Answer to Question #113413 in Complex Analysis for Mzwandile

Question

QUESTION 9

9.1 State De Moivre’s Theorem.
(2)

9.2 Express cos 5θ and sin 4θ as polynomials in terms of sin θ and cos θ.
(8)

9.3 Let w be a negative real number, z a 6th root of w.

(a) Show that z (k) = ρ 6

6th roots of w.

cos

( π+2kπ )

+ i sin

( π+2kπ )

, k = 0, 1, 2, 3, 4, 5 is a formula for the



Show all your working.
(8)

(b) Hence determine the 6th roots of −729.
(2)

(c) Given z = cos θ + i sin θ and u + iv = (1 + z)(1 + z2). Prove that v = u tan( 3θ ) and

u2 + v2 = 16 cos2( θ ) cos2(θ)
(10)

[30]

Answer 1

9.1 The Moivre’s Theorem

$z=r(\cos\theta+i\sin \theta)\\ z^n=r^n(\cos n\theta+i\sin n\theta), n\in N$

9.2 Let $r=1$

$z=\cos\theta+i\sin\theta\\ z^5=\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5=\\ =\cos^5\theta+5i\cos^4\theta\sin\theta+10i^2\cos^3\theta\sin^2\theta+\\ +10i^3\cos^2\theta\sin^3\theta+5i^4\cos\theta\sin^4\theta+i^5\sin^5\theta=\\ =\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta+\\ +i(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)\\ \cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\$

$z^4=\cos4\theta+i\sin4\theta=(\cos\theta+i\sin\theta)^4=\\ =\cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta+\\ +6i^3\cos\theta\sin^3\theta+i^4\sin^4\theta=\\ =\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta+\\ +i(4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta)\\ \sin4\theta=4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta$

$i^2=-1, i^3=-i, i^4=1,i^5=i$

9.3 a)

$z=\sqrt[6]{w}, w<0\\ w=\rho(\cos\theta+i\sin\theta)\\ z=\sqrt[6]\rho(\cos\frac{\theta+2\pi k}{6}+i\sin\frac{\theta+2\pi k}{6}), k=0,1,2,3,4,5$

b)

$\sqrt[6]{-729}=\sqrt[6]{-1\cdot3^6}=3\sqrt[6]{-1}\\ -1=\cos\pi+i\sin\pi\\ \sqrt[6]{-1}=\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}, k=0,1,2,3,4,5\\ \sqrt[6]{-729}=3(\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}),\\ k=0,1,2,3,4,5\\$

$\sqrt[6]{-729}=3(\cos\frac{\pi}{6}+\sin\frac{\pi}{6})=3(\frac{\sqrt3}{2}+i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{3\pi }{6}+\sin\frac{3\pi }{6})=3i,\\ \sqrt[6]{-729}=3(\cos\frac{5\pi }{6}+\sin\frac{5\pi }{6})=3(-\frac{\sqrt3}{2}+i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{7\pi}{6}+\sin\frac{7\pi}{6})=3(-\frac{\sqrt3}{2}-i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{9\pi}{6}+\sin\frac{9\pi}{6})=-3i,\\ \sqrt[6]{-729}=3(\cos\frac{11\pi }{6}+\sin\frac{11\pi }{6})=3(\frac{\sqrt3}{2}-i\frac{1}{2})\\$

c)

$z=\cos\theta+i\sin\theta\\ u+iv=(1+z)(1+z^2)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(1+\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\((1+\cos^2\theta-\sin^2\theta)+2i\cos\theta\sin\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(2\cos^2\theta+2i\cos\theta\sin\theta)=\\ =2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta+\\ +i(2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta)\\ u=2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta=\\ =2\cos\theta(\cos\theta+\cos2\theta)\\ v=2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta=\\ =2\sin\theta\cos\theta(1+2\cos\theta)=2\cos\theta(\sin\theta+\sin2\theta)\\ u\tan\frac{3\theta}{2}=2\cos\theta(\cos\theta+\cos2\theta)\frac{\sin\frac{3\theta}{2}}{\cos\frac{3\theta}{2}}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{\sin\frac{\theta}{2}+\sin\frac{5\theta}{2}-\sin\frac{\theta}{2}+\sin\frac{7\theta}{2}}{2}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{2\sin3\theta\cos\frac{\theta}{2}}{2}=\\ =2\cos\theta(2\sin\frac{3\theta}{2}\cos\frac{\theta}{2})=\\ =2\cos\theta(\sin\theta+\sin2\theta)=v$

$u^2+v^2=u=4\cos^2\theta(\cos\theta+\cos2\theta)^2+\\ +4\cos^2\theta(\sin\theta+\sin2\theta)^2=\\ =4\cos^2\theta(\cos^2\theta+2\cos\theta\cos2\theta+\cos^22\theta+\\ +\sin^2\theta+2\sin\theta\sin2\theta+\sin^22\theta)=\\ =4\cos^2\theta(2+2\cos\theta)=8\cos^2\theta(1+\cos\theta)=\\ =16\cos^2\theta\cos^2\frac{\theta}{2}$