Question

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

Answer 1

$z=x+iy$

Hence

$|z+3i|=\sqrt{x^2+((y+3)^2)};|z+5-2i|=\sqrt{(x+5)^2+((y-2)^2)}$

$\sqrt{x^2+((y+3)^2)}=\sqrt{(x+5)^2+((y-2)^2)}$

$x^2+(y+3)^2=(x+5)^2+(y-2)^2$

From the other equality we obtain

$|z-4i|=\sqrt{x^2+(y-4)^2};|z+2i|=\sqrt{x^2+(y+2)^2}$

$x^2+(y-4)^2=x^2+(y+2)^2$

Therefore

$(y-4)^2=(y+2)^2$

$-8y+16=4y+4$

$12y=12$

$y=1$

$x^2+16=(x+5)^2+1$

$10x+26=16$

$x=-1$

$z=-1+i$