# [SOLVED]Poisson kernel is even

#### dwsmith

##### Well-known member
For a fixed $r$ with $0\leq r < 1$, prove that $P(r,\theta)$ is an even function.

Take $-r$.
Then
\begin{alignat*}{3}
P(-r,\theta) & = & \frac{1}{2\pi}\frac{1 - (-r)^2}{1 - 2(-r)\cos\theta + (-r)^2}\\
& = & \frac{1}{2\pi}\frac{1 - r^2}{1 + 2r\cos\theta + r^2}
\end{alignat*}
I have $1 + 2r\cos\theta - r^2$. How can I get back the original denominator?

#### chisigma

##### Well-known member
For a fixed $r$ with $0\leq r < 1$, prove that $P(r,\theta)$ is an even function.
Because r is fixed the only variable remains $\theta$ and $cos \theta$ is an even function of $\theta$...

Kind regards

$\chi$ $\sigma$