Show that (z*)^n is complex differentiable using the differential quotient So, I am trying to show for which $z\in\mathbb{C}$ the function $f:\mathbb{C}\rightarrow\mathbb{C},z\mapsto\bar{z}^n; n\geq 2$ is complex differentiable.
So far I found that 
$\lim\limits_{z\rightarrow 0}\frac{\overline{z_0+z}^n-\bar{z}_0^n}{z}=\lim\limits_{z\rightarrow 0}\frac{1}{z}\sum\limits_{k=0}^{n-1}\binom{n}{k}\bar{z}_0^k\bar{z}^{n-k}=\lim\limits_{z\rightarrow 0}(\frac{\bar{z}^n}{z}+n\frac{\bar{z_0}\bar{z}^{n-1}}{z}+\cdots+n\frac{\bar{z_0}^{n-1}\bar{z}}{z})$.
Now, to show whether this limit exists I want to show that the limit $\lim\limits_{z\rightarrow 0}\frac{\bar{z}^n}{z}$ exists/ does not exist for $n\geq2$.
Any hints? Note that I must use the differential quotient for the assignment. I already tried showing that the latter limit does not exist by using the sequences $x_n=1/n,y_n=i/n$, however $\lim\limits_{m\rightarrow\infty}\frac{\bar{x}_m^n}{x_m}=\lim\limits_{m\rightarrow\infty}\Big(\frac{1}{m}\Big)^{n-1}=0=\lim\limits_{m\rightarrow\infty}(-i)^n\Big(\frac{1}{m}\Big)^{n-1}=\lim\limits_{m\rightarrow\infty}\frac{\bar{y}_m^n}{y_m}$ i.e. at least along the real and imaginary axes the limit exists.
I am out of ideas for now.
 A: I have figured it out now. 
Proposition: $f$ is not complex differentiable in $\mathbb{C}\setminus\{0\}$ but in in $0\in\mathbb{C}$.
Proof:
We need to show that the limit $\lim\limits_{z\rightarrow 0}\frac{\overline{z_0+z}^n-\bar{z}_0^n}{z}=\lim\limits_{z\rightarrow 0}(\frac{\bar{z}^n}{z}+n\frac{\bar{z_0}\bar{z}^{n-1}}{z}+\cdots+n\frac{\bar{z_0}^{n-1}\bar{z}}{z})$ does not exist for $z_0\neq0$.
This limit does not exist. Take the last term $\lim\limits_{z\rightarrow 0}n\frac{\bar{z_0}^{n-1}\bar{z}}{z}=n\bar{z_0}^{n-1}\lim\limits_{z\rightarrow 0}\frac{\bar{z}}{z}$ in the sum above. The latter limit exists, iff the limit $\lim\limits_{m\rightarrow\infty}\frac{\overline{z_m}}{z_m}$ exists for all sequences $z_m\subset\mathbb{C}$ with $\lim\limits_{m\rightarrow\infty}z_m=0$. We choose two sequences $x_m=1/m$ and $y_m=i/m$ and see that $\lim\limits_{m\rightarrow\infty}\frac{\overline{x_m}}{x_m}=1\neq-1=\lim\limits_{m\rightarrow\infty}\frac{\overline{y_m}}{y_m}$. Thus, the limit $\lim\limits_{z\rightarrow 0}\frac{\overline{z}}{z}$ does not exist and therefore neither does $\lim\limits_{z\rightarrow 0}\frac{\overline{z_0+z}^n-\bar{z}_0^n}{z}$.
However, if $z_0=0$ we find that $\lim\limits_{z\rightarrow 0}\frac{\overline{z_0+z}^n-\bar{z}_0^n}{z}=\lim\limits_{z\rightarrow 0}(\frac{\bar{z}^n}{z}+n\frac{\bar{z_0}\bar{z}^{n-1}}{z}+\cdots+n\frac{\bar{z_0}^{n-1}\bar{z}}{z})=\lim\limits_{z\rightarrow0}\frac{\overline{z}^n}{z}=0$, i.e. $f$ is complex differentiable in $0\in\mathbb{C}$, q.e.d..
