All solutions to the equation

Question

Write out in polar form all solutions to the the equation $(x^3 +8)(x^4 −i+1)=0$. I have the answer, but I have no idea how to get to it. Any help is appreciated, thanks.

Answer 1

$(x^3 +8)(x^4-i+1) = 0$ means that either

$x^3 + 8 = 0$ or $x^4 -i + 1 = 0$

which means either

$x^3 = -8$ or $x^4 = - 1 + i$.

In polar coordinates:

$-8 = 8*(-1 + 0i) = 8*(\cos (\pi + 2k \pi) + i\sin (\pi + 2k \pi))= 8e^{(\pi + 2k\pi)i}$.

And $-1 + i = \sqrt 2(-\frac {\sqrt 2}2 + i\frac {\sqrt 2} 2) = \sqrt 2(\cos (\frac {3\pi}4 + 2k\pi) + i\sin (\frac {3\pi}4 + 2k\pi)) = \sqrt 2e^{(\frac {3\pi}4 + 2k\pi)}$.

So if $x^3 = 8e^{(\pi + 2k\pi)i}$ then $x = \sqrt[3]8*e^{\frac {\pi + 2k\pi}3i}$

And if $x^4 = \sqrt 2e^{(\frac {3\pi}r + 2k\pi)}$ then $x = \sqrt[4]{\sqrt 2}e^{\frac {\frac {3\pi}4 + 2k\pi}4i}$.