If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$, then find $\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$.

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> [!TLDR]
> The expression $\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$ simplifies to $1$ when $|\beta| = 1$, regardless of $\alpha$ (as long as the denominator is non-zero). The key is that $|\beta| = 1$ implies $\beta \bar{\beta} = 1$, which lets us rewrite the numerator as $|\beta - \alpha| = |\bar{\beta} - \bar{\alpha}|$ and then factor $\beta$ cleverly.

This problem is a classic in complex number algebra — it tests your ability to use the property $|z| = 1$ to replace $1$ with $\beta \bar{\beta}$ and then factor. The trick is to see that the denominator $1 - \bar{\alpha}\beta$ is actually the conjugate of something when $|\beta| = 1$, but more directly, we can pull $\beta$ out of the numerator.

Let’s go step by step.

1. **Start with what we know.**  
   We are given $|\beta| = 1$, so $\beta \bar{\beta} = 1$. This is the only fact about $\beta$ — it lies on the unit circle. $\alpha$ is any complex number (different from $\beta$, but that doesn’t affect the algebra).

2. **Rewrite the numerator using $|\beta| = 1$.**  
   The numerator is $|\beta - \alpha|$. Since $|\beta| = 1$, we can multiply inside the modulus by $|\bar{\beta}|$ (which is also 1) without changing the value:
   $$
   |\beta - \alpha| = |\bar{\beta}| \cdot |\beta - \alpha| = |\bar{\beta}(\beta - \alpha)|
   $$
   But $\bar{\beta}\beta = 1$, so:
   $$
   \bar{\beta}(\beta - \alpha) = 1 - \bar{\beta}\alpha
   $$
   Hence:
   $$
   |\beta - \alpha| = |1 - \bar{\beta}\alpha|
   $$

3. **Compare with the denominator.**  
   The denominator is $|1 - \bar{\alpha}\beta|$. Notice that $1 - \bar{\beta}\alpha$ and $1 - \bar{\alpha}\beta$ are **conjugates** of each other? Let’s check:
   $$
   \overline{1 - \bar{\alpha}\beta} = 1 - \alpha \bar{\beta}
   $$
   That’s $1 - \alpha \bar{\beta}$, not $1 - \bar{\beta}\alpha$ — but since multiplication of complex numbers is commutative, $\alpha \bar{\beta} = \bar{\beta}\alpha$. So indeed:
   \[
   \overline{1 - \bar{\alpha}\beta} = 1 - \bar{\bet

Q.If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$ , then find $\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$ .

Concept understanding — Complex Number Transform

The Intuition: Why "Transform" a Number?

The Precise Statement

A Concrete Example

Why This Matters

Q.If α\alphaα and β\betaβ are different complex numbers with ∣β∣=1|\beta| = 1∣β∣=1, then find ∣β−α1−αˉβ∣\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|​1−αˉββ−α​​.

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Q.If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$ , then find $\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$ .

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