When I can use Cauchy's Int. Formulae Let $C$ be the circle $|z| = R$ traversed anticlockwise. Let z be any point in $\mathbb{C}$.
I'm asked to calculate: $$\frac{1}{2\pi i} \int_C \frac{e^w}{w^3}(w^2+wz+z^2) \ dw$$
I'm wondering here if I can use Cauchy's Integral formula. I'm doubting about using it as we're integrating over $|z| = R$ with respect to $w$, does this change anything? I'm just at unease with not integrating over $|w| = R$
 A: Yes, you can, but the higher powers of w suggest that you'll need the version of Cauchy's formula that expresses the nth derivative of the holomorphic function in terms of an integral.
It also seems that the notation has you confused. Keep in mind that you are integrating with respect to w, not z. (If the z distracts you, rename it to 'a' or something and change everything back in the end.)
$$
I = \frac{1}{2\pi i} \oint \frac{e^w}{w^3} \left( w^2 + wz + z^2\right) dw \\
  = \frac{1}{2\pi i} \oint \frac{e^w}{w} + \frac{ze^w}{w^2} + \frac{z^2e^w}{w^3} dw
$$
Since $e^w$ is a holomorphic function, and $b=0$  is a point that lies inside the circle of radius $R$, you can use Cauchy's formula directly for the first function. Recall
$$
\frac{1}{2\pi i} \oint \frac{f(w)}{w-b} dw = f(b)
$$
So setting $b=0$, the first integral is $e^0$ = 1.
For the second and third integrals, you'll need the formula for the nth derivative.
$$
f^{(n)}(b) = \frac{n!}{2\pi i} \oint \frac{f(w)}{(w-b)^{n+1}} dw
$$
Setting $n=1$ and $n=2$, (with $b=0$) you have
$$
f^{(1)}(0) = \frac{1}{2\pi i} \oint \frac{e^w}{w^2} dw \\
f^{(2)}(0) = \frac{2}{2\pi i} \oint \frac{e^w}{w^3} dw
$$
The first and second derivatives of $e^w$ at 0 are 1. You can now use the values of these integrals to get
$$
I= \frac{1}{2\pi i} \oint \frac{e^w}{w} + z \frac{1}{2\pi i} \oint  \frac{e^w}{w^2} + z^2 \frac{1}{2\pi i} \oint  \frac{e^w}{w^3} dw \\
= 1 + z + \frac{z^2}{2}
$$
