$Problem: $ Prove that $f(z) = \frac{1}{z(1 − z)}$ is (complex) differentiable infinitely often in $\mathbb{C}\setminus {\{0, 1}\}$. Find an expression for $f^{(n)}(z)$ for all $n ≥ 0$.
$Idea:$ Note that:
$$f(z)=\frac{1}{z(1 − z)}=\frac{1}{z}+\frac{1}{1 − z}$$ and
Let $g(z)=\frac{1}{z}$, $h(z)=\frac{1}{1 − z}$ be then
If $z=x+iy\in\mathbb{C}\setminus {\{0, 1}\}$, $g(x+iy)=\frac{1}{x+iy}=\frac{x-iy}{x^2 + y^2}$. Here, $u(x,y)=\frac{x}{x^2 + y^2}$, $v(x,y)=\frac{-y}{x^2 + y^2}$. And the Cauchy-Riemann equations are fulfilled: $u_x=\frac{y^2-x^2}{(x^2 +y^2)^2}$, $u_y=\frac{-2xy}{(x^2 +y^2)^2}$, $v_x=\frac{2xy}{(x^2 +y^2)^2}$, $v_y=\frac{y^2-x^2}{(x^2 +y^2)^2}$. Hence $g$ is holomorphic in $\mathbb{C}\setminus {\{0, 1}\}$ But how do I prove that it is (complex) infinitely differentiable often in $ \mathbb {C} \setminus {\{0, 1}\} $?
Furthermore
$h(z)=\frac{1}{1 − z}=\sum_0^\infty z^n$ then $(\frac{1}{1 − z})'=\sum_1^\infty nz^{n-1}$. Hence $$(\frac{1}{1 − z})^{(k)}=\sum_{n=k}^\infty n(n-1)\cdot...\cdot(n-k+1)z^k$$
How do I prove what $h$ is (complex) infinitely differentiable often in $ \mathbb {C} \setminus {\{0, 1}\} $? Is my idea correct?
Observe that $f(z)=\frac{1}{z(1-z)}$ is holomorphic on $\mathbb{C}-$ $\{$ $0,1$ $\}$ as 1 is holomorphic on the latter set and so is $z(1-z)$ therefore dividing the two functions yields a holomorphic map on $\mathbb{C}-$ $\{$ $0,1$ $\}$. By Cauchys differentiation formula, the $nth$ derivative always exists on the above set. Note that Cauchy's differentiation formula gives you the desired formula.