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By using cauchy's formula for derivative and for any contour $c$ such that:

$z=1$ inside $c$ and $z=\pm 2i$ out side $c$ in positive side

evalute the integral

$\int_c {{e^z}\over {{(z-1)}^2({z}^2+4)}} dz$

My try:

let $f(z)={e^z \over {z^2+4}}$

$f'(z)={{e^z(z^2+4)-e^z(2z)} \over (z^2+4)^2}$

$f'(1)={{3e}\over 25}$

$\int_c {{f(z)}\over {{(z-a)}^{n+1}}} dz={{2\pi i} \over {n!}} f^{(n)}(a)$

$\int_c {{e^z}\over {{(z-1)}^2({z}^2+4)}} dz={2\pi i ({{3e}\over 25})}$

$={{6\pi i e}\over 25}$

true?

"sorry I don't speak english well :("

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    Did you mean $\int_c \frac{e^z} {{(z-1)}^2{({z}^2+4)}} dz$? – 2017-01-16
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    I think you should use \frac{}{} – 2017-01-16
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    Slowing down on the unnecessary { and } might (also) help you to avoid the currently mistyped formulas... – 2017-01-16
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    sorry, I edited it, Thanks all. – 2017-01-16
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    @AndresMejia yes :) – 2017-01-16
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    @emka Thanks, I edited it :) – 2017-01-16
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    @did Thanks, I edited it :) – 2017-01-16
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    Now the odd thing with your solution is that you seem to prove that the integral is $2\pi i ({e\over 5})$ AND that it is also $0$... Make up your mind. – 2017-01-16
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    oh, you are right :$) , ummm what about know ? @Did – 2017-01-16
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    Sorry but why $$\int_c {{e^z}\over {{(z-1)}^2({z}^2+4)}}dz=\int_c {{{e^z}\over ({z}^2+4)}\over {{(z-1)}^2}} dz+ \int_c {{{e^z}\over {(z-1)}^2}\over {({z}^2+4)}}dz$$ would hold? This is saying that the integral $I$ to be computed is such that $$I=I+I$$ which seems absurd since you arrive at $I\ne0$. – 2017-01-17
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    @Did I edited my answer, what you think now :) , Thanks. – 2017-01-17

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