Animated Solution for Mathematics - Complex Numbers: If $z=x+iy$ and $\omega =(1-iz)/(z-i)$, then $|\omega |=1$ implies that, in the complex plane,

Imagine you are standing on the Argand plane, a vast, two-dimensional landscape where every point is a complex number $z=x+iy$. We are given the relationship:

We are also provided with the constraint that $|\omega |=1$. This constraint is not just a condition; it is a boundary, a path that $z$ must follow.

We begin with the condition $|\omega |=1$. Substituting the expression for $\omega$, we get:

Using the property that the modulus of a quotient is the quotient of the moduli, we rewrite this as:

By multiplying both sides by $|z-i|$, we arrive at the intuitive equation:

To solve this, we must manipulate the left side into the standard form $|z-z_0|$. We factor out $-i$ from the expression:

Since $\frac{1}{-i}=i$, this simplifies to $-i(z+i)$. Our equation now becomes:

Using the property that the modulus of a product is the product of the moduli, we have $|-i|\cdot |z+i|=|z-i|$. Since $|-i|=1$, the equation collapses into the symmetric form:

This is the "Aha!" moment. In the complex plane, $|z-z_1|$ represents the distance between $z$ and $z_1$.

Our equation $|z-(-i)|=|z-i|$ tells us that the distance of $z$ from the point $-i$ (which is $(0,-1)$) is exactly equal to the distance of $z$ from the point $i$ (which is $(0,1)$).

The locus of all points equidistant from two fixed points is the perpendicular bisector of the line segment joining them. The segment connects $(0,1)$ and $(0,-1)$ on the imaginary axis.

The perpendicular bisector of this vertical segment is the horizontal line passing through the origin. Therefore, for the condition $|\omega |=1$ to hold, $z$ must lie on the Real Axis, where the imaginary part is zero.