Let $\alpha$ be a complex number. Show that if $(1+z)^{\alpha}$ is taken as $e^{\alpha Log(1+z)}$, then for $|z|<1$ Let $\alpha$ be a complex number. Show that if $(1+z)^{\alpha}$ is taken as $e^{\alpha Log(1+z)}$, then for $|z|<1$ 
$(1+z)^\alpha=1+\frac{1}{\alpha}+\frac{\alpha (\alpha -1)}{1\cdot2}z^2+\frac{\alpha( \alpha-1)(\alpha-2)}{1\cdot2\cdot3}z^3+...$
I want to find Taylor series of $e^{\alpha Log(1+z)}$.
I know $\frac{d^jLog(z)}{dz^j}=(-1)^{j+1}(j-1)!z^{-j}$
So, $(e^{\alpha Log(1+z)})^{(n)}=(e^{\alpha Log(1+z)})*(\alpha Log(1+z))^{(n)}=(e^{\alpha Log(1+z)})*\alpha (-1)^{j+1}(j-1)!(1+z)^{-j}$ 
Let $z=0$, it is euqal to $\alpha(-1)^{j+1} (j-1)!$
. The coefficient of jth term is $\frac{f^{n}(0)}{j!}=\frac{\alpha (-1)^{j+1}}{j}$.   I cannot get the binomial coefficient.
 A: You are trying to deduce the binomial series. I like to derive it as follows:
Let $\alpha\in\mathbb{C}$. We want to find the Taylor series for $$(1+z)^\alpha:=e^{\alpha Log(1+z)},\ z\in\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}.$$ Here $Log$ denotes the principal branch of the complex logarithm (which is holomorphic in $\mathbb{C}\setminus\mathbb{R}_{\leq0}$). Denote
$$
f_\alpha(z)=(1+z)^\alpha,\ z\in\mathbb{D}.
$$
This function is holomorphic (in $\mathbb{D}$) so it has a Taylor series around $0$. To compute it notice that
$$
f_\alpha'(z)=\frac{\alpha}{1+z}f_\alpha(z)=\alpha f_{\alpha-1}(z),
$$
from where we deduce
$$
f^{(k)}_\alpha(z)=\alpha(\alpha-1)\cdots(\alpha-k+1)f_{\alpha-k}(z)
$$
for $k\geq1$. Evaluating at $z=0$ we obtain $f^{(k)}_\alpha(0)=\alpha(\alpha-1)\cdots(\alpha-k+1)$, and so
$$
f_\alpha(z)=(1+z)^\alpha=\sum_{k=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}z^k,\ \ z\in\mathbb{D}.
$$
It is customary to denote this coefficients by $\left(\begin{array}{c}\alpha\\ k\end{array}\right)$, so
$$
(1+z)^\alpha=\sum_{k=0}^\infty\left(\begin{array}{c}\alpha\\ k\end{array}\right)z^k
$$
accordingly with the $\alpha\in\mathbb{N}$-case.
