Hi I'm trying to solve the following complex cubic root equation for $z$ $$ -2az^2z^*+2bz^*-i\frac{c}{2}z=0\qquad a,c\in\Re \text{, } b\in\Im $$ By inspection it is easy to see that one of the roots that solve the equation is $z=0$. To find the other two roots I've been writing $z$ in polar form $z=re^{i\theta}$ and dividing by $z$, so $$ -2ar^2+2be^{-i2\theta}-i\frac{c}{2}=0 $$ From this I find that $r=\pm\Big(\frac{b}{a}e^{-i2\theta}-i\frac{c}{4a}\Big)^{1/2}$
Now I'm unsure what to do next, I would like to find an expression for $\theta$. So I was thinking of asking the question what value of $\theta$ makes $r=0$? I.e. $$ 0=\frac{b}{a}e^{-i2\theta}-i\frac{c}{4a}\Rightarrow-i2\theta=\log(i\frac{c}{b}) $$ I mean normally you would define $\theta$ as $$ \theta=arg (z)=\tan^{-1}(\frac{y}{x}) $$ if $z=x+iy$ when you transform to polar form, but is it correct what I have done thus far?