5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @MhenniBenghorbal thanks for the link and the compliment. $\endgroup$ – Gerardo Cauich Jul 6 '14 at 19:09
3
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

Why should not the value be $0?$

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

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

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.