I'm self-studying a book about complex analysis and Riemann surface and would like to get some help with the following question.
Background
On page 28 of [2], the authors have derived an equation for the inversion $I_C$ for a Euclidean circle C in $\mathbb{C}$ by employing the following definition of the inversion;
Let p be the centre and r the radius of C (a circle in $\mathbb{C}$). For each $z \in \mathbb{C} - {p}$ there is a unique point w on the line through p and z such that $|z - p| \cdot |w - p| = r^2$.
Now, in the actual derivation, they started off with the following;
We now derive an equation for $I_C$. If $z \neq p, \infty$ we have $|(\bar{z} - \bar{p}) (w - p)| = |z - p| \cdot |w - p| = r^2$, now arg(z - p) = arg(w - p), so $arg((\bar{z} - \bar{p})(w - p)) = 0$, and hence, $(\bar{z} - \bar{p})(w - p) = r^2$, ...
** Just in case, let me put the accompanying figure (Fig 2.3) from the book below;
My questions
- Where does the term $(\bar{z} - \bar{p})$ come from?
- How can we deduce to get the relationship of $(\bar{z} - \bar{p})(w - p) = r^2$?
Thank you for your support!
Reference
[2] Jones, Gareth A., and David Singerman. Complex functions: an algebraic and geometric viewpoint. Cambridge university press, 1987.
Set $\alpha = \arg(z - p) = \arg(w - p)$. Then $$ \begin{align} r^2 &= |z-p| \cdot |w-p| \\ &= |z-p| e^{-i\alpha} \cdot |w-p| e^{i\alpha}\\ &= \overline{(z-p)} \cdot (w-p) \\ & = (\bar z - \bar p) \cdot (w-p) \, . \end{align} $$