Complex integral help

**0-)** · March 14th 2009, 09:36 AM

C is the circle $|z|=a$ , 'a' is a positive real number.

$\displaystyle I=\frac{1}{2\pi i}\int_{C} \frac{z^2}{z^2+b} \ dz \ \ \ \ , \$ 'b' is a positive real number.

I'm not sure how to evaluate integrals of the form I (given a and b). It looks like I need to use Cauchy's integral formula but I'm not sure if it is possible to use it here.

Can someone help?

**HallsofIvy** · March 14th 2009, 10:53 AM

Originally Posted by 0-)

C is the circle $|z|=a$ , 'a' is a positive real number.

$\displaystyle I=\frac{1}{2\pi i}\int_{C} \frac{z^2}{z^2+b} \ dz \ \ \ \ , \$ 'b' is a positive real number.

I'm not sure how to evaluate integrals of the form I (given a and b). It looks like I need to use Cauchy's integral formula but I'm not sure if it is possible to use it here.

Can someone help?

|z|= a is a circle about the origin with radius a. $z= ae^{i\theta}$ for any point on that circle, where $\theta$ is the angle the radius to the point z makes with the positive real (x) axis. Use that substitution and integrate as $\theta$ goes from 0 to $2\pi$ .

You will have to consider the cases where $\sqrt{b}< a$ and $\sqrt{b}\ge a$ separately.

**Laurent** · March 14th 2009, 11:22 AM

Originally Posted by 0-)

C is the circle $|z|=a$ , 'a' is a positive real number.

$\displaystyle I=\frac{1}{2\pi i}\int_{C} \frac{z^2}{z^2+b} \ dz \ \ \ \ , \$ 'b' is a positive real number.

I'm not sure how to evaluate integrals of the form I (given a and b). It looks like I need to use Cauchy's integral formula but I'm not sure if it is possible to use it here.

Can someone help?

Cauchy's integral formula deals with integrals like $\frac{1}{2i\pi}\int_C \frac{f(z)}{z-z_0}dz$ , right? This integral equals $f(z_0)$ if $z_0$ is inside the circle $C$ , and 0 if $z_0$ is outside.

You can write $\frac{z^2}{z^2+b^2}=1-\frac{b^2}{z^2+b^2}= 1-\frac{b^2}{(z+i\sqrt{b})(z-i\sqrt{b})}=$ $1-\frac{b^2}{2i\sqrt{b}}\left(\frac{1}{z-i\sqrt{b}}-\frac{1}{z+i\sqrt{b}}\right)$ and use Cauchy's formula for each term.

**0-)** · March 16th 2009, 08:40 AM

Thank you Laurent!

I think your working is slightly wrong, all the $\sqrt{b}$ 's should be $b$ .

So $\displaystyle \frac{z^2}{z^2+b^2} = 1-\frac{b}{2i}\left(\frac{1}{z-ib}-\frac{1}{z+ib}\right)$

From this, I think I am right in saying that if I is the same integral in my previous post but with $b^2$ instead of $b$ then

$|b| \leq |a|\implies I=0$

Can someone confirm this for me?

**Laurent** · March 20th 2009, 09:48 AM

Originally Posted by 0-)

Thank you Laurent!

I think your working is slightly wrong, all the $\sqrt{b}$ 's should be $b$ .

So $\displaystyle \frac{z^2}{z^2+b^2} = 1-\frac{b}{2i}\left(\frac{1}{z-ib}-\frac{1}{z+ib}\right)$

Yes, there're mistakes in my post, but (for me) it was on the left-hand side: it should have been

$\frac{z^2}{z^2+b}=1-\frac{b}{z^2+b}= 1-\frac{b}{(z+i\sqrt{b})(z-i\sqrt{b})}=$ $1-\frac{b}{2i\sqrt{b}}\left(\frac{1}{z-i\sqrt{b}}-\frac{1}{z+i\sqrt{b}}\right)$ .

From this, I think I am right in saying that if I is the same integral in my previous post but with $b^2$ instead of $b$ then

$|b| \leq |a|\implies I=0$

Can someone confirm this for me?

No, this is the contrary. Or you can say that your former integral is 0 if $|b|>|a|^2$ . Indeed, it is zero if the poles $\pm\sqrt{b}$ are outside the circle of integration.

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