Laurent series expansion of $f(z)=\frac{1}{1-2^{1-z}}$ in the annulus $D_r=\{\ z\in\mathbb C\ |\ r<|z-z_0|<R\ \}$

Question

How do you compute the Laurent Series of $f(z)=\frac{1}{1-2^{1-z}}$ in the neighborhood of the point $z_0=1$ converging inside the annulus $D_r=\{\ z\in\mathbb C\ |\ r<|z-z_0|<R\ \}$? Elementary complex analysis therefore states that: $$f(z)=\frac{1}{1-2^{1-z}}=\sum \limits_{n=-\infty}^{\infty}a_n(z-z_0)^n$$ Where the coefficients are given by: $$a_n=\frac{1}{2\pi i}\int_{\gamma_\rho}\frac{f(z)}{(z-z_0)^{n+1}}dz$$ The contour of integration is given by $\gamma_{\rho}:|z-z_0|=\rho$.

Answer 1

$f(z)=\frac{1}{1-2^{1-z}}$,so $f(z)=1+\left(\frac{1}{2^{z-1}-1}\right)=1+\left(\frac{1}{e^{(z-1)\log2}-1}\right)=1+\sum_{n=1}^\infty \left(\frac{1}{(z-1)^n(\log2)^n}\right)$.
$0<|z-1|<R$,for any $R\gt0$.