Integral of g(z) around |z-i|=2: Cauchy Integral Formula

[doubleaxel195]

Homework Statement

Find the value of the integral of g(z) around the circle |z-i|=2 in the positive sense when $$g(z)=\frac{1}{z^2+4}$$. Answer: pi\2

Homework Equations

Cauchy Integral Formula
$$f(z_0)=\frac{1}{2\Pi i}\int \frac{f(z) dz}{z-z_0}$$

The Attempt at a Solution

I tried factoring the denominator by (z-2i)(z+2i), but then f is not analytic when z=+/-2i. So I'm not really sure what to do. I'm not even sure if the Cauchy Integral formula is the right way to go since it evaluates a function at a specific point, not the integral.

 

[MaxManus]

Do you know residu? or (parametrization) integration and invariance theorem?

 

[doubleaxel195]

No, I don't. Well, I know how to parametrize and integrate with respect to a real variable. The section before this homework had to do with deformation of path. The section this homework came from deals with Cauchy's Integral formula, Cauchy's Inequality, and a consequence of Cauchy's Integral formula:

$$\int \frac{f(z)dz}{(z-z_0)^{n+1}}=\frac{2\Pi i}{n!} f^{(n)}(z_0)$$

But each theorem requires the function to be analytic on and within the contour, but it's not analytic on the circle.

 

[MaxManus]

I don't remember cauchy, but

$$g(z)=\frac{1}{z^2+4} = \frac{1}{(z+2i)(z-2i)} = \frac{A}{z+2i} + \frac{B}{z-2i}$$
where B = -i/4
The first integral has no singularities inside the circle so its integral is zero.

The second part. Move the circle so that it has center in 2i. Now parametrizate z = 2i + 2e^it

$$\int \frac{B}{z-2i}dz = \int_0^{2\p}i \frac{B}{2e^{it}} 2ie^{it} dt = \int_0^{2\pi} Bidt = 2\pi B i = \frac{\pi}{2}$$
edit: I used wrong A and B
A =
B = 1/4i = -i/4

 

[gomunkul51]

As I see it, you have to solve the closed loop integral of 1/(z²+4) using the Cauchy Integral Formula (:=CIF).

This is what you need to understand about the CIF:
f(z): has to be analytic inside the closed loop of integration (i.e. to have no singularities inside the loop of integration).
z₀: is a single singular point (obviously).

Now, using the constraints above you have to take your function g(z) and basically divide it into a f(z) and a 1/z-z₀ part.
Then to get the value of the integral, as the CIF says, you just plug in z₀ inside the f(z) function you constructed before (don't forget to multiply by 2*pi*i).
** remember that in complex integration only singular points inside the loop of integration contribute.

is that clear? :)

 

[gomunkul51]

@MaxManus:
The way you solved it is correct, but you made an error in calculating A and B. They should be A = i/4; B = -i/4.
You can check it by plugging in the numbers and verifying you get the original expression back.

 

[doubleaxel195]

I just realized I kept thinking that +/- 2i was on the circle for some reason but it's not! (wow!) I got it now. Thank you for all your help though. But, gomunkul51, are you sure the CIF works for functions which are not analytic on the curve itself? I thought that the hypothesis was that a function must be analytic within and on a curve?

 

[gomunkul51]

gomunkul51, are you sure the CIF works for functions which are not analytic on the curve itself? I thought that the hypothesis was that a function must be analytic within and on a curve?

Your specific problem didn't have singularities on the curve, so it didn't matter. When I repeatedly wrote "inside" I meant that you should not take into account other singularities which lay outside the loop. I didn't trying to deal with singularities on the loop.

To the point: I'm not sure whether the function should be analytic on the loop as well. This is a question for your teacher/textbook/google :)

 

[tianaxoxo]

how to do make thread on this site

 

[MaxManus]

Press "new topic" here: https://www.physicsforums.com/forumdisplay.php?f=156 if it is math related