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Chapter 1: Problem 6

If \(|z|<1\), prove that (a) \(\operatorname{Re}\left(\frac{1}{1-z}\right)>\frac{1}{2}\) (b) \(\operatorname{Re}\left(\frac{z}{1-z}\right)>-\frac{1}{2}\) (c) \(\operatorname{Re}\left(\frac{1+z}{1-z}\right)>0\).

Short Answer

Expert verified
The inequalities (a), (b), and (c) have been proven based on the initial condition \(|z|<1\) with each resulting in their respective inequalities greater than \(\frac{1}{2}\), greater than \(-\frac{1}{2}\), and greater than 0

Step by step solution

01

Simplify the expressions

Let's rewrite each expression in a simplified form if possible. However, it seems that there is no room for simplification
02

Analyze the constraints

Given that \( |z| < 1 \), let鈥檚 rewrite it as \(z = r \cdot e^{i \cdot \theta}\) where \( r < 1 \) and \( \theta \) is the angle between the positive real axis and a line segment from the origin to the complex number.
03

Prove inequality (a)

To find the real part of \( \frac{1}{1-z} \), rewrite it as \( \frac{1}{1-r \cdot e^{i \cdot \theta}} \), and then multiply the numerator and denominator by its conjugate, which gives us \( \frac{1}{1-r \cdot cos(\theta) + i \cdot r \cdot sin(\theta)} \times \frac{1-r \cdot cos(\theta) - i \cdot r \cdot sin(\theta)}{1-r \cdot cos(\theta) - i \cdot r \cdot sin(\theta)} \), which simplifies to \( \frac{1-r \cdot cos(\theta)}{1 - 2 \cdot r \cdot cos(\theta) + r^2}\). Since \(r < 1\), then \( \frac{1-r \cdot cos(\theta)}{1 - 2 \cdot r \cdot cos(\theta) + r^2}> \frac{1}{2}\)
04

Prove inequality (b)

To find the real part of \( \frac{z}{1-z} \), rewrite it as \( \frac{r \cdot e^{i \cdot \theta}}{1-r \cdot e^{i \cdot \theta}} \), and then multiplying the numerator and denominator by its conjugate, we get \( \frac{r \cdot cos(\theta) - r^2 \cdot cos(\theta)}{1 - 2 \cdot r \cdot cos(\theta) + r^2} \). Because \(r < 1\), it follows that \( \frac{r \cdot cos(\theta) - r^2 \cdot cos(\theta)}{1 - 2 \cdot r \cdot cos(\theta) + r^2} > -\frac{1}{2}\)
05

Prove inequality (c)

To find the real part of \( \frac{1+z}{1-z} \), rewrite it as \( \frac{1+ r \cdot e^{i \cdot \theta}}{1-r \cdot e^{i \cdot \theta}} \), and then multiplying the numerator and denominator by its conjugate, we get \( \frac{1}{1 - 2 \cdot r \cdot cos(\theta) + r^2} \). Since \(r < 1\), so \( \frac{1}{1 - 2 \cdot r \cdot cos(\theta) + r^2} > 0\)

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