Find a so that $f$ is not holomorphic Let $f:\Bbb C$ $\rightarrow \Bbb C,$ $f(z)=z^2+a\overline z^2 + 4z\overline z+2z-8\overline z+1+2i$ Find $ a\in \Bbb C$ such that $f$ is not monogenic in any point. From my understanding I have to find $a$ such that $f \notin \mathcal H(\Bbb C)$. For $f$ to be holomorphic on $\Bbb C$ we have $\frac {\partial f}{\partial \overline z}(z_0)=0 $ for any $z_0 \in \Bbb C$. I have computed the following: $\frac {\partial f}{\partial \overline z}(z)=2a\overline z+4z-8$. But how do I find $ a\in \Bbb C$ from the following: $$2a\overline z+4z-8\not =0 $$ Thanks in advance!

Write $z = x + iy$. We want to find $a$ such that $2a(x - iy) + 4(x + iy) - 8$ is _never_ zero.

Looking separately at the real and imaginary parts, this happens if we never simultaneously have $2ax + 4x - 8 = 0$ and $-2ay + 4y = 0$.

The second equation is easily satisfiable regardless of $a$, by just picking $y = 0$. Can we prevent the first equation from ever holding?

> Yes, by taking $a = -2$, so that the $x$ cancel entirely.