Evaluate $\oint_C \frac{e^{2z}}{(z-1)^4} \mathrm dz$ $$I=\oint_C \frac{e^{2z}}{(z-1)^4} \mathrm dz$$
where $z$ is a complex number.
Sources of confusion:


*

*$C$ has not been given. Is such kind of a question possible?

*Considering $z=1$ is within the limits of the contour


My answer is $I=\frac{8}{3}\pi i e^2$, while my book's answer is $I=\frac{8}{3}\pi i e^{-2}$
 A: Recalling Cauchy's Integral Theorem:
$$
f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}}dz
$$
Setting $n=3$ and defining $f(z)=e^{2z}$, and $a=1$:
$$
f^{(3)}(1) = \frac{3!}{2\pi i} \oint_C \frac{e^{2z}}{(z-1)^{3+1}}dz
$$
From the definition of $f$ we have:
$$
f^{(3)}(1) = 8e^2
$$
So on one side we have the contour integral in your problem. Assuming again that $C$ is a counter-clock-wise contour that contains 1 in it's interior, we have:
$$
8e^2 = \frac{3!}{2\pi i} \oint_C \frac{e^{2z}}{(z-1)^{3+1}}dz
$$
Thus:
$$
\oint_C \frac{e^{2z}}{(z-1)^{3+1}}dz = \frac{8}{3} e^2 \pi i
$$
So... I second your result. There could be a typo somewhere in the book or maybe a printing defect on your specific copy of it.
A: Mefitico has answered comprehensively the second part of the question: however he left the first part somewhat incomplete, therefore my answer will focus on that part.

  
*
  
*$C$ has not been given. Is such kind of a question possible?
  

Yes, this question is perfectly answerable, even if you don't know the exact form of the contour, provided that the following characteristics of $C$ are known:


*

*defining the interior of $C$ as the greatest open bounded plane set $G\in\mathbb{C}$ such that $\partial{G}\subseteq C$ (the reason of higher complexity of this definition respect to the usual one will be clear in the discussion of next point) it is necessary that $1\in G$: this is due to the fact that the function $f(z)=e^{2z}/(z-1)^4$ is holomorphic in $\mathbb{C}\setminus\{1\}$ therefore by Cauchy's integral theorem,
$$
\oint_C f(z)\mathrm{d}z=0,
$$
no matter how the structure of $C$ is. This is exactly the assumption made in the second part of the question: however, even this information alone is not sufficient. We need the following one.

*It is necessary to know the widing number (or index) $\mathrm{ind}(C,a)$ of the contour $C$ respect to the point $1\in\mathbb{C}$. This is due to the fact that the most general form of Cauchy's integral formula, the homological Cauchy integral formula, involves also this topological invariant, which intuitively describes how many times the contour $C$ "turns around" the point $1$. Stated for a generic point $a\in G$
$$
\mathrm{ind}(C,a)\,f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}}\mathrm{d}z
$$
(see [1], §7.2 pp. 229-230 for an even more general version). 
If $C$ is a simple closed curve, then $\mathrm{ind}(C,a)=\pm1$ (depending on the orientation chosen on $C$) and everything goes as explained in the answer by Mefitico, otherwise you must account for the non-unity value of the index.
Note: the complexity higher than standard in the definition of $G$ is due to the fact that if $\mathrm{ind}(C,a)\neq\pm1$ then $C$ is not a simple continuous curve and therefore $\partial{G}\neq C$ since we cannot invoke Jordan's curve theorem.


[1] Burckel, Robert B. (1979), An Introduction to Classical Complex Analysis, Vol. 1, Lehrbücher und Monographien aus dem Gebiete der exakten Wissenschaften. Mathematische Reihe, Band 64, Basel–Stuttgart: Birkhäuser Verlag, p. 570, ISBN 3-7643-0989-X, MR 0555733, Zbl 0434.30001.
