Complex number equality

Question

To Prove: $\displaystyle (\cos\theta +i\sin\theta)^4(\sin\theta-i\cos\theta)=\cos 8\theta+i\sin 8\theta$

My Attempt:

$\displaystyle (\cos4\theta +i\sin4\theta)(\sin\theta-i\cos\theta)=\sin5\theta+i\cos5\theta$

Now what ?

Answer 1

You have $$(\cos\theta+i\sin\theta)^4 = e^{4i\theta}$$ and $$(\sin\theta-i\cos\theta) = -i e^{i\theta}$$ hence: $$(\cos\theta+i\sin\theta)^4 (\sin\theta-i\cos\theta) = -i e^{5i\theta}$$ that is clearly different from $$\cos 8\theta + i\sin 8\theta = e^{8i\theta}$$ unless $e^{3i\theta}=-i=e^{3i\frac{\pi}{2}}$, i.e. $\theta=\frac{\pi}{2}+\frac{2k\pi}{3}.$

Answer 2

Hint: Note that $\cos{(\theta-\frac{\pi}{2})}=\sin{\theta}$ and $\sin{(\theta-\frac{\pi}{2})}=-\cos{\theta}$. Therefore,

$$\sin{\theta}-i\cos{\theta}=\cos{(\theta-\frac{\pi}{2})}+i\sin{(\theta-\frac{\pi}{2})}=e^{i(\theta-\frac{\pi}{2})}.$$

So we have:

$$(\cos\theta +i\sin\theta)^4(\sin\theta-i\cos\theta)=e^{4i\theta}e^{i(\theta-\frac{\pi}{2})}.$$

Answer 3

The two expressions are clearly not equivalent.

Take $\theta=0$ to get $-i$ on the left but $1$ on the right.

You must have a typo in the statement of the problem.