# Finding all the solutions to a complex equation

I am asked to find all the solutions to $z^{42}=-1$. I go a head and square root both sides to produce $z^{21}= i$. Then I can write $z^{21}= r^{21} (\cos(21\ θ) + i\sin(21\ θ)) = 0+i.$ Hence $\cos(21\ θ)= 0$ and $\sin(21\ θ)=1$, as $r^{21}$ can't be equal to $0$. Therefore $21θ$ = ${π \over 2}$ and $θ = {π\over42}$. Then using $\sin(21\ θ)=1$, I can write $r^{21}=1$ and produce that $r\pm1$ $\$so my answer would be $z=\pm \cos({π\over42}) + i\sin({π\over42}).$

My answer doesn't produce the cleanest numbers so I'm a little worried but I do believe the process is correct.

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This equation has 42 solutions. – Chris Godsil Jun 3 '13 at 11:35
Try expressing the solutions in polar coordinates. – Tim Jun 3 '13 at 11:40
hehe Thanks, worked from a too basic example question. – xiA Jun 3 '13 at 11:45
"I go ahead and square root both sides to produce $z^{21} = i$" - don't do this! – Cocopuffs Jun 3 '13 at 12:14
If $z^{42}=-1$ then $z^{21}=\pm i$ – Andrea Mori Jun 3 '13 at 12:16

You could just come to my consulting hour tomorrow :-) Hint: no need to square root. We did an example like this in class.

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Hint: where $n$ is an integer, $$z^{42}=-1$$$$(re^{i\theta})^{42}=1e^{i \pi }$$ $$r^{42}e^{i 42\theta}=1e^{i(\pi+ 2 \pi n)}$$

Note $r\ge 0$ (by definition), and that $e^{i2\pi n}=1$.

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