z=x+iy
Hence
∣z+3i∣=x2+((y+3)2);∣z+5−2i∣=(x+5)2+((y−2)2)
x2+((y+3)2)=(x+5)2+((y−2)2)
x2+(y+3)2=(x+5)2+(y−2)2
From the other equality we obtain
∣z−4i∣=x2+(y−4)2;∣z+2i∣=x2+(y+2)2
x2+(y−4)2=x2+(y+2)2
Therefore
(y−4)2=(y+2)2
−8y+16=4y+4
12y=12
y=1
x2+16=(x+5)2+1
10x+26=16
x=−1
z=−1+i
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