
Complex Number Proof
For the complex equation $\displaystyle z^4 = cosx + isinx$, prove that the sum of the four solutions is always zero, no matter what size $\displaystyle x$ is.
This is what I've done so far.
$\displaystyle Z = (cis(x + 360k))^1^/^4
= cis(x/4 + 90k)$
Can someone please teach me how to do the rest? Thanks in advance (Headbang)

Rewrite it as $\displaystyle \displaystyle z^4 = e^{ix}$.
This means that the first fourth root is $\displaystyle \displaystyle z = \left(e^{ix}\right)^{\frac{1}{4}} = e^{i\frac{x}{4}} = \cos{\left(\frac{x}{4}\right)} + i\sin{\left(\frac{x}{4}\right)}$.
The other fourth roots are evenly spaced around a circle, so differ by an angle of $\displaystyle \displaystyle \frac{\pi}{4}$. What are the other solutions? What do you get when you add them together?

A little twist on ProveIt's approach:
Let $\displaystyle w = \cos x + i \sin x$, so the equation can be written
$\displaystyle z^4  w = 0$.
If the roots are $\displaystyle r_1, r_2, r_3, r_4$, then we must have
$\displaystyle z^4  w = (z  r_1) (z  r_2) (z  r_3) (z  r_4) $.
If we expand the righthand side of this equation, what can we say about the coefficient of $\displaystyle z^3$?