# Complex Number Proof

• Apr 10th 2011, 03:00 AM
womatama
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.

$\displaystyle Z = (cis(x + 360k))^1^/^4 = cis(x/4 + 90k)$

• Apr 10th 2011, 03:11 AM
Prove It
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?
• Apr 10th 2011, 05:16 AM
awkward
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 right-hand side of this equation, what can we say about the coefficient of $\displaystyle z^3$?