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- #1

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?