Determining Path Integral for Function 1/(z-z0)

[mikethemike]

Homework Statement

Notation: C=complex plane, B=ball, abs= absolute value, iff=If and only if


Given z0 in C and r>0, determine the path integral along r=abs(z-z0) of the function 1/(z-zo).



2. The attempt at a solution
It seems to me I'm being asked to find the value of a path integral (a circle) under a function. So using Cauchy's Theorem implies it is zero iff f is holomorphic. This means I must prove the function is holomorphic, which leads me to the question, what is the difference between a holomorphic function, and one that is continious, and how do I go about illustrating this?

Should this function not be continous, I could use the fact that the initial point of the path integral is equal to the final point and come up with an equation based on primatives.

Any help would be appreciated, as I feel I'm just on the cusp of understanding this stuff.

Thanks,

Mike

 

[grief]

In this case, the function is not holomorphic on the interior of the circle, since it's undefined when z=z0. So you just need to evaluate this integral without using Cauchy's Theorem, using the actual definition of a path integral.

 

[Office_Shredder]

It seems to me I'm being asked to find the value of a path integral (a circle) under a function. So using Cauchy's Theorem implies it is zero iff f is holomorphic. This means I must prove the function is holomorphic, which leads me to the question, what is the difference between a holomorphic function, and one that is continious, and how do I go about illustrating this?

A holomorphic function is a function whose complex derivative exists. A continuous function is a function that's just continuous. As an example, the map that takes z to its conjugate is continuous but not holomorphic

 

[HallsofIvy]

Homework Statement

Notation: C=complex plane, B=ball, abs= absolute value, iff=If and only if


Given z0 in C and r>0, determine the path integral along r=abs(z-z0) of the function 1/(z-zo).



2. The attempt at a solution
It seems to me I'm being asked to find the value of a path integral (a circle) under a function. So using Cauchy's Theorem implies it is zero iff f is holomorphic. This means I must prove the function is holomorphic

No, it doesn't- this function is obviously NOT holomorphic (or even continuous) inside the circle; it doesn't even exist at z= z0!

Use the parameterization $z= z_0+ re^{i\theta}$ and actually do the integration.

, which leads me to the question, what is the difference between a holomorphic function, and one that is continious, and how do I go about illustrating this?

Should this function not be continous, I could use the fact that the initial point of the path integral is equal to the final point and come up with an equation based on primatives.

Any help would be appreciated, as I feel I'm just on the cusp of understanding this stuff.

Thanks,

Mike