Show that $z^n-\frac{1}{z^n}=2i\sin n\theta$

Question

In this problem $z=\cos \theta +i\sin \theta$.

Using De Moivres Theorem and exponent laws I substitute $z^n$ for $\cos \theta +i\sin \theta$ and then expand a bit to get

$$\cos n\theta +i\sin n\theta -\cos (-n\theta) +i\sin(-n\theta) $$

How do I simplify further? Which laws can I use to help me?

Answer 1

I'm not sure how you got that. With $z= cos(\theta)+ isin(\theta)$, using "DeMoivre", $z^n= cos(n\theta)+ isin(n\theta)$ and $z^{-n}= cos(-n\theta)+ isin(-n\theta)= cos(n\theta)- isin(n\theta)$. $z^n- z^{-n}= (cos(n\theta)+ isin(n\theta))- (cos(n\theta)- isin(n\theta))= 2i sin(n\theta)$

(Remember that cosine is an "even function" and sine is an "odd function": $cos(-\theta)= cos(\theta)$ and $sin(-\theta)= -sin(\theta)$).

Answer 2

$z^n-\frac{1}{z^n}=cos(n\theta)+isin(n\theta)-\frac{1}{cos(n\theta)+isin(n\theta)}$. Now just do the usual trick-multiply and divide $\frac{1}{cos(n\theta)+isin(n\theta)}$ by the conjugate of the denominator and you will get what you need.

Answer 3

Using De-Movire's

$z^n=\cos{n\theta}+i\sin{n \theta}$

$z^{-n}=\cos{-n\theta}+i\sin{-n \theta}=\cos{n\theta}-i\sin{n \theta}$

$z^n-\frac{1}{z^{n}}=z^{n}-z^{-n}=2i\sin{n\theta}$

Answer 4

Hint:

• $\cos(-\theta)=\cos(\theta)$
• $\sin(-\theta)=-\sin(\theta)$

Also, you made a little mistake, it should be $$\cos n\theta +i\sin n\theta -\cos (-n\theta) \color{red}{-}i\sin(-n\theta)$$