- #1
stephenkeiths
- 54
- 0
The problem states: Use Morera's Theorem to verify that e^z is indeed analytic inside a circle of radius R.
Morera's Theorem states: If f(z) is continuous in a domain D and if ∫f(z)dz=0 inside any closed curve in D, then f(z) is analytic in z.
I showed that e^z is continuous inside a circle of radius R. Then I showed that ∫f(z)dz=0 over any circle of radius R inside D. I don't know how to show that this applies to ANY closed curve in D. I'm trying to do something with cross cuts. I let C be a circle for which I know the integral is zero. I then introduce cross cuts: I let Co be the curve for which I know the integral is 0. Then I cut it somewhere (call the cut L1) and L1 connects to another curve γ (the curve I'm trying to deform C to). Then there is another curve from γ back to Co. In the limit L1=-L2 so they cancel. And I'm left with Co-γ (gamma goes in the opposite direction of Co). The integral over Co is also 0 since in the limit it is the same as C.
The problem I'm having is showing that the integral over γ is 0.
Morera's Theorem states: If f(z) is continuous in a domain D and if ∫f(z)dz=0 inside any closed curve in D, then f(z) is analytic in z.
I showed that e^z is continuous inside a circle of radius R. Then I showed that ∫f(z)dz=0 over any circle of radius R inside D. I don't know how to show that this applies to ANY closed curve in D. I'm trying to do something with cross cuts. I let C be a circle for which I know the integral is zero. I then introduce cross cuts: I let Co be the curve for which I know the integral is 0. Then I cut it somewhere (call the cut L1) and L1 connects to another curve γ (the curve I'm trying to deform C to). Then there is another curve from γ back to Co. In the limit L1=-L2 so they cancel. And I'm left with Co-γ (gamma goes in the opposite direction of Co). The integral over Co is also 0 since in the limit it is the same as C.
The problem I'm having is showing that the integral over γ is 0.