Find an analytic function $w=u+iv$ such that $u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0$

Question

Find an analytic function $w=u+iv$ such that $$ u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0 $$

This is the assignment under b.) and on a.) the answer is given to use the substitutions: $$x=\frac{z+\overline{z}}{2}; y=\frac{z-\overline{z}}{2i}.$$ the answer here is suppose to be: $\frac{z}{1-z^2}$ and just cannot seem to get this. When simplifying the expression with these substitutions I get:

$$\frac{1}{2}\frac{(z+\overline{z})(1-z\overline{z})}{1-(z^2+{\overline{z}}^2)+\frac{(z^2+{\overline{z}}^2)^2}{4}-\frac{(z+\overline{z})^2(z-\overline{z})^2}{4}}$$

Answer 1

Simplifying the denominator gives $$ u = \frac{(z + \bar{z})(z\bar{z}-1)}{2(z^2-1)(\bar{z}^2-1)} $$ You can then separate the denominator to get $$ u = \frac{1}{2}\left(\frac{z}{z^2-1} + \frac{\bar{z}}{\bar{z}^2-1}\right) $$ This is clearly the real part of the analytic function $z/(z^2-1)$, as desired.