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Chapter 6: Problem 35

Show that if \(z\) is a complex number, then the real part of \(z\) is in the interval \([-|z|,|z|]\).

Short Answer

Expert verified
Given a complex number \(z = a + bi\), we found the modulus as \(|z| = \sqrt{a^2 + b^2}\). We have shown that \(a \le |z|\) and \(a \ge -|z|\), so the real part of \(z\), Re(\(z\)), lies within the interval \([-|z|, |z|]\): \(-|z| \le Re(z) \le |z|\).

Step by step solution

01

Understand complex number and its real part

Given a complex number z = a + bi, where a and b are real numbers and i is the imaginary unit, the real part of z is represented as Re(z) = a.
02

Determine the modulus of a complex number

The modulus of a complex number z = a + bi, denoted as |z|, is defined as the square root of the sum of the squares of its real and imaginary components: \(|z| = \sqrt{a^2 + b^2}\)
03

Show that the real part a is less than |z|

Now, we will show that \(a \le |z|\): Since \(b^2 \ge 0\), it follows that \(a^2 + b^2 \ge a^2\). Taking the square root of both sides, we get: \(|z| = \sqrt{a^2 + b^2} \ge \sqrt{a^2} = a\) Hence, \(a \le |z|\).
04

Show that the real part a is greater than -|z|

Similarly, we will show that \(a \ge -|z|\): Note that \(-b^2 \le 0\), so we can deduce that \(a^2 - b^2 \le a^2\). Taking the square root of the left side is not possible, as it might be negative, so we add \(2b^2\) on both sides instead: \(a^2 - b^2 + 2b^2 \le a^2 + 2b^2\) This simplifies to \(a^2 + b^2 \le a^2 + 2*|b^2|\), and taking the square root, we obtain: \(|z| = \sqrt{a^2 + b^2} \le \sqrt{a^2 + 2 |b^2|} = a + |b|\) Rearranging the inequality, we obtain: \(a \ge -|b|\) Now, since \(-|b| \ge -|z|\), it follows that: \(a \ge -|z|\)
05

Conclude the interval for the real part

As we have shown that \(a \le |z|\) and \(a \ge -|z|\), we can conclude that the real part of z, Re(z), lies within the interval [-|z|, |z|]: \(-|z| \le Re(z) \le |z|\)

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