What's the u-v equation for u and v defined as follows?

Question

I can not solve the u-v equation for $u=\frac{1-r^2}{1-2r \cos \theta + r^2}$ for $0 < r < 1$ and $v=\frac{2r\sin\theta}{1-2r \cos \theta + r^2}$ for $0 < r < 1.$ I assume that it is a circle because it is the polar form of a bilinear function $ w=\frac{i(1-z)}{1+z}.$ I have tried to equate $u^2+(v-t)^2=k$, for some t and k. And figure out what is $t$ and $k$. But I failed. Is there any suggestion ? Thank you!

Answer 1

Your expressions should read

\begin{align} w &= \frac{i(1-z)}{1+z} \\ u+vi &= \frac{2r\sin \theta+i(1-r^2)}{1+2r\cos \theta+r^2} \\ \color{red}{u} &= \frac{2r\sin \theta}{1\color{red}{+}2r\cos \theta+r^2} \\ \color{red}{v} &= \frac{1-r^2}{1\color{red}{+}2r\cos \theta+r^2} \end{align}

• $\theta=0 \implies (u,v)=\left( 0, \dfrac{1-r}{1+r} \right)$

• $\theta=\pi \implies (u,v)=\left( 0, \dfrac{1+r}{1-r} \right)$

• $2t=\dfrac{1-r}{1+r}+\dfrac{1+r}{1-r} \implies t=\dfrac{1+r^2}{1-r^2}$

• $2\sqrt{k}=\dfrac{1+r}{1-r}-\dfrac{1-r}{1+r} \implies \sqrt{k}=\dfrac{2r}{1-r^2}$

• Try to verify $$u^2+\left(v-\frac{1+r^2}{1-r^2} \right)^2=\frac{4r^2}{(1-r^2)^2}$$

Answer 2

Solve for $2r\cos \theta $ from the first equation then find $2r\sin \theta $ from the second and use the fact that $(2r\cos \theta )^{2}+(2r\sin \theta )^{2}=4r^{2}$. You don't get a circle as your answer.