<s>[INST] Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$ Find the set of complex numbers $z$ which satisfy
$$\left\lvert\frac{z-3}{z+3}\right\rvert=2\text.$$
I need help on that one. Thank you.
 [/INST] Putting  $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$\left|\frac{z-3}{z+3}\right|=2\iff |(x-3)+iy|^2=4|(x+3)+iy|^2\iff$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2\iff$$
$$x^2+10x+y^2=-9\iff (x-5)^2+y^2=16$$
Thus, the set of complex numbers fulfilling the above condition is a circle of radius four.
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