Complex Number Proof

**womatama** · April 10th 2011, 03:00 AM

For the complex equation $z^4 = cosx + isinx$ , prove that the sum of the four solutions is always zero, no matter what size $x$ is.

This is what I've done so far.

$Z = (cis(x + 360k))^1^/^4<br /> = cis(x/4 + 90k)$

Can someone please teach me how to do the rest? Thanks in advance

**Prove It** · April 10th 2011, 03:11 AM

Rewrite it as $\displaystyle z^4 = e^{ix}$ .

This means that the first fourth root is $\displaystyle z = \left(e^{ix}\right)^{\frac{1}{4}} = e^{i\frac{x}{4}} = \cos{\left(\frac{x}{4}\right)} + i\sin{\left(\frac{x}{4}\right)}$ .

The other fourth roots are evenly spaced around a circle, so differ by an angle of $\displaystyle \frac{\pi}{4}$ . What are the other solutions? What do you get when you add them together?

**awkward** · April 10th 2011, 05:16 AM

A little twist on ProveIt's approach:

Let $w = \cos x + i \sin x$ , so the equation can be written
$z^4 - w = 0$ .

If the roots are $r_1, r_2, r_3, r_4$ , then we must have
$z^4 - w = (z - r_1) (z - r_2) (z - r_3) (z - r_4)$ .

If we expand the right-hand side of this equation, what can we say about the coefficient of $z^3$ ?

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