Cauchy Integral theorem 
Let $f(z)=\sum^{\infty}_{k=0}\frac{k^3}{3^k}z^k$, compute 
  $\int_{|z|=1}\frac{f(z)}{z^4}dz$ and $\int_{|z|=1}\frac{f(z)sinz}{z^2}dz$.

I do not know how to do these problems. I know it is a closed circle centered at zero and radius $1$ so I must use Cauchy Integral theorem$$
    f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\ dz, $$ but I do not know how to use it explicitly for this question.
 A: This is related to what Alex said and your comment. Using the Cauchy differential formula, we get $$\int_{|z| = 1} \dfrac{f(z)}{z^4} = \frac{2\pi i}{3!}\left(\frac{3!}{2\pi i}\int_{|z| = 1} \frac{f(z)}{z^4} \right) = \frac{2\pi i}{3!} f^{(3)}(0),$$ which you can compute. since you can do term by term differentiation of the series and lots of terms will be zero since you're plugging in zero and $z^k$ is in the terms. For the second one, you can expand $\sin z$ as a power series and then do power series multiplication to get a new series $g(z)$. Then you can apply the same trick as above with $g$ in place of $f$. (Sorry for all the edits!)
A: Hint: The Cauchy Differential Formula tells you about the coefficients of the Taylor series for a function at a point. 
A: It should be plain that the series has a radius of convergence of 3, so that the only poles within the unit disk are at the origin in either integral.
If this helps, there is a closed-form expression for $f(z)$:
$$f(z) = \frac{\frac{z}{3} \left [\left (\frac{z}{3}\right)^2 + 4 \left (\frac{z}{3}\right ) + 1 \right ]}{\left ( 1-\frac{z}{3} \right )^4}$$
Note the $z$ factor in the numerator.  We may define $g(z) = f(z)/z$ as analytic in the unit disk, and consider
$$\oint_{|z|=1} dz \frac{g(z)}{z^3} = i \pi g''(0)$$
You can work out that
$$g''(x) = \frac{2 \left (1+\frac{x}{3}\right)\left (1+\frac{x}{27}\right)}{\left (1-\frac{x}{3}\right)^6}\implies g''(0)=2$$
Thus, the first integral is $i 2 \pi$.   The second integral is zero because there are no poles at $z=0$ due to the nature of $f$.
