
Adding: \( 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \).
If \( x=2 \), \( y=1 \); if \( x=-2 \), \( y=-1 \).
Roots: \( 2 + i \) and \( -2 - i \).
Use De Moivre’s theorem to sum the series \( \sum_{k=0}^{n} \binom{n}{k}\cos k\theta \).
(Hint: consider \( (1 + e^{i\theta})^n \).)
\( (1 + e^{i\theta})^n = \sum \binom{n}{k} e^{ik\theta} = \sum \binom{n}{k}(\cos k\theta + i\sin k\theta) \).
Real part is the required sum. Write \( 1 + e^{i\theta} = 1 + \cos\theta + i\sin\theta = 2\cos^2(\theta/2) + 2i\sin(\theta/2)\cos(\theta/2) \) (using half-angle identities).
\( = 2\cos(\theta/2)\, e^{i\theta/2} \) (when \( \cos(\theta/2) \ge 0 \)).
Thus \( (1+e^{i\theta})^n = 2^n \cos^n(\theta/2)\, e^{in\theta/2} \).
Real part: \( 2^n \cos^n(\theta/2)\cos(n\theta/2) \).
Solve \( z^5 - 1 = 0 \) and hence factorise \( z^5 - 1 \) into linear and quadratic factors with real coefficients.
Roots: \( e^{i2k\pi/5} \) for \( k=0,1,2,3,4 \). One real root \( z=1 \).
Other roots come in conjugate pairs: \( e^{\pm i2\pi/5} \), \( e^{\pm i4\pi/5} \).
\( z^5 - 1 = (z-1)(z^2 - 2z\cos(2\pi/5) + 1)(z^2 - 2z\cos(4\pi/5) + 1) \).
The complex number \( z \) satisfies \( |z| = 1 \) and \( \arg(z - 1) = \dfrac{2\pi}{3} \). Find \( z \) in the form \( e^{i\theta} \) where \( -\pi < \theta \le \pi \).
Let \( z = e^{i\theta} \). Then \( z - 1 = 2i\sin(\theta/2)\, e^{i\theta/2} \).
For \( 0 < \theta < 2\pi \), \( \sin(\theta/2) > 0 \), so \( \arg(z-1) = \dfrac{\theta}{2} + \dfrac{\pi}{2} \) (taking the principal value in \( (-\pi,\pi] \)).
Set \( \dfrac{\theta}{2} + \dfrac{\pi}{2} = \dfrac{2\pi}{3} \) ⇒ \( \dfrac{\theta}{2} = \dfrac{\pi}{6} \) ⇒ \( \theta = \dfrac{\pi}{3} \).
Check: \( e^{i\pi/3} - 1 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \), \( \arg = \dfrac{2\pi}{3} \). ✓
Hence \( z = e^{i\pi/3} \).