Let $z \in \Bbb{C}\setminus \{0\}$ and let $$f(z)=z+\frac{1}{z}-\left(\frac{\overline{1+z}}{z}\right)$$Prove that $$f(z)=z-1$$
My attempt
$$f(z)=z+\frac{1}{z}-\left(\frac{\overline{1+z}}{z}\right)$$ $$f(z)=\frac{z^2+1}{z}-\frac{1-z}{z}$$ $$f(z)=\frac{z^2+1-1+z}{z}$$ $$f(z)=\frac{z^2+z}{z}$$ $$f(z)=\frac{z(z+1)}{z}$$ $$f(z)=z+1$$ But $$z+1\neq z-1$$ Am I wrong or what the problem is asking me to prove can't be proven?
If conjugation is applied only to the numerator, you can rewrite the function as $$ f(z)=z+\frac{1}{z}-\frac{1}{z}-\frac{\bar{z}}{z} $$ which equals $z-1$ only if $z=\bar{z}$.
It's false even if you apply conjugation to the whole second term, so that it would become $$ f(z)=z+\frac{1}{z}-\frac{1}{\bar{z}}-1 $$ and this is not equal to $z-1$ unless $z=\bar{z}$.