There is one really obvious solution to the equation which I ignored because it's lazy $$z=\pm \sqrt{3+2i}$$
When trying to answer the question I tried many different approaches but never got an answer. I tried thinking of a general case where $z=x+iy$ and expanding to see what $z^2$ would be. I then compared the expansion to $3+2i$ and got $x^2-y^2=3$ and $2xy=2$. This yields two quartic equations with ugly roots : $y^4-3y^2+1=0$ and $x^4-3x^2-1=0$. This method does tie into polar coordinates because if $x=cos(\theta)$ and $y=sin(\theta)$ then $2xy=2cos(\theta)sin(\theta)=sin(2\theta)=2$ and $x^2-y^2=cos^2(\theta)-sin^2(\theta)=cos(2\theta)=3$. You could then setup $tan(2\theta)$ and find $r$. I used a different approach though.
To approach this using polar coordinates I did the following: Let $u=z^2$ so that $u=3+2i$. I assumed $u$ is a complex number so that $u=re^{i\theta}$ where $r=\sqrt{3^2+2^2}$ and $\theta=tan^{-1}(\frac{2}{3})$. Thus $u=\sqrt{13}e^{i{tan}^{-1}(\frac{2}{3})}$. Thus $z^2=u=\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))$. And indeed $$\frac{3+2i}{\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))}=1$$Therefore, $z=\sqrt{u}$. This means that$$z=\pm\biggr(\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))\biggr)^{\frac{1}{2}}$$or$$z=\pm\Bigr(\sqrt{13}e^{itan^{-1}(\frac{2}{3})}\Bigr)^\frac{1}{2}$$I didn't apply de Moivre's theorem to the sqrt to simplify it because I learned that it doesn't always work as intended for fractional powers. I really can't see how this could be the answer that they are looking for. Because in the end its just the polar coordinate version of $\pm\sqrt{3+2i}$.
Best Answer
Your quartic equations are not so ugly, because they are really just quadratic equations in $x^2$ and $y^2$ respectively. So for instance $x^4-3x^2-1$ has solutions $$x^2=\frac32\pm\frac12\sqrt{13}$$ And $x$ is real, so we can discard the solution $x^2=\frac32-\frac12\sqrt{13}$. Therefore $$x=\pm\sqrt{\frac32+\frac12\sqrt{13}}$$ and similarly for $y$.
Having said that, I'm not really sure what your question is, so this might or might not be helpful.