Cauchy Integral Formula

  • #1

Homework Statement


Evaluating using CIF.

|z| = 4

Integral [itex]{\frac {{{\rm e}^{2\,iz}}{\it dz}}{ \left( 3\,z-1 \right) ^{2}}}[/itex]



The Attempt at a Solution



So the singularity here is z = 1/3 which is inside the circle.

Therefore using the formula [itex]2\,i\pi \,f[/itex] and substituting in the z = 1/3

We get f(1/3) = exp(2/3i)

So I get the answer of 2*Pi*i(exp(2/3i))

However the answer given is (-4Pi/9)(exp(2/3i))

I'm thinking it has something to do with the repeated root, but I'm not sure.
 

Answers and Replies

  • #2
Found out that when dealing with powers, a generalised CIF has to be used. Now however I get -4Pi(exp(2/3i))

Where did that 1/9 come from?
 

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