Determine the singularities in $f(z)=\sin(\frac3{2-z})$ and thus find the radius of convergence of the Maclaurin Series for $f(z)$.
So in this case $z=2$ is a singularity.
Also, I'm not entirely sure as to how this ties in with the Maclaurin Series.
Any help is appreciated.
Since $f$ is holomorphic in $D = \{z \in \mathbb{C}: 0<|z+2|\}$ you can represent it using a Laurent series.
The Taylor series for $\sin(z)$ is \begin{equation} \sin(z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} \end{equation} Therefore \begin{equation} \sin\left(\frac{3}{z-2}\right) = \frac{3}{z-2} - \frac{1}{3!} \left(\frac{3}{z-2}\right)^3 + \frac{1}{5!} \left(\frac{3}{z-2}\right)^5 - \ldots \end{equation} Since the principal part of the Laurent series has an infinite number of terms, $z_0 = 2$ is an essential singularity of $f$.
Moreover, let $G = \{z\in \mathbb{C}: |z| < R\}$ be an open disk. Since a Maclaurin series which converges in $G$ is holomorphic in $G$ we obtain that $G$ cannot contain $z_0 = 2$ and therefore the radius of convergence of the Maclaurin series is 2.