(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.
