If $$f(z) = \frac{\exp ( iz )}{z^2+1},$$ then how do we observe that the singularities $\pm i$ are simple poles of $f(z)$? I understand that we can express $f(z)$ as $$f(z) = \frac{\exp ( iz )}{(z-i)(z+i)}$$ and observe that the powers of $z-i$ and $z+i$ are $1$. But I don't think this is enough justification for that.
Would someone please clarify this matter?
By observing that $$ \begin{align} \left|\lim_{z \to i}(z-i)\cdot f(z)\right|&=\left|\lim_{z \to i}\frac{\exp\{ iz \}}{(z+i)}\right|=\left|\frac{\exp\{-1 \}}{2i}\right|=\left|-\frac{e^{-1}}2i\right|<\infty, \\\\ \left|\lim_{z \to -i}(z+i)\cdot f(z)\right|&=\left|\lim_{z \to -i}\frac{\exp\{ iz \}}{(z-i)}\right|=\left|\frac{\exp\{+1 \}}{-2i}\right|=\left|\frac{e}2i\right|<\infty \end{align} $$ one deduces the two poles are simple.