Is the function holomorphic? Consider the function $f(z) = z^m(\overline z)^n$
for non-negative integers $m$ and $n$. Is $f$
holomorphic in any open subset of $ \Bbb C$?
I know that for a function to be holomorphic I need to show $\lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}$ exists. So should I consider cases where $m=0, n=0$ and $m=1,n=1$ separately or should I try to work with $f(z)$ using $z=x+iy$ directly?
 A: If $n=0$, we have $f(z)=z^m$ which is trivially holomorphic in $\mathbb C$. So let suppose $n \gt 0$ in what follows.
You know that $g_m(z)=z^m$ is holomorphic in $\mathbb C$. As the ratio of two holomorphic maps is holomorphic if the denominator doesn't vanish, if $f$ was holomorphic in $\mathbb C \setminus \{0 \}$, $z \mapsto \overline{z}^n$ would also be. As it isn't the case for $n \gt 0$, $f$ is not holomorphic in any open subset $U \subseteq \mathbb C \setminus \{0\}$.
Now at $z_0=0$, you have $\vert f(z)/z \vert \le \vert z \vert$ for $\vert z \vert \le 1$ and $m \ge 2$. Hence $f$ is complex differentiable in that case at $0$. And you'll see that for $m \in \{0,1\}$ $f$ is differentiable at $0$ if and only if $n=0$.
Conclusion
$f$ is holomorphic in $\mathbb C$ for $n=0$. Otherwise, for any open subset $U \subseteq \mathbb C$, $f$ is not holomorphic in $U$.
A: I am assuming that the claim you are trying to show is: "$f$ is holomorphic for all $n$ and all $m$ in $\mathbb Z$".
If a function is holomorphic it will satisfy the Cauchy Riemann equations.
Does your $f$ satisfy the Cauchy Riemann equations for all $n$ and all $m$?
Take for example $n=m=1$. If in this case your $f$ does not satisfy the Cauchy Riemann equations then you have shown that the claim is false because if $f$ does not satisfy the Cauchy Riemann equations then $f$ is not holomorphic.
To show $A$ implies $B$ you can show that not $B$ implies not $A$ (logical contraposition).
