Why is $i$ a removable singularity of $\frac{\sin(z-i)}{z^2+1}$?

Question

Why is $i$ a removable singularity of $f(z)=\frac{\sin(z-i)}{z^2+1}$? We can find the Taylor expansions of the $\sin(z-i)$ and $\frac{1}{z+i}$ to get the Laurent series (actually Taylor series since $i$ is removable) of $f(z)$ for $0<|z-i|<2$, but is there another way to show that $f$ is removable?

Answer 1

In general, a discontinuity of a function is removable if the limit of the function exists at that point. So any method of showing that $$\lim_{z\to i}\frac{\sin(z-i)}{z^2+1}$$ exists will do the trick.

As Mattos notes in a comment, one way to do so is write $z^2+1=(z+i)(z-i)$ and then use the fact that $\lim_{z\to 0}\frac{\sin z}{z}=1$.

Answer 2

Use the fact that$$\lim_{z\to i}\frac{\sin(z-i)}{z^2+1}=\lim_{z\to i}\frac{\sin(z-i)}{z-i}\times\frac1{z+i}=\frac1{2i}$$and Riemann's theorem on removable singularities.