If $\alpha\in\mathbb{C}$ show that the complex function $z^{\alpha}$ is holomorphic and that $(z^{\alpha})'=\alpha z^{\alpha-1}$ without chain rule If $\alpha\in\mathbb{C}$ show that the complex function $z^{\alpha}$ is holomorphic and that $$(z^{\alpha})'=\alpha z^{\alpha-1}$$
We don't have the chain rule for complex functions yet.
I showed that it is holomorphic using that it's a composition of holomorphic functions, since $z^{\alpha}=e^{\alpha \log (z)}$.
I know that for a complex function $f(z)=u((x,y))+iv((x,y))$, where $z=x+iy$, we have $f'=u_x+iv_x$ and i was trying to prove that way and i got to $$z^{\alpha}=|z|^{\alpha}(\cos(\arg(z)+i\sin(\arg(z))=(x^2+y^2)^{\frac{\alpha}{2}}\left(\cos\left(\arctan\left(\frac{y}{x}\right)\right)+i\sin\left(\arctan\left(\frac{y}{x}\right)\right)\right)$$
$$=\displaystyle\frac{(x^2+y^2)^{\frac{\alpha}{2}}}{\sqrt{\frac{y^2}{x^2}+1}} + i\frac{y}{x}\displaystyle\frac{(x^2+y^2)^{\frac{\alpha}{2}}}{\sqrt{\frac{y^2}{x^2}+1}}$$
but i couldn't move from here and it feels like the
wrong path, so any help would be appreciated.
 A: Let $\,\alpha\,$ be a complex number $\,\big(\,\alpha\in\mathbb C\,\big)\,$.
The function $\,z^\alpha\,$ is defined for every $\,z\in\mathbb C\setminus\{0\}\,$ in the following way :
$z^\alpha=e^{\alpha\ln z}\quad$ for any $\,z\in\mathbb C\setminus\{0\}\,.$
There are two cases :
$\alpha=0\quad$ or $\quad\alpha\ne0\,.$
If $\,\alpha=0\,,\,$ then $\,z^\alpha=1\;$ for any $\,z\in\mathbb C\setminus\{0\}\,,$
consequently ,
$D_{z=z_0}\,z^\alpha=\lim\limits_{z\to z_0}\dfrac{z^\alpha-z_0^\alpha}{z-z_0}=\lim\limits_{z\to z_0}\dfrac{1-1}{z-z_0}=0=\alpha z_0^{\alpha-1}$
for any $\,z_0\in\mathbb C\setminus\{0\}\,.$
If $\,\alpha\ne0\,,\,$ then
$D_{z=z_0}\,z^\alpha=\lim\limits_{z\to z_0}\dfrac{z^\alpha-z_0^\alpha}{z-z_0}=\lim\limits_{z\to z_0}\dfrac{z_0^\alpha\left[\left(\dfrac z{z_0}\right)^\alpha-1\right]}{z_0\left(\dfrac z{z_0}-1\right)}=$
$\quad=\lim\limits_{z\to z_0}\dfrac{z_0^{\alpha-1}\left[\left(\dfrac z{z_0}\right)^\alpha-1\right]}{\dfrac z{z_0}-1}=$
$\underset{\overbrace{\text{by letting }\,w=\frac z{z_0}-1}}{=}\lim\limits_{w\to0}\dfrac{z_0^{\alpha-1}\left[\left(1+w\right)^\alpha-1\right]}w=$
$\quad=z_0^{\alpha-1}\lim\limits_{w\to0}\dfrac{e^{\alpha\ln(1+w)}-1}w=$
$\quad=z_0^{\alpha-1}\lim\limits_{w\to0}\dfrac{e^{\alpha\ln(1+w)}-1}{\alpha\ln(1+w)}\lim\limits_{w\to0}\dfrac{\alpha\ln(1+w)}w=$
$\underset{\overbrace{\text{by letting }\,u=\alpha\ln(1+w)}}{=}\alpha z_0^{\alpha-1}\lim\limits_{u\to0}\dfrac{e^u-1}u\lim\limits_{w\to0}\dfrac{\ln(1+w)}w=$
$\quad=\alpha z_0^{\alpha-1}\cdot1\cdot1=\alpha z_0^{\alpha-1}$
for any $\,z_0\in\mathbb C\setminus\{0\}\,.$
Hence , in any case , we get that
$D_{z=z_0}\,z^\alpha=\alpha z_0^{\alpha-1}\quad$ for any $\,z_0\in\mathbb C\setminus\{0\}\,.$
