Find $\oint\exp((z-1)^{-1})(z+3)^{-1}\,dz$

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:

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

And:

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

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?

• What is the path of integration? – saulspatz Apr 5 at 13:37
• 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. – MPW Apr 5 at 13:37
• @MPW Oh definetely, I forgot about that, I will add it right away! – C. Cristi Apr 5 at 13:40
• @saulspatz Look at the edit – C. Cristi Apr 5 at 13:40
• @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? – C. Cristi Apr 5 at 13:53