Complex numbers

(a) Because $z\bar{z} = |z|^2$, $z\bar{z} = 1^2$. This means $\overline z= \dfrac1z$.

Similarly, with the exact same steps, $\overline w= \dfrac1w$.

(b) We write z and w in polar form. Let $z = \cos \theta + i\sin \theta$ and $w = \cos \phi + i\sin\phi$.

$\dfrac{z + w}{zw + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta \cos \phi - \sin \theta \sin \phi + 1) + i(\sin \theta \cos \phi + \cos \theta \sin \phi)}$

Simplifying, $\dfrac{z + w}{zw + 1} = \dfrac{((\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi))(\cos(\theta + \phi) + 1 - i\sin(\theta + \phi))}{(\cos(\theta + \phi) + 1)^2+\sin^2 (\theta + \phi)}$

This means $\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = \dfrac{(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) - (\cos \theta + \cos \phi)\sin(\theta + \phi)}{(\cos(\theta + \phi) + 1)^2 + \sin^2(\theta + \phi)}$

Notice that 

$(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) \\= \sin \theta + \sin \phi + \sin \theta \cos \theta \cos \phi -\sin^2 \theta \sin \phi + \sin \phi \cos \theta \cos \phi - \sin \theta \sin^2 \phi\\=\sin \theta \cos^2 \phi + \sin \phi \cos^2 \theta+\sin \theta \cos \theta \cos \phi + \sin \phi \cos \theta \cos \phi$

Also, 

$(\cos \theta + \cos \phi)\sin(\theta + \phi)\\ = \cos \theta \sin \theta \cos \phi + \cos \theta \sin \phi \cos \theta+\cos \phi \sin \theta \cos \phi + \cos \phi \sin \phi \cos \theta\\ = \cos \theta \sin \theta \cos \phi + \cos^2 \theta \sin \phi+\cos^2 \phi \sin \theta + \cos \phi \sin \phi \cos \theta$

This means $(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) = (\cos \theta + \cos \phi)\sin(\theta + \phi)$.

Then it immediately follows that $\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = 0$, which means $\dfrac{z + w}{zw + 1}$ is a real number.