Solve for $x$ : $(2-3\mathbb i)x^6 + 1 + 5\mathbb i = 0$ $x^6 = \frac{ 1 + 5\mathbb i}{-2+3\mathbb i}$
How do I convert the right hand side into polar form to find 6th roots?
For numerator we have argument $= \arctan(5) + \pi$
And for denominator argument $= \arctan(-3/2) + \pi$
But $\arctan(5)$ and $\arctan(-3/2)$ are not in standard $\arctan$ ratios 

I prefer to put z instead of x in the equation.
any way 
$$Z^6=1-i\\z^6=\sqrt{((-1)^2+(1^2)}.exp(arctan(\frac{-1}{1}))=\\ z^6=\sqrt 2.e^{-\frac{\pi}{4}i}\\
 z^6=\sqrt 2.e^{(-\large\frac{\pi}{4}i+2k\pi)}\\$$ now go to the power of $\frac{1}{6}$ 
$$\sqrt[6]{z^6}=(\sqrt 2.e^{\large(-\frac{\pi}{4}i+2k\pi)})^{\frac{1}{6}}\\z=
\sqrt[12]{2}.e^{\frac{\large(-\frac{\pi}{4}i+2k\pi)}{6}}$$ now put 
$$k=0,1,2,3,4,5$$ to find six root of this equation
$$k=0 \to z_0=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i)}{6}}\\
k=1 \to z_1=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i+2\pi)}{6}}\\
k=2 \to z_2=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i+4\pi)}{6}}\\...\\k=5 \to 
\sqrt[12]{2}.e^{\frac{(\large-\frac{\pi}{4}i+10\pi)}{6}}$$
