Past Paper Question:
a) State the generalized form of Cauchy’s integral theorem
b)Evaluate $$\displaystyle f(z)=\int_{\gamma}\frac{z^2}{\biggr(z-\dfrac{\pi}{4}\biggl)^3} dz$$
where $\gamma$ is a path traversed in the counter clockwise direction with vertices ${1,2i,−1,−2i}$.
Attempt:
a) Since my professor is terrible at labelling his definitions and theorems in his lecture notes, I am assuming that the generalised integral formula is $\displaystyle f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-z_0} dz$.
b)This is why I'm here, the diagram makes a slanted rectangular path, and the area of the shape is to the left of the traversed path, but apart from that, I don't know what the method is to evaluate contour integrals.
Unfortunately I don't have an example in my notes where this would be explained. How would I use answer this question?
The "generalised" Cauchy formula is $$f^{(n-1)}(z_0)= \frac{(n-1)!}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-z_0)^{n}} dz$$
Therefore we have that $z_0= \frac{\pi}{4}$ is a pole of order 3. Therefore we simply place this into our formula.
$$ \frac{(2\pi i)}{(n-1)!}f^{(n-1)}(z_0)= \int_{\gamma}\frac{f(z)}{(z-z_0)^{n}} dz$$ So, the integral is equal to $$\frac{(2\pi i)}{(3-1)!}.f^{''}(\frac{\pi}{4})=2\pi i$$