How does the function $f(z)=\dfrac{\sin z}{z}$ have isolated singularity at $z=0$? I want to ask how $f(z)=\dfrac{\sin z}{z}$ has a isolated  singularity at $z=0$? please someone tell.

A function $f$ has an isolated singularity at $z=a$ if there is an $R>0$ such that $f$ is defined and analytic in $B(a;r)\setminus \{a\}$ but not in $B(a;R)$. [Conway, pg. 103]
What values of $z \in \mathbb{C}$ make your function non-analytic?
