Consider the general equation of a circle in $(x,y)$-plane and use the transformation $w = \frac{1}{z}$

Question

Consider the general equation of a circle in $(x,y)$-plane and use the transformation $w = \frac{1}{z}$ where $w = u + iv$ and $z=x+iy$.

I understand that

\begin{align} u = \frac{x}{x^2+y^2}\\ v = \frac{-y}{x^2+y^2}\\ |w|^2= \frac{1}{x^2+y^2}\\ \end{align}

but i don't understand how the equation transforms to

\begin{align} a+bu-cv+d(u^2+v^2)=0 \end{align}

could someone please xplain this to me?

Answer 1

Hint. Call the given circle $$(x-x_0)^2+(y-y_0)^2=r^2\ .$$ Note that $$x+iy=z=\frac{1}{w}=\frac{1}{u+iv}\ .$$ Now you can find $x,y$ in terms of $u,v$, substitute into the equation of the circle, and simplify.

Good luck!

Answer 2

Hint: See also wiki on generalised circle. Basically, as shown in the wiki: $$ A z \bar z + B z + C \bar z + D = 0$$ goes to

$$D \bar w w + C w + B \bar w + A = 0. $$