Solving integral using residue theorem

**4Math** · November 11th 2010, 04:05 AM

I have the following integral:

$\displaystyle\int_{0}^{2\pi}\frac{1}{(4sinx+3i)^{2 }}= \left [{\begin{array}{ll}z=e^{ix} \,\,\\\frac{dz}{iz}=dx\end{array}\right]=$

since:

$\displaystyle sin x=\frac{1}{2i}(z-z^{-1})$

I rewrite the integral and get the following:

$\displaystyle=\int_{0}^{2\pi}\frac{1}{((\frac{2z^2-2}{iz})+3i)^{2}}\frac{dz}{iz}=$

$\displaystyle=\int_{0}^{2\pi}\frac{1}{(\frac{2z^2-2-3z}{iz})^{2}}\frac{dz}{iz}=$

$\displaystyle=\int_{0}^{2\pi}\frac{-z^{2}}{(2z^2-2-3z)^{2}}\frac{dz}{iz}=$

$\displaystyle=\int_{0}^{2\pi}\frac{-z}{(2z^2-2-3z)^{2}}\frac{dz}{i}$

How do I get rid off the minus in the numerator in the last integral? Is this correct so far? I need some help and suggestions to continue the solution.

When I solve for $z$ I get the following:

$\displaystyle{z_{1}=2$

$\displaystyle{z_{2}=-\frac{1}{2}$

I don't know if that is correct.

Thank you

**FernandoRevilla** · November 11th 2010, 04:35 AM

You should write $\int_{|z|=1}\ldots dz$ instead of $\int_0^{2\pi}\ldots dz$ . Now find the residue for the pole $z_2,\;\;(|z_2|<1)$

Regards.

---
Fernando Revilla

**4Math** · November 11th 2010, 01:19 PM

Originally Posted by FernandoRevilla

You should write $\int_{|z|=1}\ldots dz$ instead of $\int_0^{2\pi}\ldots dz$ . Now find the residue for the pole $z_2,\;\;(|z_2|<1)$

Regards.

---
Fernando Revilla

I am trying to find the residue using the following:
$\displaystyle\mathrm{Res}_{z \to z_2} \frac{f_1(z)}{f_2(z)}=\frac{f_1(z_2)}{f{'}_2(z_2)}$

In other words I just derive the denominator in the last integral in my previous post. But the answer I got is incorrect. Is there any other way finding the residue?

Kind regards

**FernandoRevilla** · November 11th 2010, 09:32 PM

The residue at $x_2=-1/2$ is:

$\begin{array}{rcl} \textrm{Res}[f,-1/2]= \\ \displaystyle\lim_{z \to -1/2}{\dfrac{1}{i}\cdot}\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)'= \\ -\dfrac{1}{4i}\displaystyle\lim_{z \to -1/2}{\left(\dfrac{z}{(z-2)^2}\right)'= \\\ldots\;= \\ -\dfrac{3}{125i} \end{array} $

Then,

$I=2\pi i\textrm{Res}[f,-1/2]=-\dfrac{6\pi}{125}$

Regards.

Fernando Revilla

**4Math** · November 12th 2010, 09:56 AM

Originally Posted by FernandoRevilla

$\displaystyle \left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)$

How does the following:

$\displaystyle\frac{-z}{(2z^2-2-3z)^{2}}$

becomes:

$\displaystyle\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right) $

Thank you for your response.

Kind regards

**FernandoRevilla** · November 12th 2010, 01:30 PM

Originally Posted by 4Math

How does the following:

$\displaystyle\frac{-z}{(2z^2-2-3z)^{2}}$

becomes:

$\displaystyle\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right) $

Using two theorems:

(i) If $z_1,\ldots,z_n$ are the roots of the polinomyal of degree $n$ :

$f(z)=a_nz^n+\ldots+a_1z+a_0\in{\mathbb{C}}[z]$

then:

$f(z)=a_n(z-z_1)\cdot\ldots\cdot(z-z_n)$

(ii) If $z_0$ is a pole of $f$ with multiplicity $k$ , then:

$\textrm{Res}[f,z_0]=\dfrac{1}{(k-1)!}\displaystyle\lim_{z \to z_0}\dfrac{d^{k-1}}{dz^{k-1}}{\left(f(z)(z-z_0\right)^k})} $

Regards.

**4Math** · November 13th 2010, 01:42 PM

I follow the two applied theorems that you are refering to.

But I still can't understand how you factor:

$\displaystyle{(2z^2-2-3z)^{2}$

to the following:

${4(z-2)^2(z+(1/2))^2}$

Kind regards

**FernandoRevilla** · November 13th 2010, 11:00 PM

Using (i) :

$(2z^2-2-3z)^{2}=[\;2(z-2)(z+(1/2))\;]^2=4(z-2)^2(z+(1/2))^2$

Regards.

**4Math** · November 14th 2010, 01:15 AM

Originally Posted by FernandoRevilla

Using (i) :

$(2z^2-2-3z)^{2}=[\;2(z-2)(z+(1/2))\;]^2=4(z-2)^2(z+(1/2))^2$

Regards.

I can now clearly see the equality of these two expressions. I have to blame it that it was late night when I looked at it last time.

Thank you very much indeed.

Kind regards

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