How to solve $(-19w + 93\overline w)^4=-1$ , if $w\in \mathbb C$

I really have no direction where to solve this question or at least a hint, can someone help?

  • $\begingroup$ Looks like your best shot is breaking it up and writing out expressions for the real and imaginary parts. $\endgroup$ – Adam Hughes Oct 29 '14 at 0:46
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    $\begingroup$ I would go to the other extreme. The four fourth roots of $-1$ are not hard to find. Take each one, call one $r_1,$ then solve $-19 w + 93 \bar{w} = r_1,$ with $w = x + i y$ and $x,y$ real. $\endgroup$ – Will Jagy Oct 29 '14 at 0:56
  • $\begingroup$ Is $\overline{w}$= $|w|$? $\endgroup$ – Joao Oct 29 '14 at 0:58
  • $\begingroup$ @Joao $\bar{w}$ is $x-iy$, where $w = x+iy$ and $w, y, \in \mathbb{R}$ $\endgroup$ – MCT Oct 29 '14 at 1:05
  • $\begingroup$ @Soke you mean conjugate? thanks $\endgroup$ – Joao Oct 29 '14 at 1:08

HINT: Let $z=-19w+93\bar{w}$. Then $z=\dfrac{1+i}{\sqrt2}$ or $\dfrac{-1+i}{\sqrt2}$, or $\dfrac{-1-i}{\sqrt2}$, or $\dfrac{1-i}{\sqrt2}$ (roots of $-1$). Let $w=a+bi$. Then in the first case $$ -19a-19bi+93a-93bi=\dfrac{1+i}{\sqrt2} $$ and the rest should be obvious.

  • $\begingroup$ you need to divide your roots by $\sqrt 2$ $\endgroup$ – Will Jagy Oct 29 '14 at 0:59
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    $\begingroup$ @WillJagy Thanks! Corrected. $\endgroup$ – Przemysław Scherwentke Oct 29 '14 at 1:04

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