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Chapter 1: Problem 20

Consider \(z \in \mathbb{C}\) with \(\operatorname{Re}(z)>1\). Prove that $$ \left|\frac{1}{z}-\frac{1}{2}\right|<\frac{1}{2} $$

Short Answer

Expert verified
Question: Prove that if the real part of a complex number \(z\) is greater than 1, then the absolute value of the complex number \(\frac{1}{z}-\frac{1}{2}\) is strictly less than \(\frac{1}{2}\). Solution: We first simplified the expression inside the absolute value and then used the properties of absolute values to find inequalities involving real numbers. We showed that if \(\operatorname{Re}(z) > 1\), then \(\left|\frac{1}{z}-\frac{1}{2}\right| < \frac{1}{2}\).

Step by step solution

01

Simplify the expression inside the absolute value

We want to simplify the expression: $$ \left|\frac{1}{z}-\frac{1}{2}\right| $$ First, we'll get a common denominator for the two fractions: $$ \left|\frac{2-z}{2z}\right| $$
02

Use the properties of absolute values

Using the properties of absolute values, specifically the fact that \(|ab|=|a||b|\), we can rewrite the expression as: $$ \left|\frac{2-z}{2z}\right| = \frac{|2-z|}{|2z|} $$
03

Analyze the real part of \(z\)

Since we know that \(\operatorname{Re}(z) > 1\), without loss of generality, we can say that \(z = x + yi\), where \(x > 1\) and \(y \in \mathbb{R}\). Now we have: $$ \frac{|2-z|}{|2z|} = \frac{|2 - (x + yi)|}{|2(x + yi)|} $$
04

Find the absolute value of the numerator

Let's find the absolute value of the numerator: $$ |2 - (x + yi)| = |(2 - x) - yi| = \sqrt{(2 - x)^2 + y^2} $$ Since \(x > 1\), we know that \((2-x) < 1\), and therefore: $$ (2 - x)^2 \le (2 - 1)^2 = 1 $$ As a result: $$ |2 - (x + yi)| = \sqrt{(2 - x)^2 + y^2} \le \sqrt{1 + y^2} < \sqrt{2 + y^2} $$
05

Find the absolute value of the denominator

Now let's find the absolute value of the denominator: $$ |2(x + yi)| = 2|x + yi| = 2\sqrt{x^2 + y^2} $$ Since \(x > 1\), we know that $$ 2\sqrt{x^2 + y^2} > 2\sqrt{1^2 + y^2} = 2\sqrt{1 + y^2} $$
06

Prove the inequality

Now we can use the inequalities we found in Steps 4 and 5 to prove the desired inequality: $$ \frac{|2-z|}{|2z|} \le \frac{\sqrt{1 + y^2}}{2\sqrt{1 + y^2}} < \frac{\sqrt{2 + y^2}}{2\sqrt{1 + y^2}} < \frac{1}{2} $$ From this, we have shown that: $$ \left|\frac{1}{z}-\frac{1}{2}\right| < \frac{1}{2} $$

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Most popular questions from this chapter

Compute the following: (a) \(i^{2000}+i^{1999}+i^{201}+i^{82}+i^{47} ;\) (b) \(E_{n}=1+i+i^{2}+i^{3}+\cdots+i^{n}\) for \(n \geq 1\) (c) \(i^{1} \cdot i^{2} \cdot i^{3} \cdots i^{2000}\) (d) \(i^{-5}+(-i)^{-7}+(-i)^{13}+i^{-100}+(-i)^{94}\).

Solve in \(\mathbb{C}\) the following equations: (a) \(z^{2}=i\); (b) \(z^{2}=-i ;\) (c) \(z^{2}=\frac{1}{2}-i \frac{\sqrt{2}}{2}\).

Find all complex numbers \(z\) such that $$ z^{\prime}=(z-2)(\bar{z}+i) $$ is a real number.

Solve in \(\mathbb{C}\) the equations: (a) \(z^{2}+z+1=0 ;\) (b) \(z^{3}+1=0\).

Find the geometric images of the complex numbers \(z\) such that the triangle with vertices at \(z, z^{2}\), and \(z^{3}\) is a right triangle.
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