How to compute $\int_C {e^{3z}-z\over (z+1)^2z^2}$?

Question

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?

There are easier techniques to find the residue instead of deriving Laurent series. Good job by the way. — Mhenni Benghorbal, Jul 06 '14 at 19:03
They're either trying to fool you, or they are themselves wrong. Your result and your way to the result are both correct. — Daniel Fischer, Jul 06 '14 at 19:07
Note: $$\frac{e^{3z}-z}{(z+1)^2} = \frac{1+2z + O(z^2)}{1+2z + O(z^2)} = (1+2z + O(z^2))(1-2z + O(z^2)) = 1 + O(z^2).$$ — Daniel Fischer, Jul 06 '14 at 19:09
@DanielFischer Sadly, I don't understand what you just wrote, but I guess it has something to do with complexity. — Gerardo Cauich, Jul 06 '14 at 19:14
Should I leave the question posted, delete it or choose an answer? — Gerardo Cauich, Jul 06 '14 at 19:16
@Gerardo Definitely don't delete it. (You can't delete it now anyway). There's no rush in accepting an answer either, but you should do it eventually. — Git Gud, Jul 06 '14 at 19:17
That way you can quickly see that the residue of $\frac{e^{3z}-z}{(z+1)^2z^2}$ in $0$ is zero. You can ignore anything with a power $z^2$ or higher, only the constant and linear terms of $e^{3z}-z$ and $(z+1)^2$ are relevant. — Daniel Fischer, Jul 06 '14 at 19:17
@DanielFischer I think Gerardo just doesn't understand (know) the notation. — Git Gud, Jul 06 '14 at 19:18
@DanielFischer I will look into the subject, thanks for sharing. — Gerardo Cauich, Jul 06 '14 at 19:29

score 3 · Accepted Answer · answered Jul 06 '14 at 19:14

3

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

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

answered Jul 06 '14 at 19:14

Git Gud

31,356

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. – Gerardo Cauich Jul 06 '14 at 19:17
@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. – Git Gud Jul 06 '14 at 19:25

score 0 · Answer 2 · answered Jul 06 '14 at 19:07

0

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.

answered Jul 06 '14 at 19:07

Marm

3,371
1
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I thought the partial fractions algebra would be too time-consuming. But I'll consider it next time. Thanks. – Gerardo Cauich Jul 06 '14 at 19:10

How to compute $\int_C {e^{3z}-z\over (z+1)^2z^2}$?

2 Answers2