I'm stuck on the way to prove the following statement:

Show that the circumference equation or straight line equation on the complex plane has this form $$\alpha z\overline{z}+ \beta z + \overline{\beta}\overline{z} + \gamma =0, $$ where $\alpha$ and $\gamma$ are real constants and $\beta$ can be a complex constant.

We start from the circumference equation of radius $r$ and center at $(b,c)$ on $\mathbb{R}^2$: $$(x-b)^2 + (y+c)^2= r^2$$ $$\qquad \Longleftrightarrow x^2 - 2xb +b^2 + y^2 -2yc + c^2 -r^2 =0 $$

If we took $z= x+iy$ and $a=b+ic$ as a fixed point. Then: $$z\overline{z} + a\overline{a} - z\overline{a} - a\overline{z}-r^2=0$$

So apparently, I'm close to the final answer but i don't know how to get those constants.


1 Answer 1


$\alpha = 1, \beta = -\bar{a}$ and $\gamma = a\bar{a} -r^2$. Note that $a\bar{a}$ will be a real quantity.

Another approach:

Consider $c, R$ to be the center and radius of circle and $z$ represents any point on the circumference. Note that $c$ will be complex constant and $R$ will be a real constant. Then:

$|z-c| = R $

Square both sides and apply $z\bar{z} = |z|^2$. You will get the answer.

This approach also matches with your one but the point is that instead of going to the $x, y$ co-ordinates and then converting them back into complex system is quite painful. Instead try to think in terms of complex co-ordinates only.

  • $\begingroup$ You really clear my mind from all the x-y coordinate. But now what happend with the straight line equation? Probably the same way? $\endgroup$ Sep 21, 2018 at 15:51

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