$P(z)=a_0+a_1z+a_2z^2+...+a_nz^n , a_k\in\mathbb{R}$ a complex polynomial prove: $a_k= \frac{1}{2\pi i}\oint_{|z|=1}z^{k-1}\overline{P(z)}dz$.

Question

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.

Answer 1

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. $OE\Delta$.