How to know if final equation is a circle (complex plane)

Question

in my proof of line transformation in a circle i get and equation: $z'(\frac{\bar{w}}{c})+\bar{z'}(\frac {w}{c})+\bar {z'}{z'}=0$. How can i show that it is a circle equation?

Answer 1

Let $z'\bar{w}=q\in\mathbb{C},c\in \mathbb{R}$;

$$z'(\frac{\bar{w}}{c})+\bar{z'}(\frac {w}{c})+\bar {z'}{z'}=z'(\frac{\bar{w}}{c})+\overline{{z'}(\frac {\bar{w}}{c})}+|{z'}|^2=c^{-1}(q+\bar q)+|{z'}|^2=2c^{-1}\Re(q)+|{z'}|^2.$$

Rearranging terms, the following equation is obtained:

$|z'|^2\equiv(\sqrt{x^2+y^2})^2=-2c^{-1}\Re(q)\in\mathbb{R}$

$\therefore x^2+y^2=-2c^{-1}\Re(q)$

Thus, the equation is a circle with a radius of $({2c^{-1}\Re(q)})^{\frac{1}{2}}$ iff $-2c^{-1}\Re(q)>0$. If $c^{-1}\Re(q)=0$, it is the zero point and if RHS $<0$, the equation is undefined.