Im(z) and Re(z) of $\bar z^2+ \frac {1}{z^2}$ Given:
$\bar z^2+ \frac {1}{z^2}$, $z \in \mathbb{C}, z \neq 0$
Task:
Determine Im(z) and Re(z).
So my approach:
$$\bar z^2+ \frac {1}{z^2} = \bar z^2+ \frac {\bar z^2}{(z \bar z)^2} = \frac {\bar z^2(z \bar z)^2}{(z \bar z)^2}+\frac {\bar z^2}{(z \bar z)^2} = \frac {\bar z^2 + \bar z(z \bar z)^2}{(z \bar z)^2}$$
Since z is a complex number it´s of the shape $z = x + iy$, but if I insert this definition, the hole calculation becomes a huges mess. Does anyone see an easier way to solve this problem?
 A: Let's assume $z=re^{i\varphi}$. We have then
$$ \bar{z}^2 + \frac{1}{z^2} = e^{-2i\varphi} (r^2+ \frac{1}{r^2})$$
Knwing $\bar{z}^2 + \frac{1}{z^2}$ you can find its module and argument, and from them you can calculate $r$ and $\varphi$ respectively. Then you have ${\rm Im}(z) = r\sin\varphi$ and ${\rm Re}(z) = r\cos\varphi$.
A: You are in the right track for finding the Im and Re of $\bar z^2+ \frac {1}{z^2}$. Just note that $z\bar z=x^2+y^2\in \mathbb R$ and this simplifies your calculations hugely.
A: You are given $\bar z^2 +1/z^2$ and need to find $z$. Write $z$ as $\rho(\cos\theta+i\sin\theta)$, $\rho$ is the length, $\theta$ is the argument, both to be found. Then $\bar z=\rho(\cos\theta-i\sin\theta)$, $\bar z^2=\rho^2(\cos2\theta -i\sin2\theta)$,
$$\bar z^2+1/z^2=\rho^2(\cos2\theta -i\sin2\theta)+1/(\rho^2)(\cos(2\theta)-i\sin(2\theta)))=(\rho^2+1/\rho^2)(\cos2\theta-i\sin2\theta).$$ So the length of $\bar z^2 +1/z^2$ is $\rho^2+1/\rho^2$ from which you find $\rho$ and the argument of  $\bar z^2 +1/z^2$ is $-2\theta$ which gives $\theta$. This determines $z$.
