How to factorize $zz^*-4z-4z^*+12=0$ (where $z^*$ is the complex conjugate of $z$)

Question

I'm trying to factorize this: $$zz^*-4z-4z^*+12=0$$ to get this: $$|z-4|^2 - 4 = 0$$ where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.

I'm trying to factorise this using the completed square method but had no luck so far.

Could use some help.

Answer 1

\begin{align} zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\\ &= z(z^* -4) -4(z^* -4) -4\\ &= (z-4)(z^* -4) -4\\ &= (z-4)(z-4)^* -4\\ &= |z-4|^2 -4. \end{align}

Answer 2

Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write

$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$

Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.