Prove that if \(w,z\) are complex numbers such that \(|w|=|z|=1\) and \(wz\ne -1\), then \(\frac{w+z}{1+wz}\) is a real number.

Could I get a thorough explaination for each step?

Guest Apr 15, 2020

#1**+1 **

Let w = a + bi and z = c + di. Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real. The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real. Therefore, the complex number (w + z)/(1 + wz) is real.

Guest Apr 15, 2020