11TH•Mathematics•Complex Numbers•Question 13

Solution

Solve the equation $|z-2|=|z+2|$ for $z=x+iy$.

Let $z=x+iy$ .

|z-2|=|z+2|

|(x-2)+iy|=|(x+2)+iy|

Square both sides:

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

Cancel $y^2$ and expand:

\begin{aligned} x^2-4x+4&=x^2+4x+4\\ -8x&=0\\ x&=0 \end{aligned}

So $z$ is purely imaginary:

\boxed{z=iy\ \text{(i.e. }x=0\text{)}}

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