Animated Solution for Mathematics - Complex Numbers: The number of complex numbers $z$ such that $|z-1|=|z+1|=|z-i|$ equals

Welcome, fellow traveler of the mathematical landscape! Today, we are going to unravel a problem that might look like a dry algebraic exercise, but is actually a beautiful dance of geometry in the Argand plane.

We are tasked with finding the number of complex numbers $z$ that satisfy the triple equality:

At first glance, this looks like a system of equations, but let us pause and look at the soul of the expression. Each part of this equality represents a distance. Specifically, $|z-z_0|$ is the distance between the complex number $z$ and the fixed point $z_0$.

So, we are looking for a point $z$ that is equidistant from three distinct points: $1$, $-1$, and $i$.

Before we dive into the algebra, let us sharpen our geometric intuition. Recall that the equation $|z-z_1|=|z-z_2|$ is the classic definition of a perpendicular bisector. It is the set of all points that are perfectly balanced between two anchors, $z_1$ and $z_2$.

If we have three points, $A(1)$, $B(-1)$, and $C(i)$, our condition $|z-1|=|z+1|=|z-i|$ means that $z$ must lie on the perpendicular bisector of $AB$ AND the perpendicular bisector of $BC$.

This is the geometric definition of the circumcenter of a triangle! Since any three non-collinear points form a unique triangle, and every triangle has exactly one circumcenter, we should expect exactly one solution.

Let us translate this geometry into the language of algebra. We assume $z=x+iy$, where $x$ and $y$ are real numbers. Our first condition is $|z-1|=|z+1|$.

Substituting $z=x+iy$, we get:

Squaring both sides to remove the modulus, we have:

Expanding these squares, we get $x^2-2x+1+y^2=x^2+2x+1+y^2$. Notice the elegance of the cancellation! The $x^2$, $y^2$, and $1$ terms vanish from both sides, leaving us with $-2x=2x$, which simplifies to $4x=0$, or $x=0$.

This confirms our geometric intuition: the perpendicular bisector of $1$ and $-1$ is indeed the imaginary axis, where the real part $x$ is zero.

Now that we know $z$ must be of the form $z=iy$, we turn to the second part of our equality: $|z+1|=|z-i|$. Substituting $z=iy$ into this equation, we get $|iy+1|=|iy-i|$.

We can rewrite the right side as $|i(y-1)|$. Since the modulus of a product is the product of the moduli, and $|i|=1$, this becomes:

Squaring both sides again, we get $1^2+y^2=(y-1{)}^2$. Expanding the right side, we obtain $1+y^2=y^2-2y+1$.

Again, the $y^2$ and $1$ terms cancel out beautifully, leaving us with $0=-2y$. This forces $y=0$.

Substituting $y=0$ back into our form $z=iy$, we find $z=0$. We have arrived at the origin! This is the unique point that is equidistant from $1$, $-1$, and $i$.

Thus, there is exactly one complex number satisfying the condition. The beauty of this problem lies in how the algebra and geometry converge to a single, elegant point.