Complex Integral $\int_{C} \frac{e^{iz}}{z^3} dz$ using Cauchy Integral Formula of Derivatives

Question

I am trying to find $\int_{C} \frac{e^{iz}}{z^3} dz$ on circle of $|z| = 2$ traversing once on positive direction.

My approach was using Cauchy derivative formula $f^{(n)}(a) = \frac{n!}{2\pi i} \int_{C} \frac{f(z)}{(z-a)^{n+1}} dz$.

This is a case with $a = 0$ and $n = 2$. Second derivative of $f(z) = e^{iz}$ is $-e^{iz}$ so our integral is simply $\frac{2 \pi i}{2} (-e^{iz})$ at $z=0$ which gives me $-\pi i$.

However, I am not sure if this is correct mostly because of the fact that the circle is not a unit circle. Can somebody verify and if I am wrong, can somebody tell me which part I have done wrong? Thanks.

Answer 1

As an addendum to the previous answer, you actually do not need to compute any derivative since the Maclaurin series of the exponential function is fairly well-known: $$ e^{iz} = 1 + iz -\frac{z^2}{2} + O(z^3) $$ directly leads to $$ \oint_{|z|=\rho}\frac{e^{iz}}{z^3}= 2\pi i\cdot\operatorname*{Res}_{z=0}\left(\frac{e^{iz}}{z^3}\right)=2\pi i\cdot\left(-\frac{1}{2}\right)=\color{red}{-\pi i} $$ for any $\rho>0$.

Answer 2

The circle in question need not be a unit circle.

Indeed, the key fact is that it's a closed curve of counterclockwise orientation, enclosing the one singularity of the integrand, $z = 0$, which is being traversed one time. Strictly speaking, it could be any sort of crazy curve so long as those facts are maintained, even if it's not even circle! (If you can continuously deform/reshape the curve, the integral remains the same.)

That nuance aside - yup, your solution is correct!