<s>[INST] Simplifying$\left|\frac{z-3}{z+3} \right|=2$ I want to graph the following, but simplifying is the question here:
$$\left|\frac{z-3}{z+3} \right|=2$$
Now I can do this : $$\frac{|z-3|}{|z+3|}=2 $$
$$|z-3|=2|z+3|$$
$$|x+iy-3|=2|x+iy+3|$$
What manipulation do I use here? I have the answer is a circle of radius $4$ centered at $(-5,0)$
 [/INST] $$|x+iy-3|^2=4|x+iy+3|^2$$
$$|x+iy-3|^2=(x-3)^2+y^2$$
$$4|x+iy+3|^2=4(x+3)^2+4y^2$$
So:
$$(x-3)^2+y^2=4(x+3)^2+4y^2$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2$$
$$0=3x^2+30x+27+3y^2$$
$$0=x^2+10x+9+y^2$$
$$0=(x+5)^2-16+y^2$$
$$16=(x+5)^2+y^2$$
It's an equation of circle of radius $4$ centered at $(-5,0)$.
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