Let w and z be complex numbers such that \(Let $w$ and $z$ be complex numbers such that $|w|=|z|=1$ and $wz\ne -1.$ Prove that $\frac{w+z}{1+wz}$ is a real number. \)
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.
