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 ${1,2i,−1,−2i}$.


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?

  • 1
    $\begingroup$ Some professors like to assume that you will do the research to find more advanced questions, but considering the is a past paper question, this is shocking. The generalized integral formula is exactly the same, but i think you have $(z-z_0)^n$ as the denominator, someone correct me if I'm wrong. $\endgroup$
    – HELP
    Commented May 22, 2016 at 11:17
  • 1
    $\begingroup$ I guess the "generalized" form is the one with a power in the denominator, and a factorial in front, so that the answer is $f^{(k)}(a)$. $\endgroup$
    – GEdgar
    Commented May 22, 2016 at 12:00

1 Answer 1


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 $$\frac{(2\pi i)}{(3-1)!}.f^{''}(\frac{\pi}{4})=2\pi i$$

  • $\begingroup$ After reading a few comments on stack exchange, i have one question, is the power term in the denominator of the top of your post $n$ or $n+1$, does it matter, or is it just a case of shifting indices. $\endgroup$
    – UniStuffz
    Commented May 22, 2016 at 21:29
  • $\begingroup$ Just a case of shifting indices $\endgroup$ Commented May 22, 2016 at 21:44

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