Exercise 831Algebra › Complex numbersL1Added 3 Nov 2026

The modulus equation $|z-a|=|z-b|$

#perpendicular bisector#modulus#locus

Statement

Describe the set of $z\in\mathbb{C}$ such that $|z-2|=|z+2i|$ .

Hint

Points equidistant from $A(2,0)$ and $B(0,-2)$ .

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Solution

Interpretation. $|z-2|=|z+2i|$ means $MA=MB$ with $A(2,0)$ and $B(0,-2)$ .

Perpendicular bisector. Midpoint $(1,-1)$ , direction of $AB$ is $(-2,-2)$ , perpendicular direction $(1,-1)$ . The line has equation $y+1=(x-1)$ , i.e. $y=x-2$ .

Conclusion. The locus is the line $y=x-2$ .

The modulus equation $|z-a|=|z-b|$

Statement

Solution

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