Solution Let z=x+iyz=x+iyz=x+iy. ∣z−2∣=∣z+2∣|z-2|=|z+2|∣z−2∣=∣z+2∣ ∣(x−2)+iy∣=∣(x+2)+iy∣|(x-2)+iy|=|(x+2)+iy|∣(x−2)+iy∣=∣(x+2)+iy∣ Square both sides: (x−2)2+y2=(x+2)2+y2(x-2)^2+y^2=(x+2)^2+y^2(x−2)2+y2=(x+2)2+y2 Cancel y2y^2y2 and expand: x2−4x+4=x2+4x+4−8x=0x=0\begin{aligned} x^2-4x+4&=x^2+4x+4\\ -8x&=0\\ x&=0 \end{aligned}x2−4x+4−8xx=x2+4x+4=0=0 So zzz is purely imaginary: z=iy (i.e. x=0)\boxed{z=iy\ \text{(i.e. }x=0\text{)}}z=iy (i.e. x=0)