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How would I find all of the complex roots to $(x + yi)^5 + 16(x - yi) = 0$?

I am lost as to where to start. Binomial expansion seems like it would take too long. I'm guessing it has to do with conversion to polar form, but without being given the $x$ and $y$ values, how would I find $\theta$?

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    $\begingroup$ How about ($re^{i\theta})^5+ 16re^{-i\theta}=0$? $\endgroup$ – user99680 Dec 8 '13 at 22:50
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Conversion to polar form is a good idea. Right now, you have two unknowns: $x$ and $y$, but writing that equation out to set real and imaginary parts equal gives you terrible fifth degree polynomials. If you convert to polar form, you won't know what $\theta$ is, and you won't know what $r$ (the magnitude of the complex number is), but facts about the polar form will make it much easier to solve for $r$ and $\theta$ than the original $x,y$ equations.

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How about ($re^{i\theta})^5+ 16re^{-i\theta}=r^5e^{i5\theta}+16re^{-i\theta}=0$ ? Then multiply through by $e^{i\theta}....$

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Set $x+iy=ru$, where $r=|x+iy|$. Then, unless $x+iy=0$, which is a solution and that we can exclude from now on, $x-iy=r\bar{u}$ and your equation is $$ r^4u^4+16r\bar{u}=0. $$

Since $|u|=1$ by construction, $\bar{u}=u^{-1}$, so we have (from $r\ne0$),

$$ r^4u^5+16=0 $$

or

$$r^4u^5=-16$$

that gives $r=2$ and $u^5=-1$

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