- #1
oddiseas
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1. I am having trouble grasping cauchys theorm. To my understanding the integral for any closed contour in which f(z) is path connected and analytic =0.
Example 1:
Now the integral around the unit circle for 1/z. In the unit circle 1/z is anlaytic for all z, but not path connected because when z=0 1/z is not defined. Therefore the integral does not equal zero and is equal to 2πi. I am confused about the following.
1) is my reasoning correct?
2) the book says that this integral is equal to 2πi. However when i integrate using γ(t)=e^it i get the right answer. When i apply the fundamental theorm and use the antiderivative of 1/z = Lnz and evaluate from 0 to 2π. I get Ln(1)+2πi+Kπi-Ln(1)-0πi+πKi. Now since o and 2π give the same argument doesn't this equal zero?
3) if indeed the integral of any analytic function in a path connected set is zero. Applying this reasoning to f(z) =1/z^2. With the unit circle. F(z) is analytic in the unit circle but not path conneceted because it is not defined at z=0. To my understanding it should thus not equal zero. But it Does equal zero when i compute it. These apparent contradictions are doing my head in so if anyone understands this i would like some help.
My concern is : I can solve these things no problem. I know that for 1/z^2+1 i can factorise the denominator and integrate in two regions where z+i, and z-i are analytic, but the above two examples seem to contradict my reasoning. So now i am confused and i am worried because if i am confused now i will be totally confused in an exam!
Homework Equations
Example 1:
Now the integral around the unit circle for 1/z. In the unit circle 1/z is anlaytic for all z, but not path connected because when z=0 1/z is not defined. Therefore the integral does not equal zero and is equal to 2πi. I am confused about the following.
1) is my reasoning correct?
2) the book says that this integral is equal to 2πi. However when i integrate using γ(t)=e^it i get the right answer. When i apply the fundamental theorm and use the antiderivative of 1/z = Lnz and evaluate from 0 to 2π. I get Ln(1)+2πi+Kπi-Ln(1)-0πi+πKi. Now since o and 2π give the same argument doesn't this equal zero?
3) if indeed the integral of any analytic function in a path connected set is zero. Applying this reasoning to f(z) =1/z^2. With the unit circle. F(z) is analytic in the unit circle but not path conneceted because it is not defined at z=0. To my understanding it should thus not equal zero. But it Does equal zero when i compute it. These apparent contradictions are doing my head in so if anyone understands this i would like some help.
My concern is : I can solve these things no problem. I know that for 1/z^2+1 i can factorise the denominator and integrate in two regions where z+i, and z-i are analytic, but the above two examples seem to contradict my reasoning. So now i am confused and i am worried because if i am confused now i will be totally confused in an exam!