$P(z)=a_0+a_1z+a_2z^2+...+a_nz^n , a_k\in\mathbb{R}$ a complex polynomial prove: $a_k= \frac{1}{2\pi i}\oint_{|z|=1}z^{k-1}\overline{P(z)}dz$. I got this question:
Given $P(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ , $a_k\in\mathbb{R}$ a complex polynomial 
prove: $a_k= \frac{1}{2\pi i}\oint_{|z|=1}z^{k-1}\overline{P(z)}dz$.
P(z) being a polynomial makes it  its own Maclaurin series 
then we can assume that:
$a_k=\frac{P^{(k)}(z)}{k!}$ (P is polynomial $\Longrightarrow$ P is a holomorphic function )
from Cauchy integral formula we can say that:
$a_k=\frac{P^{(k)}(z)}{k!}=\frac{1}{2\pi i}\oint_{|z|=1}\frac{P(z)}{z^{k+1}}dz$
lets define $1/t=z \Longrightarrow -1/t^2dt=dz $
$\frac{1}{2\pi i}\oint_{|z|=1}\frac{P(z)}{z^{k+1}}dz=\frac{1}{2\pi i}\oint_{|\frac{1}{t}|=1}\frac{P(\frac{1}{t})}{(\frac{1}{t})^{k+1}}(-\frac{1}{t^2}dt)=\frac{1}{2\pi i}\oint_{|\overline{t}|=1}-t^{k-1}\overline{P(t)}dt$
(the third passage is right since $|z|=1 \Longrightarrow z=\frac{1}{\overline{z}}$)
I got very close and I don't know if i can assume that
$\frac{1}{2\pi i}\oint_{|\overline{t}|=1}-t^{k-1}\overline{P(t)}dt=\frac{1}{2\pi i}\oint_{|t|=1}t^{k-1}\overline{P(t)}dt$
since $|\overline{t}|$ is oriented counter-$|t|$-wise (I guess).
I wanted to know if I am right in assuming the last passage and if not I would appreciate some help.
 A: With
$P(z) = a_0 + a_1 z + a_2 z^2 + \ldots + a_n z^n = \displaystyle \sum_0^n a_j z^j, \; a_j \in \Bbb R, \tag 1$
we have
$\overline{P(z)} = \displaystyle \sum_0^n \overline{a_j} \bar z^j, 
 \tag 2$
whence
$z^{k - 1} \overline{P(z)} = \displaystyle \sum_0^n \overline{a_j} z^{k - 1} \bar z^j; \tag 3$
with $\vert z \vert = 1$ we may take
$z = e^{i \theta}, \; dz = ie^{i\theta} \; d\theta, \tag 4$
and thus (3) may be written
$z^{k - 1} \overline{P(z)} = \displaystyle \sum_0^n \overline{a_j} (e^{i \theta})^{k - 1} (e^{-i \theta})^j = \sum_0^n \overline{a_j} e^{i(k - j - 1) \theta}, \tag 5$
and
$\displaystyle \oint_{\vert z \vert = 1} z^{k - 1} \overline{P(z)}  \; dz = \oint_{\vert z \vert = 1} 
\left (\sum_0^n \overline{a_j} e^{i(k - j - 1) \theta} \right ) i e^{i\theta} \; d\theta$
$= \displaystyle \oint_{\vert z \vert = 1} 
\left ( \sum_0^n \overline{a_j} e^{i(k - j ) \theta} \right ) i \; d\theta = \sum_0^n \overline{a_j} \oint_{\vert z \vert = 1} e^{i(k - j) \theta} i \; d\theta; \tag 6$
the integrals may be evaluated as follows:  for $j \ne k$,
$\displaystyle \oint_{\vert z \vert = 1} e^{i(k - j) \theta} i \; d \theta = \int_0^{2\pi} e^{i(k - j) \theta} i \; d \theta = i \dfrac{1}{i(k - j)} (e^{2\pi i (k - j)} - e^{0 \cdot i(k - j)})$
$= \dfrac{1}{(k - j)}(e^{2\pi i (k - j)} - 1) = \dfrac{1}{(k - j)}(1 - 1) = 0;  \tag 7$
whereas if $j = k$,
$\displaystyle \oint_{\vert z \vert = 1} e^{i(k - j) \theta} i \; d \theta = \int_0^{2\pi} e^{i(0)\theta} i \; d\theta = \int_0^{2\pi} 1 i \; d\theta = 2\pi i; \tag 8$
when these evaluations are substituted back into (6) we find 
$\displaystyle \oint_{\vert z \vert = 1} z^{k - 1} \overline{P(z)}  \; dz = 2 \pi i  \overline{a_k}, \tag 9$
whence
$\overline{a_k} = \dfrac{1}{2 \pi i} \displaystyle \oint_{\vert z \vert = 1} z^{k - 1} \overline{P(z)}  \; dz; \tag{10}$
lastly, since $a_k \in \Bbb R$ (cf. (1)),
$\overline{a_k} = a_k, \tag{11}$
we immediately conclude that (10) yields
$a_k = \dfrac{1}{2 \pi i} \displaystyle \oint_{\vert z \vert = 1} z^{k - 1} \overline{P(z)}  \; dz, \tag{12}$
the requisite formula.  $OE\Delta$.
