<s>[INST] Express $\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i$ in the form of $f(z)$ I need to express $f(z)$ from the form $\color{blue}{u+vi}$ to the form $\color{blue}z$ for example if: $g(z)=\frac{1}{x+yi}$ so $ =g(z)=\frac{1}{z}$

$$f(z)=\underbrace {\frac{1-x}{(x-1)^2+y^2}}_{=u(x,y)}+\underbrace {\bigg(-\frac{y}{(x-1)^2+y^2}\bigg)}_{=v(x,y)}i$$

My try:
$$f(z)=\frac{1-x-yi}{(x-1)^2+y^2}$$
$$=\frac{1-\overbrace{x-yi}^{=\bar z}}{\underbrace{x^2+y^2}_{=|z|^2}-2x+1^2}$$
$$-\frac{1-\bar z}{|z|^2-2x+1}$$

I'm stuck here

 [/INST] Hint ( to prove the answer of @RonGordon):
$$
f(z)=\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i=
$$
$$
=\frac{1}{(1-x)^2+y^2}\left(1-x-i y \right) = \frac{\bar u}{|u|^2}=\frac{1}{u}
$$
for $u=1-x +iy=1-(x-iy)=1-\bar z$
</s>