Integrating complex numbers

**synclastica_86** · September 5th 2009, 01:28 PM

I was asked to evaluate the following:
$\oint _C \frac{dz}{z^2-1}$
where C is the circle $\left | z \right |=2$
Well I figure that is the function $f(z)= \frac{1}{z^2-1}$ is analytical then according to Cauchy's intergral theorem:
$\oint _C f(z)=0$
This is somewhat complicated since it's difficult to separate the function into the real and imaginary part to take their derivatives. I pluged it into the calculator and it seems like it's not analytical. So, I went about trying to do it.
$z=r e^{it}$ and $dz=ire^{it}dt$
then it becomes:
$\oint _C \frac{dz}{z^2-1}=\int_0^{2\pi} \frac{ire^{it}dt}{r^2 e^{2it}-1}=\frac{i}{r}\int_0^{2\pi} \frac{e^{it}dt}{e^{2it}-\frac{1}{r^2}}$
let:
$u=e^{2it}-\frac{1}{r^2}$ and $dv=e^{it}dt$
then:
$du=2ie^{2it}$ and $v=\frac{e^{it}}{i}$
And using by parts:
$=(e^{2it}-\frac{1}{r^2})\frac{e^{it}}{i}-\int\frac{e^{it}}{i}2ie^{2it}dt$
$=\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\int e^{3it}dt$
$=\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\frac{e^{3it}}{3i}$
$=\frac{3e^{3it}-2e^{3it}-\frac{3e^{it}}{r^2}}{3i}$
since $\left | z \right |=2$ r=2, and we have:
$=\frac{e^{3it}-\frac{3e^{it}}{4}}{3i}$
Now pluging in the limits $2\pi$ and 0 we have:
$=\frac{1-\frac{3}{4}}{3i}-\frac{1-\frac{3}{4}}{3i}=0$
How can this be? It shouldn't be 0 since it's not analytical. What am I missing here?

**mr fantastic** · September 5th 2009, 02:11 PM

Originally Posted by synclastica_86

I was asked to evaluate the following:
$\oint _C \frac{dz}{z^2-1}$
where C is the circle $\left | z \right |=2$
Well I figure that is the function $f(z)= \frac{1}{z^2-1}$ is analytical then according to Cauchy's intergral theorem:
$\oint _C f(z)=0$
This is somewhat complicated since it's difficult to separate the function into the real and imaginary part to take their derivatives. I pluged it into the calculator and it seems like it's not analytical. So, I went about trying to do it.
$z=r e^{it}$ and $dz=ire^{it}dt$
then it becomes:
$\oint _C \frac{dz}{z^2-1}=\int_0^{2\pi} \frac{ire^{it}dt}{r^2 e^{2it}-1}=\frac{i}{r}\int_0^{2\pi} \frac{e^{it}dt}{e^{2it}-\frac{1}{r^2}}$
let:
$u=e^{2it}-\frac{1}{r^2}$ and $dv=e^{it}dt$
then:
$du=2ie^{2it}$ and $v=\frac{e^{it}}{i}$
And using by parts:
$=(e^{2it}-\frac{1}{r^2})\frac{e^{it}}{i}-\int\frac{e^{it}}{i}2ie^{2it}dt$
$=\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\int e^{3it}dt$
$=\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\frac{e^{3it}}{3i}$
$=\frac{3e^{3it}-2e^{3it}-\frac{3e^{it}}{r^2}}{3i}$
since $\left | z \right |=2$ r=2, and we have:
$=\frac{e^{3it}-\frac{3e^{it}}{4}}{3i}$
Now pluging in the limits $2\pi$ and 0 we have:
$=\frac{1-\frac{3}{4}}{3i}-\frac{1-\frac{3}{4}}{3i}=0$
How can this be? It shouldn't be 0 since it's not analytical. What am I missing here?

Just because f(z) is not analytic at all points in the region enclosed by the contour $C$ doesn't mean that $\oint_{C} f(z) \, dz$ can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

$\oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0$

where $C_1$ and $C_2$ are circles inside $C$ centred at $z = -1$ and $z = 1$ .

**synclastica_86** · September 5th 2009, 09:05 PM

Originally Posted by mr fantastic

Just because f(z) is not analytic at all points in the region enclosed by the contour $C$ doesn't mean that $\oint_{C} f(z) \, dz$ can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

$\oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0$

where $C_1$ and $C_2$ are circles inside $C$ centred at $z = -1$ and $z = 1$ .

Forgive me, my math background isn't that solid. I'm a physics major and tend t o learn math on need to know basis, but what do you mean by dumbell contour? Why does:
$\oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz$
How do know to redefine the new paths?

**mr fantastic** · September 5th 2009, 09:28 PM

Originally Posted by synclastica_86

Forgive me, my math background isn't that solid. I'm a physics major and tend t o learn math on need to know basis, but what do you mean by dumbell contour? Why does:
$\oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz$
How do know to redefine the new paths?

http://www.mathhelpforum.com/math-he...tegration.html might help.

**synclastica_86** · September 9th 2009, 08:56 AM

Originally Posted by mr fantastic

Just because f(z) is not analytic at all points in the region enclosed by the contour $C$ doesn't mean that $\oint_{C} f(z) \, dz$ can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

$\oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0$

where $C_1$ and $C_2$ are circles inside $C$ centred at $z = -1$ and $z = 1$ .

Ok, so did you use Cauchy's integral theorem here? and assume $f(z)=1\frac{1}{z-1}$ for $C_1$
and
$f(z)=1\frac{1}{z+1}$ for $C_2$
But doesn't this required $f(z)$ to be analytical?

**synclastica_86** · September 9th 2009, 09:14 AM

I worked on it a bit more. Please tell me if I am on the right track.
Using partial fraction I constructed 2 circles around z=1 and -1.
$\oint_C\frac{dz}{z^2-1}=\oint_{C_1}\frac{dz}{2(z-1)}-\oint_{C_2}\frac{dz}{2(z+1)}$ where $C_1$ is the circle around z=1 and $C_2$ is a circle around z=-1
Letting $f(z)=\frac{1}{2}$ and using the Cauchy's Integral Formula $\oint_C\frac{f(z)dz}{z-z_0}=2\pi if(z_0)$
Here, f(z) must be analytical because all the partial derivatives are indeed 0. Is this right?
This gives:
$\oint_{C_1}\frac{dz}{2(z-1)}-\oint_{C_2}\frac{dz}{2(z+1)}=\frac{2\pi i}{2}-\frac{2\pi i}{2}=0$
Is this correct?

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