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 $(-5,0)$
 A: Try writing out the definition of $|a + ib | = \sqrt{a^2 + b^2}$ on both sides, squaring both sides, and then simplifying. 
For the left hand side, your $a$ will be $x-3$ and $b = y$. 
A: $$|x+iy-3|^2=4|x+iy+3|^2$$
$$|x+iy-3|^2=(x-3)^2+y^2$$
$$4|x+iy+3|^2=4(x+3)^2+4y^2$$
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 $(-5,0)$.
A: From where you left, after squaring both sides :
$$(x-3)^2+y^2=4(x+3)^2+4y^2\iff -9=x^2+10x+y^2\iff $$
$$\iff (x+5)^2+y^2=16$$
Indeed a circle centered at $\;(-5,0)\;$ and radius $\;4\;$ .
A: Square both sides of your equation and using $\left| z \right|^2 = z z^*$
$$\left|\frac{z-3}{z+3} \right|^2=4$$
$$\frac{z z^* - 6 z z^* +9}{z z^* + 6 z z^* +9} =4$$
$$z z^* - 6 z z^* +9 =4 (z z^* + 6 z z^* +9)$$
$$ 0  =3 z z^* + 30 z z^* + 27$$
$$ 0  =3 (z z^* + 10 z z^* + 9)$$
Cancel the $3$ factor
$$ 0  =z z^* + 10 z z^* + 9$$
$$ 0  = \left| z + 5 \right|^2 -16$$
$$ 4  = \left| z + 5 \right|$$
which is nothing else than the equation of a circle centred at 5 of radius 4.
