Integrate a sum (geometric series) round |z| = 1

Question

This is a question from a text book (Saff and Snider, Complex analysis for mathmatics science and engeneering, page 203).

Let $$ f(z) = \sum_{k=0}^\infty (k^3/3^k)z^k $$ Evaluate $$ \oint_{|z|=1} \frac{f(z)}{z^4} dz$$

My attempt to solve this has been

The sum is over k and the intergral over z, so we can take the sum (and other k dependent terms out of the integral; $$ \sum_{k = 0}^\infty (k^3/3^k)\oint_{|z|=1} \frac{z^k}{z^4} dz$$ $$ \sum_{k = 0}^\infty (k^3/3^k)\oint_{|z|=1} z^{k-4} dz$$ we can them paramaterise the integral $z = e^{it}$ from $0$ to $2\pi$ $$ \sum_{k = 0}^\infty (k^3/3^k)\int_{0}^{2\pi} e^{i(k-4)t} dt$$ This intergral will evaluate to zeros for every term except the term k = 4, for this term; $$ (4^3/3^4)\int_{0}^{2\pi} e^{0} dt = (4^3/3^4)\times 2\pi = \frac{128\pi}{81}$$

Apparently this is wrong, the answer given in the back of the book is $2\pi i$. Where do I go wrong?

Answer 1

Note that since you have made the change of variable as $z = {e^{it}}$, then $dz = i{e^{it}}dt$ so that $$\int\limits_0^{2\pi } {\sum\limits_{k = 0}^\infty {\frac{{{k^3}}}{{{3^k}}}{e^{it(k - 4)}}i{e^{it}}dt} } = \sum\limits_{k = 0}^\infty {\frac{{i{k^3}}}{{{3^k}}}\int\limits_0^{2\pi } {{e^{it(k - 3)}}dt} } = 2\pi i$$

Answer 2

The only term that will not vanish is the $k=3$ term. That way, the answer is indeed $i 2 \pi$.

In parametrizing, you forgot that $dz = i e^{i t} dt$.