Show that, if $f(z)$ is a polynomial with $$f(z)=\sum_{n=0}^{k} a_{n}z^{n}$$ for some $k \in \mathbb{N}$ that $$\int_{C_{R}^{+}(0)} \frac{f(z)}{z^{j}} dz=2\pi i(a_{j-1})$$ (For the case of $j-1 > k$ then assume $a_{j-1}=0).$
I started out with parameterizing $$z(t)=Re^{it}\ \ \ \ \ z'(t)=iRe^{it}dt \ \ \ \ \ \ \ 0\leq t\leq 2\pi$$ So $$f(z(t))=\sum_{n=0}^{k} a_{n}(Re^{it})^{n}$$
$$\int_{0}^{2\pi} \frac{\sum_{n=0}^{k} a_{n}(Re^{it})^{n}}{(Re^{it})^{j}} iRe^{it}dt$$ From there I'm a little confused at what to do next. If anyone can help me solve this problem that would be help me greatly. Also I'm not sure if I tagged this correctly, so please don't be completely upset if I used the wrong one tag.
By the residue theorem, your integral is $2\pi i$ times the coefficient of $\frac1z$ in the expression that you are integrating. This coefficient is precisely $a_{j-1}$. In fact$$\frac{\sum_{n=0}^ka_nz^n}{z^j}=\sum_{n=0}^na_nz^{n-j}$$and it is clear that $n-j=-1$ if and only if $n=j-1$; in other words, the coefficient of $z^{-1}$ is (as I said) $a_{j-i}$.
You can use Cauchy's integral formula instead of the residue theorem to compute this integral. Note that the integral is the sum of all integrals of the type $\int_{C_R^+(0)}a_nz^{n-j}\,dz$. If $n\neq j-1$, this integral is $0$, because $a_nz^{n-j}$ has a primitive: $\frac{a_n}{n-j+1}z^{n-j+1}$. All that remains is the integral $\int_{C_R^+(0)}\frac{a_{j-1}}z\,dz$, which is equal to $2\pi ia_{j-1}$, by Cauchy's integral formula.