I have the following question which i am trying to solve using Cauchy’s Integral Formula (nothing else)
Let $γ$ denote the positively oriented circular contour of radius 1 about 0.
Evaluate the contour integral $\int_γ\frac{(z^2+1)}{z(2z-1)^2} dz $
I find that $\frac{(z^2+1)}{z(2z-1)^2}$ has singularites at $z=0$ and $z=1/2$
So i subdivide the interior of $\gamma$ as follows:
Then we have $\int_γ\frac{(z^2+1)}{z(2z-1)^2} dz =\int_{γ_1}\frac{(z^2+1)}{z(2z-1)^2} dz +\int_{γ_2}\frac{(z^2+1)}{z(2z-1)^2} dz $ .
So i let $f(z)=\frac{z^2+1}{z4(z-1/2)} $
Thus we have: $\int_{γ_1}\frac{(z^2+1)}{z(2z-1)^2} dz= \int_{γ_1}\frac{f(z)}{(z-1/2)} dz =2\pi i f(1/2) $
However this would give my $f$ has being undefined.
I was wondering if this question can be solve by Cauchy’s Integral Formula, the questions actually says to use Cauchy’s formula for Derivatives but i wanted to see if both formula gives the right answer.
So my question is how would i use Cauchy’s Integral Formula when there is the "same singularites twice" (in this case $z=1/2$).
When you have higher order poles, you cannot apply the Cauchy integral formula directly, simply because the assumptions of the theorem are not satisfied; particularly, your function $f$ is not holomorphic in any simply connected subset of $\mathbb{C}$ containing the contour.
If you proceed with the Cauchy integral formula, you would end up with following the steps of proving the higher order Cauchy integral formula, which uses the derivatives of the function $f$.