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There are two complex numbers, namely z and w. They satisfy |z| = |w| = 1 and zw =/= -1.

(a) Prove that \(\overline{z} = \frac{1}{z}\) and \(\overline{w} = \frac{1}{w}\)

(b) Prove that \(\frac{z + w}{zw + 1}\) is a real number.

 

The lines indicate the magnitude (ex. |z|).

I began by expressing z and w as complex numbers- a + bi and c+di. The magnitude of z would then be \(|z| = \sqrt{a^2+b^2} = 1\) and the magnitude of w would then be \(|w| = \sqrt{c^2+d^2}=1\). Based on that, a^2+b^2 and c^2+d^2 would both be equal to 1, but it's not really getting me anywhere. Could someone help me out?

 Jan 18, 2021
 #1
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(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.

 Jan 19, 2021

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