I am asked to compute the integral $$ \int_C {e^{3z}-z\over (z+1)^2z^2} $$ where $C$ is a circle with the center at the origin and radius ${1 \over 2}$.

My approach was to separate the integral as a differentiation of 2 contour integrals:

$$ \int_C {e^{3z}-z\over (z+1)^2z^2} = \int_C {e^{3z}\over (z+1)^2z^2} - \int_C {1\over (z+1)^2z} $$

Then I calculated the residue of each contour integral with a Laurent series around $z_0 = 0$:

$$ {e^{3z}\over (z+1)^2z^2} = {1\over (z+1^2)}\ .\ e^{3z}\ .\ {1\over z} $$

$$ {e^{3z}\over (z+1)^2z^2} = \sum_{n=0}^\infty {3^nz^{n-2}\over n!}\ .\ (1-2z+3z^2+...) $$

$$ {e^{3z}\over (z+1)^2z^2} = {a_{-2}\over z^2}+{-2+3\over z}+a_0+... $$

So the residue for this contour integral is $1$ and the final result is $2\pi i$

I did the same with the other countour integral:

$$ {1\over (z+1)^2z} = {1\over z}\ .\ (1-2z+3z^2+...) $$

$$ {1\over (z+1)^2z} = {1\over z}-2+3z^2+... $$

So the residue for this contour integral is also $1$ and the final result is $2\pi i$

Then I substitute my results in the original contour integral:

$$ \int_C {e^{3z}-z\over (z+1)^2z^2} = 2\pi i - 2\pi i $$

And this is where my problem is (I get zero), can someone point to me what I did wrong?

  • $\begingroup$ There are easier techniques to find the residue instead of deriving Laurent series. Good job by the way. $\endgroup$ – Mhenni Benghorbal Jul 6 '14 at 19:03
  • $\begingroup$ Why do you think you did something wrong? $\endgroup$ – Daniel Fischer Jul 6 '14 at 19:03
  • $\begingroup$ The final answer is indeed zero. $\endgroup$ – lemon Jul 6 '14 at 19:04
  • $\begingroup$ @DanielFischer because I was told by my classmates... $\endgroup$ – Gerardo Cauich Jul 6 '14 at 19:05
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    $\begingroup$ @MhenniBenghorbal thanks for the link and the compliment. $\endgroup$ – Gerardo Cauich Jul 6 '14 at 19:09

Alternatively use Cauchy's differentiation formula on the function $f\colon \mathbb C\setminus\{0\}\to \mathbb C, z\mapsto \dfrac{e^{3z}-z}{(z+1)^2}$ which gives

$$\int _C \dfrac{e^{3z}-z}{z^2(z+1)^2}\mathrm dz=2\pi if'(0)=2\pi i\left[\dfrac{z+e^{3z}(3z+1)-1}{(z+1)^3}\right]_{z=0}=0.$$

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  • $\begingroup$ I also thought it would be too time-consuming but it actually seems faster than my approach. I will take it into account next time, thanks. $\endgroup$ – Gerardo Cauich Jul 6 '14 at 19:17
  • $\begingroup$ @Gerardo It's as time consuming at it is differentiating $f$. Annoying to do it, but not one of the worse functions that can appear. $\endgroup$ – Git Gud Jul 6 '14 at 19:25

Why should not the value be $0?$

But an easier way to solve this is the usage of the following:


Then the integral of the 2 last addend will be zero (Cauchy's integral theorem)

And the other 2 integrals can be easily computed with Cauchy's integral formula.

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  • $\begingroup$ I thought the partial fractions algebra would be too time-consuming. But I'll consider it next time. Thanks. $\endgroup$ – Gerardo Cauich Jul 6 '14 at 19:10

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