Determining the form of the Complex valued function

Exercise:

Let $z=x+iy$ and $f\left( z \right)=1-\frac{2{{y}^{2}}-2xyi}{{{x}^{2}}+{{y}^{2}}}$

Determine the form of $f\left( z \right)$

Solution: we have \(f\left( z \right)=1-\frac{2{{y}^{2}}-i2xy}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}-2{{y}^{2}}+i2xy}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}-{{y}^{2}}+i2xy}{{{x}^{2}}+{{y}^{2}}}\)

\(=\frac{{{x}^{2}}+{{\left( iy \right)}^{2}}+i2xy}{{{x}^{2}}+{{y}^{2}}}=\frac{{{\left( x+iy \right)}^{2}}}{{{x}^{2}}-{{\left( iy \right)}^{2}}}=\frac{{{\left( x+iy \right)}^{2}}}{\left( x-iy \right)\left( x+iy \right)}=\frac{x+iy}{x-iy}=\frac{z}{{\bar{z}}}\) thus $f\left( z \right)=\frac{z}{{\bar{z}}}$