How do I solve this by hand?

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>>9126673
I know this is the answer but my substitution is not working

>>9126676
you solve it using the residue theorem

>>9126682
Haven't lean't that in my course yet, any other ways?

>>9126685
no, not analytically

>>9126685
Cauchy Integral Theorem. Look it up brainlet, your only singularity is in the [math]z^2[/math] term, so put the rest as your function [math]f[/math]

>>9126701
Yeah I am working on it now. I don't quite understand what to do with the z squared term though

>>9126703
What do you not understand, brainlet? It says here that for all [math]z[/math] such that [math]|z-z_0|<r[/math], we have:

[eqn] f^{(n)}(z) = \frac{n!}{2\pi \mathrm{i}} \oint_{\alpha} \frac{f(\zeta)}{(\zeta-z)^{n+1}} \mathrm{d} \zeta [/eqn]

Where [math]\alpha[/math] is some fucking circle. If we now set [math]z = 0 [/math] and [math]f(\zeta) = \frac{\mathrm{e}^{4\zeta}}{(\zeta +1)^2}[/math]. Then we get:

[eqn] \oint_{\alpha} \frac{f(\zeta)}{(\zeta+0)^2} \mathrm{d} \zeta = \frac{2 \pi \mathrm{i}}{1!} f'(0) [/eqn]

>>9126724
Note here that the last integral is exactly the one you want to calculate.

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>>9126731
Hey thanks man much appreciated, thanks for walking me through that. I'll be able to do it alone next time and I understand it more now.

Have a lovely day :)

>>9126760
That's alright mate, good luck with it.