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

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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marked as duplicate by Jonas Meyer, JimmyK4542, Bookend, Mike Miller, Micah Dec 28 '14 at 4:26

This question was marked as an exact duplicate of an existing question.

2  
what is the equation? – Nameless Nov 11 '12 at 19:58
    
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 $-\frac{3\sqrt{2}}2\pm i\frac{3\sqrt{2}}2$.

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