If |z-1|=|z+1| find the locus of z

Question

If $|z - 1| = |z + 1|$ , find the locus of the complex number $z$ .

Solution — Step by Step

Let $z = x + iy$ where $x, y \in \mathbb{R}$ .

Substituting, the condition becomes $|x + iy - 1| = |x + iy + 1|$ , which simplifies to $|(x-1) + iy| = |(x+1) + iy|$ .

The modulus of a complex number $a + ib$ is $\sqrt{a^2 + b^2}$ .

So we get: $\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$

Squaring both sides (both sides are non-negative, so this is safe):

(x-1)^2 + y^2 = (x+1)^2 + y^2

The $y^2$ terms cancel on both sides, leaving:

(x-1)^2 = (x+1)^2

Expanding: $x^2 - 2x + 1 = x^2 + 2x + 1$

The $x^2$ and constant terms cancel: $-2x = 2x$

Therefore $4x = 0$ , which gives us $x = 0$ .

Since $x = 0$ means the real part of $z$ is zero, $z$ lies on the imaginary axis (the $y$ -axis in the Argand plane).

The locus is the equation $\text{Re}(z) = 0$ , or geometrically, the perpendicular bisector of the segment joining $(1,0)$ and $(-1,0)$ .

Why This Works

The condition $|z - 1| = |z + 1|$ has a beautiful geometric meaning. In the Argand plane, $|z - 1|$ represents the distance of the point $z$ from the point $(1, 0)$ , and $|z + 1|$ represents the distance from $(-1, 0)$ .

So we’re asking: where is the set of all points that are equidistant from $(1, 0)$ and $(-1, 0)$ ? The answer is the perpendicular bisector of the line segment joining those two points. Since $(1,0)$ and $(-1,0)$ are symmetric about the $y$ -axis, their perpendicular bisector is exactly the $y$ -axis.

This geometric reasoning is far faster than algebra in an exam — you can write the answer in one line once you recognize the pattern.

Alternative Method

We can use the property of modulus directly. Notice that $|z-1|^2 = (z-1)\overline{(z-1)} = (z-1)(\bar{z}-1)$ .

Setting $|z-1|^2 = |z+1|^2$ :

(z-1)(\bar{z}-1) = (z+1)(\bar{z}+1)

z\bar{z} - z - \bar{z} + 1 = z\bar{z} + z + \bar{z} + 1

-2(z + \bar{z}) = 0 \implies z + \bar{z} = 0 \implies 2\text{Re}(z) = 0

Since $z + \bar{z} = 2\text{Re}(z) = 2x$ , we again get $x = 0$ .

In JEE, locus problems often have elegant geometric interpretations. Always ask: “What does this modulus condition mean as distances in the Argand plane?” That insight can save 2-3 minutes per question.

Common Mistake

Many students forget to cancel the $y^2$ terms after squaring, or make sign errors expanding $(x \pm 1)^2$ . The most frequent error is writing $(x-1)^2 = x^2 - x + 1$ instead of $x^2 - 2x + 1$ . Always expand carefully: $(x-1)^2 = x^2 - 2(x)(1) + 1^2$ .

Another trap: students write the locus as ” $z$ is purely imaginary” — but $z = 0$ also satisfies $x = 0$ , and $0$ is neither purely real nor purely imaginary. The correct statement is ” $z$ lies on the imaginary axis” or " $\text{Re}(z) = 0$ ".

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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