Determine the domain-sets of $$f(z)=\frac{z\bar{z}}{z^2+\bar{z}^2}$$
Here's my attempt:
Consider z = x+iy,
$z^2+\bar{z}^2 = (x+iy)^2 +(x-iy)^2 = 2x^2$
we know that $z^2+\bar{z}^2 \neq 0$ and so $x \neq 0$
The solution is $\Bbb{C}$ \ {z = x + iy | x = y or x = -y}
I don't really understand the solution or the steps that will lead to it. Any help is appreciated.
You made a slight error:$$\begin{align}z^2+\bar{z}^2=&(x+iy)^2+(x-iy)^2 \\=&(x^2+2ixy+i^2y^2)+(x^2-2ixy+i^2y^2) \\=&2x^2+2i^2y^2 \\=&2(x^2-y^2) \\2(x^2-y^2)\ne & 0 \\x^2\ne&y^2 \\x\ne&\pm y\end{align}$$ so the solution should be: $$\mathbb{C}\backslash\{z=x+iy\,|\,x=\pm y\}$$ you could also word it as: $$\left|\Re(z)\right|\ne\left|\Im(z)\right|$$