The question is following.
Suppose $P(x)$ is a polynomial of order $n(n\geqslant1)$ and all of its roots $(P(x)=0)$ are included in the domain $D$. $\partial{D}$ is a simple closed curve.$f(x)$ is holomorphic on in the closure $\overline{D}=D \cup \partial{D}.$
1 Suppose $$R(z)=\frac{1}{2\pi i} \oint_{\partial{D}} \frac{f(t)}{P(t)} \frac{P(t)-P(z)}{t-z} dt,\quad z\in D,$$ $$Q(z) = \frac{1}{2 \pi i} \oint_{\partial{D}} \frac{f(t)}{P(t)} \frac{dt}{t-z}, \quad z\in D.$$Prove: $R(z)$ is a polynomial of order no more than $n-1$ and $Q(z)$ is holomorphic in domain $D$.
2 Prove:$$f(z) = P(z)Q(z)+R(z), \quad x\in D.$$ If $Q_{1}(x)$ holomorphic in the domain $D$ and polynomial $R_{1}(x)$ of order no more $n-1$ satisfying $$f(z) = P(z)Q_{1}(z)+R_{1}(z), \quad z\in D,$$ then $$Q(x) \equiv Q_{1}(x), \quad R(x) \equiv R_{1}(x). $$
This is an exercise after my textbook. This chapter involves Cauchy-Gousant Theorem, Cauchy Integral Formula and Morera's Theorem. And we haven't touched with the power series expansion. I guess the solution doesn't have to use any knowledge of series expansion (both Taylor and Laurent), but I also welcome solutions with that.
I can somewhat feel that $R(z)$ is the part deleted or decomposed from our original $f(x)$ by $P(z)$ if I seen $R(z)$ as $$R(z) = \frac{1}{2 \pi i} \oint_{\partial{D}} \Biggl( \frac{f(t)}{t-z} - \frac{f(t)P(z)/P(t)}{t-z} \Biggr) dt$$ whose first part is precisely what we usually see in the Cauchy Integral Formula. Thus $Q(z)$ also looks similar with the second part of the formula above.
I have no idea about this question. I've tried to use simple polynomials to replace $P(z)$ and $f(z)$ and finally get nothing after annoying calculation.