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 $0$, $\operatorname{Res}(f,-i)=1$.

  • $\begingroup$ 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 $\frac{1}{z}$? $\endgroup$
    – offline
    Apr 25 '20 at 18:15
  • $\begingroup$ 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 $-i$. $\endgroup$ Apr 25 '20 at 18:19
  • $\begingroup$ 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? $\endgroup$
    – offline
    Apr 25 '20 at 18:30
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ Apr 25 '20 at 18:43

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