Animated Solution for Mathematics - Complex Numbers: The set of all $\alpha \in R$, for which $w=\frac{1+(1-8\alpha )z}{1-z}$ is a purely imaginary number, for all $z\in C$ satisfying $|z|=1$ and $\text{Re }z\ne 1$, is :

Welcome, student. Today, we are not just solving an equation; we are exploring the elegant dance between geometry and algebra. We are given a complex number

and told that $w$ is purely imaginary for all $z$ on the unit circle ($|z|=1$). This is a classic JEE Advanced setup. It looks intimidating, but it is actually a beautiful puzzle waiting to be unlocked.

First, let us look at our playground: the unit circle. We are told $|z|=1$. In the complex plane, this is a circle of radius $1$ centered at the origin.

There is a crucial constraint: $\text{Re } z eq 1$. If $\text{Re }z=1$ and $|z|=1$, then $z$ must be $1$. If $z=1$, our denominator $1-z$ becomes zero, and the expression for $w$ explodes into infinity.

So, we are working on the entire unit circle except for the point $(1,0)$. This exclusion is our safety net; it ensures our math remains well-defined.

Now, what does it mean for $w$ to be purely imaginary? Geometrically, it means $w$ sits on the vertical imaginary axis.

Algebraically, this is equivalent to saying that the real part of $w$ is zero. The most powerful way to express this is the condition:

This is our golden key. It allows us to work with the conjugate $\bar{w}$ without having to decompose $w$ into its real and imaginary parts manually.

Let us find $\bar{w}$. We take the conjugate of the entire expression:

Here, we use the magic of the unit circle. Since $|z|=1$, we know $z\bar{z}=1$, which implies $\bar{z}=\frac{1}{z}$.

Substituting this into our expression for $\bar{w}$, we get:

To clean this up, we multiply the numerator and denominator by $z$, yielding:

Notice the denominator is $z-1$, while our original $w$ had $1-z$. By factoring out a negative sign, we get:

Now, we are ready for the final act. We set $w+\bar{w}=0$:

Combining the numerators, we get:

Watch closely as the terms cancel: $1$ and $-1$ vanish, $z$ and $-z$ vanish. We are left with:

Since $z eq 1$, we can safely divide by $(1-z)$, leaving us with $8\alpha =0$. Thus, $\alpha =0$.

The elegance of this result is breathtaking. The complex variable $z$ completely disappears, leaving us with a single, definitive value for $\alpha$. You have successfully navigated the trap and found the truth.