Question

Complex Numbers / De Moivre's Theorem help. Will give medal and fan. Question attached below.

Answer 1

Can I have some help with how to do this question, please?

Answer 2

\[1-z^{10} = 1 - e^{i10x} = 1 - (\cos 10x +i \sin 10x)\]
using De Moivres thm
\[= 1 - (\cos 5x +i \sin 5x)^2\]
\[= 1-(\cos^2 5x -\sin^2 5x +2i \sin 5x \cos 5x)\]
\[=2 \sin^2 5x - 2i \sin 5x \cos 5x\]
\[= -2i \sin 5x (i \sin 5x + \cos 5x)\]
\[= -2i \sin 5x e^{i5x}\]

Answer 3

For 1-z^2 , you replace the 10 with a 2
\[1-z^2 = -2i \sin x e^{ix}\]

Answer 4

Oh wow! You made this look so simple :O Thank you soooooo much! It's so much clearer now (Y)

Answer 5

yw glad to help