Problem: Let $f$ be a rational function of a complex variable $z$ given by
\[
f(z) = \frac{z^3 + 2z -4}{z}.
\]
The radius of convergence of the Taylor series of $f$ at $z = 1$ is
Solution: Note that the above function can be written as
\begin{align*}
f(z) & = \frac{z^3 + 2z - 4}{z} \\
& = z^2 + 2 - \frac{4}{z} \\
& = (z-1)^2 +2z + 1 + 2 - \frac{4}{(z - 1) + 1} \\
& = (z - 1)^2 + 2(z - 1) + 5 - \frac{4}{(z - 1) + 1} .
\end{align*}
Therefore, the radius of convergence will be the same as for the $\frac{4}{(z - 1) + 1}$.
\begin{align*}
\frac{4}{(z - 1) + 1} = 4 \sum_{n=0}^{\infty} (-1)^n(z-1)^n.
\end{align*}
Thus, the radius of convergence will be $1$.