Calculating Contour Integrals with Cauchy Theorem on Annulus/Donut Boundaries

[mrjohns]

A complex analysis question.

Homework Statement

Verify the Cauchy theorem by calculating the contour integrals.

Where ω is the appropriately orientated boundary of the annulus/donut defined by 1/3 ≤ IzI ≤ 2 for the following analytic functions:

i. f(z)=z^2
ii. f(z)=1/z

Homework Equations

closed ∫(ω) f(z)dz=0

closed ∫(ω) 1/(z-a) dz = 0 (a outside ω), 2∏i (a inside ω)

The Attempt at a Solution

The first bit that is confusing me is how ω can be considered a closed contour when it includes both the inner and outer boundaries of the annulus, which aren't connected. I'm not sure how it can be considered one closed contour and not two.

My plan of attack for the first one would be to just take the closed contour of the big circle, then subtract the little circle. Obviously both would be zero. But I'm not sure if this is the correct way to do things.

The second one obviously trying the same strategy would cause problems due to the discontiuity at z=0. So can I 'deform' the contours of each circle to exclude that central point? In that case I would get 2∏i - 2∏i = 0.

Is this the correct strategy or do I need to connect the outer and inner boundaries with a line?

 

[pasmith]

A complex analysis question.

Homework Statement

Verify the Cauchy theorem by calculating the contour integrals.

Where ω is the appropriately orientated boundary of the annulus/donut defined by 1/3 ≤ IzI ≤ 2 for the following analytic functions:

i. f(z)=z^2
ii. f(z)=1/z

Homework Equations

closed ∫(ω) f(z)dz=0

closed ∫(ω) 1/(z-a) dz = 0 (a outside ω), 2∏i (a inside ω)

The Attempt at a Solution

The first bit that is confusing me is how ω can be considered a closed contour when it includes both the inner and outer boundaries of the annulus, which aren't connected. I'm not sure how it can be considered one closed contour and not two.

My plan of attack for the first one would be to just take the closed contour of the big circle, then subtract the little circle. Obviously both would be zero. But I'm not sure if this is the correct way to do things.

The second one obviously trying the same strategy would cause problems due to the discontiuity at z=0. So can I 'deform' the contours of each circle to exclude that central point? In that case I would get 2∏i - 2∏i = 0.

Is this the correct strategy or do I need to connect the outer and inner boundaries with a line?

Yes, you need to connect the outer and inner boundaries with a line. But this is equivalent to taking the integral over the outer boundary and subtracting the integral over the inner boundary, because the contribution from the line cancels (it's traversed once in each direction).

 

[mrjohns]

Thanks. Does this mean I should really do it the first way you've said (rather than subtracting) as this would satisfy the lecturer better - or does it not really matter.

And if I do it that way, does that mean the f(z)=1/z case won't cause any issues anywhere?

 

[pasmith]

Thanks. Does this mean I should really do it the first way you've said (rather than subtracting) as this would satisfy the lecturer better - or does it not really matter.

And if I do it that way, does that mean the f(z)=1/z case won't cause any issues anywhere?

The question does tell you to "verify Cauchy's theorem", so you have actually to calculate all the path integrals and show that their sum is the expected result.

 

[mrjohns]

Ok excellent, thanks for the help, will post if I have any issues.