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Chapter 1: Problem 8

If \(|z|=1, z \neq-1\), show that \(z\) may be expressed in the form $$ z=\frac{1+i t}{1-i t}, $$

Short Answer

Expert verified
For a complex number with modulus 1 and argument different from \(\pi\), both representations \(z = e^{i\theta}\) and \(z=\frac{1+it}{1-it}\) are possible.

Step by step solution

01

Initial expression

Given \(z\) is a complex number such that \(|z|=1\) and \(z \neq -1\). We want to show that \(z\) can be expressed in the form \(z = \frac{1+it}{1-it}\) for some real number \(t\).
02

Complex number in polar form

Recall that a complex number of modulus 1 can be written in the polar form as \(z = e^{i \theta}\), where \(\theta\) is the argument of \(z\). Since \(z \neq -1\), it must be that \(\theta \neq \pi\).
03

Convert given form into polar notation

Now let's work with the form we want to prove. We write \(z = \frac{1+it}{1-it}\) in terms of \'e\'. For that, write \(t = \tan(\alpha/2)\) which gives \(z = \frac{1+i\tan(\alpha/2)}{1-i\tan(\alpha/2)}\). Multiplying numerator and denominator with their conjugate we can simplify this to \(z = e^{i\alpha}\).
04

Matching forms

For the two ways to express \(z\) to be the same, the arguments must coincide modulo \(2\pi\), i.e., \(\theta = \alpha + 2 \pi n\) for some integer \(n\). Because \(z \neq -1\), it must be that \(\alpha\neq \pi\) as well. Hence, for \(\theta \neq \pi\), it is possible to choose \(n\) such that the equality holds.

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