Integrating a Multiply Connected Domain Using Cauchy's Theorem

[FanofAFan]

Homework Statement

Let P(z) = (z-z₁)^M₁(z-z₂)^M₂...(z-z_n)^M_n
Let $$\varsigma$$ be a positively directed simple closed curve surrounding all of the z_k, k = 1, ... n verify directly that $$\int$$_{$$\varsigma$$} ($$\frac{P'(z)}{P(z)}$$dz = 2$$\pi$$i(M₁+...+M_n)

Homework Equations

The Attempt at a Solution

did product rule, this is an extension of the Cauchy Gorsat Theorem for multiply connected domains and evaluation of integral of 1/(z-z_k) dz where $$\varsigma$$_k

 

[Dick]

Well, ok. It asked you to verify the equation. And if you used the integral of 1/(z-zk) is 2*pi*i then you did it correctly if you got that answer. What's the question? If you don't know how to get that answer then why not? What's P'(z)/P(z)? Use the product rule.

 

[FanofAFan]

First, is the P'(z) = M₁(z-z₁)^M₁-1M₂(z-z₁)^M₂-1... etc?

Second, is the intergral of 1/(z-z_k) just ln(z-z_k)? I just not sure how you got 2pi*i?

 

[Dick]

First, is the P'(z) = M₁(z-z₁)^M₁-1M₂(z-z₁)^M₂-1... etc?

Second, is the intergral of 1/(z-z_k) just ln(z-z_k)? I just not sure how you got 2pi*i?

No, that's not P'(z). You didn't use the product rule, now did you? Furthermore the integral of 1/(z-zk) around a counterclockwise contour enclosing zk is 2*pi*i. Isn't that Cauchy's integral theorem?

 

[FanofAFan]

P'(z) = M₁(z-z₁)^M₁-1(z-z₂)^M₂...(z-z_n)^M_n +M₂(z-z₁)^M₁(z-z₂)^M₂-1...(z-z_n)^M_n +...+ M_n(z-z₁)^M₁(z-z₂)^M₂...(z-z_n)^M_n-1

 

[Dick]

P'(z) = M₁(z-z₁)^M₁-1(z-z₂)^M₂...(z-z_n)^M_n +M₂(z-z₁)^M₁(z-z₂)^M₂-1...(z-z_n)^M_n +...+ M_n(z-z₁)^M₁(z-z₂)^M₂...(z-z_n)^M_n-1

Right! Now divide by P(z). It's a lot harder to write all that out than to say what the answer is.

 

[FanofAFan]

M₁(z-z₁)^M₁-1 + M₂(z-z₂)^M₂-1 +...+M_n(z-z_n)^M_n-1

and the integral of that is just (z-z₁)^M₁ + (z-z₂)^M₂ .. etc?

 

[Dick]

M₁(z-z₁)^M₁-1 + M₂(z-z₂)^M₂-1 +...+M_n(z-z_n)^M_n-1

and the integral of that is just (z-z₁) + (z-z₂).. etc?

Wrong! Isn't it M1/(z-z1)+M2/(z-z2)...+Mn/(z-zn)?

 

[FanofAFan]

Wrong! Isn't it M1/(z-z1)+M2/(z-z2)...+Mn/(z-zn)?

wait the integral is M1/(z-z1)+M2/(z-z2)...+Mn/(z-zn)...

 

[Dick]

wait the integral is M1/(z-z1)+M2/(z-z2)...+Mn/(z-zn)...

If you can get behind me and believe that's what P'(z)/P(z) is then yes, that's the integrand. Tell me again why that is what P'(z)/P(z) is.

 

[FanofAFan]

so i got the integral of M₁(z-z)^M₁-1/(z-z₁)^M₁ + ... + etc

which by my attempt in wolfram is M₁/(z-z₁)
http://www.wolframalpha.com/input/?i=y%28x-1%29^%28y-1%29+%2F%28x-1%29^y

 

[Dick]

so i got the integral of M₁(z-z)^M₁-1/(z-z₁)^M₁ + ... + etc

which by my attempt in wolfram is M₁/(z-z₁)
http://www.wolframalpha.com/input/?i=y%28x-1%29^%28y-1%29+%2F%28x-1%29^y

Ok, so you want the integral of M1/(z-z1)+...+Mn/(z-zn) around a contour surrounding all of the zi, right? Integrate it term by term using Cauchy.