# Find two complex numbers which satisfy the equation $\frac{3+2i}{z^2}=1$

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 $$\pm\sqrt{3+2i}$$.

• Your quartic for $y$ should be $y^4+3y^2-1=0$ with real solutions of $y =\pm \sqrt{\frac{\sqrt{13}}2-\frac32}$ Commented Oct 2, 2021 at 17:27

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 $$y$$.