# How to solve complex equation, all values?

$(-81)^{1/4}$

The answer, I got was -3,3..but when I input it in the webwork, it does not work. What is the method to solve this?

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what is the equation? –  Nameless Nov 11 '12 at 19:58
Hi, Saraphine. Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  amWhy Nov 11 '12 at 20:23
What is $3^4$; similarly, what is $(-3)^4$? Each is equal to $81$, not $(-81)$. No need to use webwork to see that your "answers" do not work. –  amWhy Nov 11 '12 at 20:26

I'll assume you mean that you want to find all complex fourth roots of $-81$--that is, all $z\in\Bbb C$ such that $z^4=-81$, or all complex roots of the polynomial $z^4+81$.
Before we do that, there are a few important observations. It's a fourth-degree polynomial, so there are at most $4$ complex roots. If there's some $\zeta$ such that $\zeta^4=1$ (a fourth root of unity), and $z^4+81=0$, then $(z\zeta)^4+81=z^4\zeta^4+81=z^4+81=0$. Thus, noting that the solutions to $\zeta^4=1$ are $1,i,-1,-i$ (or $i^k$ for $k=0,1,2,3$)--then we need only find some $z$ such that $z^4+81=0$, and at least (and so exactly) four solutions will be $zi^k$ ($k=0,1,2,3$).
Our job will be easier if we write in polar form $z=r(\cos\theta+i\sin\theta)$. DeMoivre's formula gives us $-81=z^4=r^4(\cos 4\theta+i\sin 4\theta)$. Taking the modulus of both sides yields $81=r^4\sqrt{\cos^24\theta+\sin^24\theta}=r^4,$ so we'll need $r=3$. We'll need $\sin 4\theta=0$, too, so $4\theta$ is an integer multiple of $\pi$. We may as well take $4\theta=\pi$, so that $z=3(\cos\frac{\pi}4+i\sin\frac{\pi}4)=\frac{3\sqrt{2}}2+i\frac{3\sqrt{2}}2.$
Thus, our solutions will be $\frac{3\sqrt{2}}2\pm i\frac{3\sqrt{2}}2$ and $-\frac{3\sqrt{2}}2\pm i\frac{3\sqrt{2}}2$.