Complex analysis 1: $P \in \mathcal{P}(E,F)$ I have a problem:
Suppose that $P \in \mathcal{P}(E,F)$ be a polynomial (continuous) of degree $m$.
Prove that: 
$$\int_{|\zeta |=r}\dfrac{P(a+\zeta t)}{\zeta^{k+1}}\mathrm{d} \zeta=0 $$,
$\forall a,t \in E$, $r>0$ and $k>m$.
Where:
$1/$. We denote by ${\mathcal P}(^mE,F)$ the vector space of all $m$ - homogeneous (continuous) polynomials from $E$ into $F$.
http://en.wikipedia.org/wiki/Homogeneous_polynomial
$2/$. The map $P : E \to F$ is called polynomial of degree $m$, if:
$$P=P_0+\ldots+P_m, \forall P_j \in \mathcal{P}_a(^jE,F), j=0, \ldots , m$$
$3/$.  $ \mathcal{P}(E,F)$ be a vector space of all polynomials (continuous) from $E$ into $F$.
 A: Assuming I understand the somewhat confusing notation, for fixed $a$ and $t$, $P(a+\zeta t)$ is a polynomial of degree $m$, i.e.
$$
P(a+\zeta t) = a_0 + a_1 \zeta + \cdots + a_m \zeta^m. $$
Put
$$
g(\zeta) = \frac{P(a+\zeta t)}{\zeta^{k+1}} = \frac{a_0}{\zeta^{k+1}} + \frac{a_1}{\zeta^k} + \cdots + \frac{a_m}{\zeta^{k+1-m}}. $$
Then $g$ has a primitive function on $\mathbb{C} \setminus \{0\}$ (since $k+1-m > 1$), namely
$$
G(\zeta) = -\frac{a_0/k}{\zeta^{k}} - \frac{a_1/(k-1)}{\zeta^{k-1}} - \cdots - \frac{a_m/(k-m)}{\zeta^{k-m}},$$
so the interal of $g$ along $|\zeta| = r$ must be $0$.
A: Ops :)
Ok, here's my solution:
We need only prove in this case:
$$P \in \mathcal{P}^m(E,F)$$.
Suppose $A \in \mathcal{L}(^mE,F)$ such that:
$$\widehat{A}=P,\ A(x^m)=P(x)$$.
Applying the formula Newton's binomial theorem
$$P(x)=A(x^m)=A(a+x-a)^m=\sum_{j=0}^{m}\binom{m}{j}(Aa^{m-j})(x-a)^j=\sum_{j=0}^{\infty}P^jP(a)(x-a)^j$$
Hence, 
$$P^jP(a)= \left\{\begin{matrix}
 & \binom{m}{j}(Aa^{m-j}),\ \text{if}\ j=\overline{0,m}\\ 
 & 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \text{if}\ j \ge m+1
\end{matrix}\right.$$ 
Therefore: 
$P^kP(a)(t) =\dfrac{1}{2 \pi i} \int_{|\zeta |=r}\dfrac{P(a+\zeta t)}{\zeta^{k+1}}\mathrm{d} \zeta=0$ $\blacksquare $
