Answer to Question #117472 in Complex Analysis for Joseph Ocran

Question

(1) Express -1+i in polar form. Hence show that (-1+i)^16 is real and that 1/(-1+i)^6 is purely imaginary, giving the value for each.
(2) Express sin5α and cos5α/cosα in terms of sinα

Answer 1

(1)

$-1+i =\sqrt 2\big(\frac{-1}{\sqrt 2}+i \frac{1}{\sqrt 2}\big)=\sqrt 2(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})$

Using de Moivre's formula, we have:

$(-1+i)^{16}=(\sqrt 2)^{16} (\cos (16\times \frac{3\pi}{4})+i \sin(16\times \frac{3\pi}{4}))=256(\cos 12\pi+i\sin 12\pi)=256$

$(-1+i)^{16}=256$ is real number

$\frac{1}{(-1+i)^6}=(-1+i)^{-6}=(\sqrt 2)^{-6} (\cos (-6\times \frac{3\pi}{4})+i \sin(-6\times \frac{3\pi}{4}))=1/8(\cos -4,5\pi+i\sin -4,5\pi)=-i /8$

$\frac{1}{(-1+i)^6}=-i /8$ is purely imaginary

(2)

Let $z=\cos \alpha +i \sin \alpha$

Using de Moivre's formula, we have that $z^5=\cos 5 \alpha +i \sin 5 \alpha$

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

$\cos 5 \alpha = \cos ^5 \alpha -10\cos ^3 \alpha \sin^2 \alpha +5\cos \alpha \sin^4 \alpha =\cos \alpha (\cos^4 \alpha -10\cos^2 \alpha \sin^2 \alpha +5\sin^4 \alpha )=\cos \alpha ((1-\sin^2 \alpha )^2-10(1-\sin^2 \alpha )\sin^2 \alpha +5\sin^4 \alpha )=\cos \alpha (16\sin^4 \alpha -12 \sin^2 \alpha +1)$

$\frac{\cos 5 \alpha }{\cos \alpha }=16\sin^4 \alpha -12\sin^2 \alpha +1$

$\sin 5 \alpha =5\cos^4\alpha \sin \alpha -10\cos^2 \alpha \sin^3 \alpha +\sin ^5 \alpha =5(1-\sin^2 \alpha )^2\sin \alpha -10(1-\sin^2 \alpha )\sin^3 \alpha +\sin ^5 \alpha =16\sin ^5 \alpha -20\sin^3 \alpha +5\sin\alpha$

$\sin 5\alpha= 16\sin ^5 \alpha -20\sin^3 \alpha +5\sin \alpha$