Compute $\oint\exp((z-1)^{-1})(z+3)^{-1}\,dz$ I have to calculate this. On the contour : $|z+3| = 7$.

First I want to use the residues formula for this, so I calculated the residue at the simple pole $z=-3$ and got that, if we denote the function under the integral as $f(z)$, $Res(f, -3) = e^{-\frac 1{4}}$.

Now I have to compute the residue at the essential singular point $z = 1.$ For that I want to use Laurent series expansion at $z = 1$, right? Thus, I have:

$$\frac 1{z+3}=\frac 1{z-1}\frac 1{1+\frac 4{z-1}}=\sum_{n\geq0}\frac {(-1)^n4^n}{(z-1)^n}, \space\space\space for\space\space\space |\frac{4}{z-1}|<1.$$


$$e^{\frac 1{z-1}}=\sum_{n\geq0} \frac 1{n!}\frac 1{(z-1)^n}.$$

then $f(z)$ must be $$f(z)= \sum_{n\geq0}\frac {(-1)^n4^n}{(z-1)^n}\sum_{n\geq0} \frac 1{n!}\frac 1{(z-1)^n}$$

But to find the residue at $z = 1$. I only have to find the coefficient of $\frac 1{z-1}$, right? Then I have to multiply all the members that would give me $\frac 1{z-1}$, and if I do that I get $1$. And I don't think that's correct... what's wrong with my thinking?

  • $\begingroup$ What is the path of integration? $\endgroup$ – saulspatz Apr 5 at 13:37
  • $\begingroup$ If this is a contour integral, what is the contour? That's critical, because you must know which singularities it encircles in order to compute the integral. $\endgroup$ – MPW Apr 5 at 13:37
  • $\begingroup$ @MPW Oh definetely, I forgot about that, I will add it right away! $\endgroup$ – C. Cristi Apr 5 at 13:40
  • $\begingroup$ @saulspatz Look at the edit $\endgroup$ – C. Cristi Apr 5 at 13:40
  • $\begingroup$ @saulspatz Why not? I thought that if $z$ was an essential point then I have to do the Laurent series expansion of the function at that point and find the coefficient, what are you saying now? $\endgroup$ – C. Cristi Apr 5 at 13:53

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