Animated Solution for Mathematics - Complex Numbers: If $\frac{z-\alpha }{z+\alpha }$ ($\alpha \in R$) is a purely imaginary number and $|z|=2$, then a value of $\alpha$ is :

My dear student, welcome to a beautiful journey through the complex plane. Today, we are not just solving an equation; we are uncovering a hidden geometric truth.

Imagine you are standing on the complex plane, looking at the condition $|z|=2$. What do you see? You see a perfect circle, centered at the origin, with a radius of exactly $2$. Every point $z$ on this circle satisfies this condition.

Now, we are presented with a mysterious expression:

Here, $\alpha$ is a real number. We are told this expression is purely imaginary. This is not just a random constraint; it is a geometric signal. Let us peel back the layers of this problem together.

The Purely Imaginary Condition

When we say a complex number $w$ is purely imaginary, we are saying its real part is zero. But how do we translate this into the language of algebra?

The most powerful tool in our arsenal is the identity $w+\bar{w}=0$. Think of this as the "Real Part Killer."

If $w=x+iy$, then $\bar{w}=x-iy$. Adding them gives $w+\bar{w}=2x$. If this sum is zero, then $x$ must be zero. This is our master key. We don't need to perform messy division; we just need to set the sum of the expression and its conjugate to zero.

Let us substitute our expression into this condition:

Now, here is where many students stumble. We must apply the conjugate property to the fraction. Remember, the conjugate of a quotient is the quotient of the conjugates.

So, we get:

Since we are given that $\alpha \in \mathbb{R}$, we know that $\bar{\alpha }=\alpha$. This simplifies our equation beautifully to:

Take a deep breath—the hard part is almost over.

Now, let us clear the denominators. By taking the common denominator and cross-multiplying, we get:

Let us expand these brackets with surgical precision. The first part gives us $z\bar{z}+\alpha z-\alpha \bar{z}-{\alpha }^2$. The second part gives us $\bar{z}z+\alpha \bar{z}-\alpha z-{\alpha }^2$.

When we add them together, look at the magic! The terms $\alpha z$ and $-\alpha z$ cancel out. The terms $-\alpha \bar{z}$ and $\alpha \bar{z}$ also vanish. We are left with:

This is the elegance of symmetry in action.

We know the fundamental property that $z\bar{z}=|z{|}^2$. Substituting this into our simplified equation, we get $2|z{|}^2-2{\alpha }^2=0$, which simplifies to $|z{|}^2={\alpha }^2$.

We were given that $|z|=2$, so $|z{|}^2=4$. Therefore, $4={\alpha }^2$, which means $\alpha =\pm 2$.

But wait, there is a deeper geometric insight here. The points $\alpha$ and $-\alpha$ are the endpoints of the diameter of our circle. A classic theorem in geometry states that a diameter subtends a $90^{\circ }$ angle at any point on the circle.

This is exactly why the expression $\frac{z-\alpha }{z+\alpha }$—which represents the ratio of vectors—becomes purely imaginary. It is a perfect harmony of algebra and geometry. You have just mastered a classic JEE Advanced problem. Keep this intuition, and no problem will ever be too difficult for you.