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I've been trying to solve this problem but have a few questions regarding the final steps. I've been using this formula here $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint _{C} \frac{f(z)}{(z-a)^{n+1}}dz$$

and finding $\oint _{C} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{2\pi i}{n!}f^{(n)}(a) =$ was relatively easy. I am stuck on how to find $f(a)$ however. When integrating do I still include constants so that $f(a)=\frac{1}{27}e^{3a}+c_1a^{2}+c_2a+c_3$? If so how do I find what the constants are? C is the square with vertices $\pm1 \pm i$.

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    The answer is precisely what the formula gives. Which constants are you referring to? The contour description comes in for the formula, not for finding additional constants. – 2017-01-20

1 Answers 1

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You need $\oint _{C} \frac{e^{3z}}{(z-\ln{2})^{4}}dz$. You have the formula $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint _{C} \frac{f(z)}{(z-a)^{n+1}}dz.$$ Here $a= \ln 2$, $n=3$ and $f(z)=e^{3z}$. Since $f^{(3)}(z)=27e^{3z}$, then $$\oint _{C} \frac{e^{3z}}{(z-\ln{2})^{4}}dz=\frac{2\pi i}{3!} \cdot f^{(3)}(\ln 2)=\frac{\pi i}{3}\cdot(27e^{3(\ln 2)})=\frac{\pi i}{3}\cdot(27\cdot 2^3)=72 \pi i.$$