Question

Determine the complex number z which satisfies the equations |z + 3i| = |z + 5 − 2i| and |z − 4i| = |z + 2i| simultaneously.

Answer 1

Let $z=x+iy$, then $|z+3i|^2=|x+(y+3)i|^2=x^2+(y+3)^2$, $|z+5-2i|^2=|(x+5)+(y-2)i|^2=(x+5)^2+(y-2)^2$, $|z-4i|^2=x^2+(y-4)^2$, $|z+2i|^2=x^2+(y+2)^2$, so we obtain the system $\begin{cases} x^2+(y+3)^2=(x+5)^2+(y-2)^2\\ x^2+(y-4)^2=x^2+(y+2)^2 \end{cases}$

From the second equation we obtain $(y-4)^2=(y+2)^2$, that is $y^2-8y+16=y^2+4y+4$, so $y=1$.

Substitute $y=1$ to the first equation. We obtain $x^2+16=(x+5)^2+1$, that is $x^2+16=x^2+10x+26$, so $x=-1$.

Answer: $z=-1+i$