= Solution
The <Taylor series> of the complex sine is
$$
\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots.
$$
Therefore, for $z\ne0$,
$$
\frac{\sin z}{z}
=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots,
$$
and the right side defines a <holomorphic function> at $z=0$ with value $1$. Thus the apparent singularity is a <removable singularity>.
The resulting power series is
$$
\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}.
$$
The <ratio test> shows convergence for every $z\in\mathbb C$, so its <radius of convergence> is
$$
\boxed{\infty}.
$$
Back to article page
