Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$

Question

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.

Answer 1

Take square both sides then we get $$|z-3|^2=4|z+3|^2.$$ Evaluate it (using the fact $|z|^2=z\overline{z}$ and $2\operatorname{Re}z=z+\overline{z}$) then $$3|z|^2+30\operatorname{Re}z+27=0$$

Let $z=x+iy$. From above equation, we get

$$x^2+10x+y^2+9=0$$

and is equivalent to $(x+5)^2+y^2=4^2$

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Details: From $|z-3|^2=4|z+3|^2$, we get $$(z-3)\overline{(z-3)}=4(z+3)\overline{(z+3)}$$ and is equivalent to $$(z-3)(\overline{z}-3)=4(z+3)(\overline{z}+3).$$ Expand the above equality then $$z\overline{z}-3(z+\overline{z})+9=4(z\overline{z}+3(z+\overline{z})+9)$$ Note that $z\overline{z}=|z|^2$ and $z+\overline{z}=2\operatorname{Re}(z)$ (you can check it easily - take $z=x+iy$ and calculate it) so $$3|z|^2+30\operatorname{Re}z+27=0$$

Answer 2

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.

Answer 3

$$\begin{align} \left\lvert\frac{z-3}{z+3}\right\rvert=2\\ \left\lvert1-\frac{6}{z+3}\right\rvert=2\\ \left\lvert\frac16-\frac{1}{z+3}\right\rvert=\frac13\\ \left\lvert\frac16-Q\right\rvert=\frac13 \end{align}$$

where $Q=\frac{1}{z+3}$. This last equation shows us $Q$ is along a circle of radius $\frac13$ centered at $\frac16$. The reciprocal of a circle in the complex plane is another circle. The circle $Q$ is on has a diameter passing through $-1/6$ and $1/2$ on the real line. So after we take the reciprocal, we have a circle with a diameter on the real line passing through $-6$ and $2$, and so its center is at $-2$ and radius is $4$.

So $z+3$ is on this circle. And then finally, $z$ is on a circle of radius $4$ centered at $-5$. That is, the solution set is the same as for $\left\lvert z+5\right\rvert=4$.