Animated Solution for Mathematics - Complex Numbers: If $z_1$ and $z_2$ are two complex numbers such taht $|z_1|<1<|z_2|$ then prove that $|\frac{1-z_1{\bar{z}}_2}{z_1-z_2}|<1$.

Imagine you are standing on the Argand plane, looking at the unit circle centered at the origin. This circle is the boundary of destiny for our complex numbers.

We are given two points: $z_1$, which is trapped strictly inside this circle ($|z_1|<1$), and $z_2$, which roams free in the vast expanse outside ($|z_2|>1$).

We are tasked with proving that the modulus of the fraction $\left|\frac{1-z_1{\bar{z}}_2}{z_1-z_2}\right|$ is strictly less than $1$. When you see a modulus inequality involving a fraction, your intuition should immediately scream: "Get rid of the square roots!"

The most elegant way to handle this is to compare the squares of the numerator and the denominator. If we can prove that $|1-z_1{\bar{z}}_2{|}^2<|z_1-z_2{|}^2$, then the result follows naturally.

To execute this, we need our most powerful tool in complex analysis: the identity $|z{|}^2=z\bar{z}$. This identity is the key that unlocks the door to the solution.

We define an expression $E=|1-z_1{\bar{z}}_2{|}^2-|z_1-z_2{|}^2$. Our goal is to show that $E<0$.

Let's expand these terms. For the first term, we have $(1-z_1{\bar{z}}_2)\overline{(1-z_1{\bar{z}}_2)}$.

Remember that the conjugate of a product is the product of the conjugates, and the conjugate of a conjugate is the original number. So, $\overline{1-z_1{\bar{z}}_2}=1-{\bar{z}}_1{z}_2$.

Expanding this product gives us:

Now, let's turn our attention to the second term: $|z_1-z_2{|}^2$. This expands to $(z_1-z_2)({\bar{z}}_1-{\bar{z}}_2)$, which simplifies to:

Now, take a deep breath. This is the moment where the complexity melts away. When we subtract the second expansion from the first, the cross-terms $-{\bar{z}}_1{z}_2$ and $-z_1{\bar{z}}_2$ appear in both and cancel out completely.

We are left with:

This is a masterpiece of factorization. We can group the terms as follows:

Factoring out $(1-|z_1{|}^2)$, we arrive at the final, stunning form:

Now, we return to our geometric constraints. Because $|z_1|<1$, the term $(1-|z_1{|}^2)$ is strictly positive. Because $|z_2|>1$, the term $(1-|z_2{|}^2)$ is strictly negative.

A positive number multiplied by a negative number is always negative. Therefore, $E<0$.

This implies that $|1-z_1{\bar{z}}_2{|}^2<|z_1-z_2{|}^2$. Taking the square root of both sides, we confirm that $|1-z_1{\bar{z}}_2|<|z_1-z_2|$.

Dividing by the modulus of the denominator, we reach our destination:

You have just navigated the complex plane and emerged victorious. This is the beauty of mathematics—taking a seemingly impossible problem and revealing the simple, elegant truth hidden beneath the surface.