Animated Solution for Mathematics - Complex Numbers: Let $a,b\in \mathbb{R}$ and $a^2+b^2\ne 0$. Suppose $S=\{z\in \mathbb{C}:z=\frac{1}{a+ibt},t\in \mathbb{R},t\ne 0\}$, where $i=\sqrt{-1}$. If $z=x+iy$ and $z\in S$, then $(x,y)$ lies on

Unveiling the Locus

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful problem in complex numbers.

Often, when we see an expression like $z=\frac{1}{a+ibt}$, our instinct is to panic. It looks like a mess of variables and parameters.

But I want you to pause and breathe. In the complex plane, this is not just an algebraic expression; it is a geometric transformation. We are looking for the path, or the locus, of a point $z$ as the parameter $t$ dances across the real line.

Our first hurdle is the imaginary unit $i$ trapped in the denominator. We cannot easily see the real and imaginary parts of $z$ while they are held hostage in a fraction.

Our standard toolkit for this is rationalization. We multiply the numerator and the denominator by the complex conjugate of the denominator, which is $a-ibt$.

Look at that denominator! It has become purely real: $a^2+b^2{t}^2$. This is the magic of the conjugate.

Now, we can split the fraction into its real and imaginary components:

By comparing this to the standard form $z=x+iy$, we have successfully extracted our parametric equations: $x=\frac{a}{a^2+b^2{t}^2}$ and $y=\frac{-bt}{a^2+b^2{t}^2}$.

Now, we need to eliminate the parameter $t$. This is where many students get stuck, trying to solve for $t$ in terms of $x$ and plugging it into $y$.

Don't do that! It is a trap. Instead, look for symmetry. Let's calculate $x^2+y^2$:

Do you see it? The numerator $a^2+b^2{t}^2$ cancels out one power of the denominator!

We are left with the incredibly clean result:

We are almost there. Look back at our expression for $x$. We can rewrite it as $x=a\cdot \left(\frac{1}{a^2+b^2{t}^2}\right)$.

Since we just discovered that $\frac{1}{a^2+b^2{t}^2}=x^2+y^2$, we can substitute this directly into our equation for $x$:

Rearranging this gives us $x^2+y^2-\frac{x}{a}=0$. This is the equation of a circle!

By completing the square for $x$, we get:

This tells us the center is at $(\frac{1}{2a},0)$ and the radius is $|\frac{1}{2a}|$. Whether $a$ is positive or negative, the geometry holds firm.

You have just mapped a line into a circle through the power of complex inversion. Keep practicing this, and you will start to see the geometry behind every equation! The locus is a circle with center $(\frac{1}{2a},0)$ and radius $|\frac{1}{2a}|$.