Animated Solution for Mathematics - Complex Numbers: If $|z|=1$ and $z\ne \pm 1$, then all the values of $\frac{z}{1-z^2}$ lie on

Welcome, fellow traveler of the complex plane! Today, we are going to unravel a problem that, at first glance, might look like a daunting algebraic mess.

We are given a complex number $z$ that lives on the unit circle, defined by $|z|=1$, with the caveat that $z eq \pm 1$. Our mission is to find the locus of the transformation:

This is not just about solving an equation; it is about seeing the hidden geometry beneath the symbols.

When you see $|z|=1$, do not just think of it as a circle. Think of it as a powerful constraint. In the Argand plane, this circle is the set of all points at unit distance from the origin.

More importantly, for any complex number $z$ on this circle, we have the beautiful property $z\bar{z}=|z{|}^2=1$. This means that for any $z$ on the unit circle, its reciprocal is simply its conjugate:

This is the key that will unlock our transformation.

Now, let us look at our expression: $w=\frac{z}{1-z^2}$. If we try to substitute $z=x+iy$ directly, we will end up in a swamp of algebra.

Instead, let us use a classic JEE maneuver: divide both the numerator and the denominator by $z$. This gives us:

Suddenly, the expression looks much cleaner. We have transformed a complex fraction into a simple difference in the denominator.

Recall our earlier insight: $\frac{1}{z}=\bar{z}$. Let us substitute this into our expression for $w$:

This is where the magic happens. For any complex number $z=x+iy$, the conjugate is $\bar{z}=x-iy$.

When we subtract $z$ from $\bar{z}$, the real parts $x$ cancel out, leaving us with:

Since $y$ is the imaginary part of $z$, we can write this as $-2i\text{Im}(z)$.

Substituting this back into our expression for $w$, we get:

To make this look standard, we multiply the numerator and denominator by $i$. Since $i^2=-1$, the negative signs cancel out, and we are left with:

Look at this result! The numerator is $i$, and the denominator is a real number, $2\text{Im}(z)$. This means $w$ has no real part. It is purely imaginary.

The Locus Revealed

A complex number with no real part must lie on the imaginary axis. In the Argand plane, the imaginary axis is the y-axis.

Therefore, the locus of $w$ is the y-axis. We have navigated the complex algebra and arrived at a simple, elegant geometric truth.

Remember, in JEE Advanced, the most complex-looking problems often yield to the most elegant properties. Keep exploring, keep visualizing, and most importantly, keep falling in love with the math!