# $\operatorname{Res}(f,-i)$ for $f(z) = \frac{1}{z} + \exp\left(\frac{z-i}{z^2+1} \right)$ at $z = -i$

How do I show that $$z = -i$$ is an essential singularity of $$f(z) = \frac{1}{z} + \exp\left(\frac{z-i}{z^2+1} \right)$$ and find its Residue $$\operatorname{Res}(f,-i)$$?

My idea was to show that the principle part of the Laurent series around $$z = -i$$ of $$f$$ has infinitely many terms but somehow I don't know how to deal with the $$\frac{1}{z}$$.

And of course I do know that \begin{align} \exp\left(\frac{z-i}{z^2+1} \right) = \exp\left(\frac{1}{z+i} \right) = \sum_{k=0}^{\infty} \frac{1}{k! (z+i)^k}. \end{align}

Would anyone be so kind to help me out?

It follows from your computations that the residue at $$-i$$ of $$\exp\left(\frac{z-i}{z^2+1}\right)=1$$. Since the residue at $$-i$$ of $$\frac1z$$ is $$0$$, $$\operatorname{Res}(f,-i)=1$$.
• Maybe I should refine my question. What is the Laurent series of $f$ around $z = -i$? Is it simply $f(z) = \frac{1}{z} + \sum_{k=0}^{\infty}\frac{1}{k!(z+i)^k}$ or do I need to somehow deal with the $\frac{1}{z}$? Apr 25 '20 at 18:15
• No. It's$$i+(z+i)-i (z+i)^2-(z+i)^3+i (z+i)^4+(z+i)^5+\cdots$$It's just the Taylor series of $\frac1z$ centered at $-i$. Apr 25 '20 at 18:19
• So the Laurent series is $\sum_{k=0}^{\infty} i(-i)^k(i+z)^k + \sum_{k=0}^{\infty} \frac{1}{k!(z+i)^k}$ and therefore $-i$ has to be an essential singularity and the Residue is $1$. Correct? Apr 25 '20 at 18:30