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[h] IB Mathematics AA HL Flashcards- Complex roots & de Moivre’s theorem
[q] Complex roots & de Moivre’s theorem
EXPLORATION
If \( z = r(\cos \theta + i \sin \theta) \), find \( z^2 \) & \( z^3 \):
\( \mathbin{\blacktriangleright} z^2 = (r(\cos \theta + i \sin \theta)) (r(\cos \theta + i \sin \theta)) = r^2((\cos \theta \cos \theta – \sin \theta \sin \theta) + i(\sin \theta \cos \theta + \cos \theta \sin \theta)) \)
= \( r^2((\cos 2\theta – \sin 2\theta) + i(2 \sin \theta \cos \theta)) = r^2(\cos 2\theta + i \sin 2\theta) \)
\( \mathbin{\blacktriangleright} z^3 = z \cdot z^2 = r(\cos \theta + i \sin \theta) \cdot r^2(\cos 2\theta + i \sin 2\theta) \)
= \( r^3(\cos (\theta + 2\theta) + i \sin (\theta + 2\theta)) = r^3(\cos 3\theta + i \sin 3\theta) \)
[q]
DE MOIVRE’S THEOREM// This extends to any real \( n \):
\[
z^n = r^n (\cos (n\theta) + i \sin (n\theta)) = r^n e^{in\theta} = r^n \text{cis}(n\theta) \quad \text{→ in F.B.}
\]
[a]
PROOF// Use induction (see [1.15]):
BASIS STEP: If \( P(n) \) is the statement: \( z^n = r^n (\cos (n\theta) + i \sin (n\theta)) \). We need to show that \( P(1) \) is true, but \( z^1 = r^1 (\cos \theta + i \sin \theta) \) is given as part of the theorem.
[q]
INDUCTIVE STEP: Assume \( P(k) \) is true: i.e. \( z^k = r^k (\cos (k\theta) + i \sin (k\theta)) \). We need to show that \( P(k+1) \) then holds: i.e. \( z^{k+1} = r^{k+1} (\cos ((k+1)\theta) + i \sin ((k+1)\theta)) \Rightarrow \)
[a]
\(\Rightarrow z^{k+1} = z \cdot z^k = r (\cos \theta + i \sin \theta) \cdot r^k (\cos (k\theta) + i \sin (k\theta))\)
\(\Rightarrow r^{k+1} \left[ (\cos \theta \cos k\theta – \sin \theta \sin k\theta) + i (\sin \theta \cos k\theta + \cos \theta \sin k\theta) \right]\)
\(\Rightarrow r^{k+1} (\cos ((k+1)\theta) + i \sin ((k+1)\theta))\) using compound angle formulae.
[q]
STATEMENT: Since it is true for \( P(1) \), and whenever \( P(k) \) is true, then \( P(k+1) \) is proven to be true… By induction, \( P(n) \) is true for all \( Z^+ \).
E.G.1 // Find \( (1+i)^6 \): Convert to polar form:
[a]
1. \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \), \( \theta = \tan^{-1} (1) = \frac{\pi}{4} \)
\(\Rightarrow \sqrt{2} \text{cis} (\frac{\pi}{4})^6 = (\sqrt{2})^6 \cdot \text{cis} (6 \cdot \frac{\pi}{4}) = 8 \text{cis} \frac{3\pi}{2} = 8(0 + i(-1)) = -8i \)
[q]
APPLICATIONS// The first arrives from looking at \( z^n \), then \( Z^{-n} \). We get the formula:
\[
Z^{-n} = (\cos n\theta – i \sin n\theta) \cdot r^{-n} = \frac{1}{Z^n} = 2 \cos (n\theta) / Z^n – Z^{-n} = 2i \sin (n\theta) \quad \text{(when } r = 1 \text{)}
\]
[a]
E.G.2 // Express \( \cos^3 \theta \) in terms of the first powers of cosine:
1. Starting with \( (z + \frac{1}{z})^3 = (\cos \theta + i \sin \theta + \cos \theta – i \sin \theta)^3 = (2 \cos \theta)^3 \)
[q]
Expand:
\[
z^3 + 3z + 3 \cdot \frac{1}{z} + \frac{1}{z^3} = 8 \cos^3 \theta = \left(z^3 + \frac{1}{z^3} \right) + 3 \left(z + \frac{1}{z}\right)
\]
\(\Rightarrow 8 \cos^3 \theta = 2 \cos 3\theta + 6 \cos \theta \Rightarrow \cos^3 \theta = \frac{1}{4} \cos 3\theta + \frac{3}{4} \cos \theta\)
[a] Conjugate roots
[q]
COMPLEX n{th} ROOTS// Due to the cyclical nature of cos & sin, we get multiple results for roots of complex numbers. From de Moivre’s, we get:
\[
Z_k = \sqrt[n]{r} \left( \cos \left( \frac{\theta}{n} + \frac{2k\pi}{n} \right) + i \sin \left( \frac{\theta}{n} + \frac{2k\pi}{n} \right) \right) \quad \text{where } k = 0, 1, 2, 3, \dots, n-1
\]
[a]
E.G. 3// Find the cube roots of \( z = 8(\text{cis } \frac{3\pi}{2}) \):
\[
\omega_k = \sqrt[3]{r} (\text{cis } \frac{\theta}{3} + \frac{2k\pi}{3}) = \sqrt[3]{8} (\text{cis } \frac{3\pi}{2} + \frac{2k\pi}{3}) = 2 \text{cis } (\frac{\pi}{2} + \frac{2k\pi}{3}) \quad ; k = 0, 1, 2
\]
[q]
Meaning:
\[
\omega_1 = 2 \text{cis } \frac{\pi}{2}, \quad \omega_2 = 2 \text{cis } \frac{11\pi}{12}, \quad \omega_3 = 2 \text{cis } \frac{9\pi}{12}
\]
[a]
1// You can also find the n{th} roots of 1 by considering \( 1 = 1 (\cos 0 + i \sin 0) \).
COMPLEX CONJUGATE ROOTS// You’ll see in [2.7] that quadratics sometimes don’t have real roots. This is when you are doing the square root of a -ve in the quadratic formula. But we now know how to do these square roots.
[q]
E.G. 4// Solve \( x^2 + 4x + 5 = 0 \):
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 – 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 + i \quad \& \quad -2 – i
\]
[a]
E.G. 5// Solve \( x^3 – 7x^2 + 41x – 87 = 0 \), where \( (x – 3) \) is a factor:
1. Factor out \( (x – 3) \):
\[
x^3 – 7x^2 + 41x – 87 = (x – 3)(x^2 – 4x + 29)
\]
[q]
So now, we solve \( x^2 – 4x + 29 = 0 \) to get the other roots. Using the quadratic formula, we get:
\[
x = \frac{4 \pm \sqrt{16 – 116}}{2} = \frac{2 \pm \sqrt{100}}{2} = 2 \pm 5i
\]
[a]
So we have 3 roots = 2 complex, 1 real:
\[
2 + 5i, \quad 2 – 5i, \quad 3
\]
NOTE// We see that we have complex conjugate roots here.
[q]
RULES//
Polynomials of degree \( n \) (\( n > 0 \)) have exactly \( n \) complex zeros.
(Reals are included in complex nos.) & (the degree is the highest power of \( x \))
\[
\mathbin{\blacktriangleright} \text{If } z \text{ is a root, then } z^* \text{ is also a root.}
\]
\[
\mathbin{\blacktriangleright} \text{If the polynomial has odd degree, it has at least 1 real root.}
\]
(This is because the complex conjugate roots only come in pairs)
[a]
E.G. 6// Given that \( 2 – 3i \) is a zero of \( 5x^3 – 19x^2 + 61x + 13 \), find the others:
1. If \( 2 – 3i \) is a zero, then \( 2 + 3i \) is… we get:
\[
5x^3 – 19x^2 + 61x + 13 = (x – (2 – 3i))(x – (2 + 3i))(ax + b)
\]
\[
= (x^2 – 4x + 13)(ax + b)
\]
[q]
Solve, and get:
\[
5x – 1
\]
So the final zero is:
\[
x = \frac{1}{5}
\]
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