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I'm trying to find the residue of $$f(z) = \frac{4}{(z-2)(e^z-e^2)} + \frac{z+4}{z^4 + 2^4} $$ in $$z=2$$

I know that the second part of the function doesn't affect the residue, what I don't understand is which formula to apply, since for both $(z-2)$ and $(e^z-e^2)$, $z=2$ is a simple pole, but when evaluating one or the other I get that the residue is equal to $\infty$.

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They both affect the residue, hence, we must multiply by $(z-2)^2$ and take the derivative:

$$\text{Residue}=\frac d{dz}\frac{4(z-2)^2}{(z-2)(e^z-e^2)}\bigg|_{z=2}=4\frac d{dz}\frac{z-2}{e^z-e^2}\bigg|_{z=2}=\dots$$

Can you take it from here?

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    Perhaps I'm doing something wrong, but I get $$4\cdot\frac{-e^2 + 3e^z - ze^z}{(e^z - e^2)^2}$$ so I still get $\infty$ if I substitute $z=2$... – 2017-02-26
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    @R.Gallo Check your work on that one. You should get $\frac00$, which should tell you something... – 2017-02-26
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    The answer in advance is $-\frac2{e^2}$ by the way. – 2017-02-26
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    Thank you. I feel so stupid for not thinking about De l'Hopital theorem before. $$4\cdot\frac{e^z(2-z)}{2e^z(e^z-e^2)} \rightarrow 2\cdot\frac{-1}{e^z}$$ which then becomes $$-2e^{-2}$$ Thank you! – 2017-02-26
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    :-) No problem. – 2017-02-26
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    @R.Gallo It is generally recommended that you accept answers that are the most helpful to you. So it would be better if you would accept this answer, seeing as you thanked him. – 2017-02-26
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    You're right, sorry, I was in a hurry and forgot yesterday -- accepted it now. – 2017-02-27