What is the principal part of the Laurent series expansion of $f(z)= \frac{2}{z^5 (1-2z)}$ at the origin?
Attempt: Rewrite the function as $\frac{2}{z^5}\cdot \frac{1}{1-2z}$. The Laurent series expansion of $\frac{1}{1-2z}$ is given by:
\begin{align*} \frac{1}{1-2z} &= 1 + 2z + 4z^2 + 8z^3 + \ldots \ &= \sum_{n=0}^{\infty} 2^n z^n \end{align*}
So the Laurent series expansion of $f(z)$ is:
$$f(z) = \frac{2}{z^5}\left(\sum_{n=0}^{\infty} 2^n z^n\right) = \sum_{n=0}^{\infty} 2^{n+1} z^{n-5}$$
Hence, principal part is
$$2 z^{-5} + 4 z^{-4} + 8 z^{-3} + 16 z^{-2} + 32 z^{-1}$$
Is this correct?