# Circumference and straight line equation in complex plane

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.

$$\alpha = 1, \beta = -\bar{a}$$ 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.

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