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Chapter 6: Problem 17

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

Short Answer

Expert verified
For any complex number \(z = a + bi\), we have the modulus \(|z| = \sqrt{a^2 + b^2}\). We showed that \(-|z| \leq b \leq |z|\) by proving the inequalities on both sides, which demonstrates that the imaginary part of \(z\) falls within the interval \([-|z|, |z|]\).

Step by step solution

01

Write the complex number in the form of a + bi

A complex number \(z\) can be represented in the form \(a+bi\), where \(a\) and \(b\) are real numbers and \(i\) is an imaginary unit, denoted by \(i^2 = -1\). Here, \(a\) is the real part and \(b\) is the imaginary part.
02

Calculate the modulus of z

The modulus (or magnitude) of a complex number \(z = a + bi\) is given by the formula: \(|z| = \sqrt{a^2 + b^2}\)
03

Analyze the given interval

We are given that the imaginary part of \(z\) should fall within the interval \([-|z|, |z|]\). This can be rewritten as: \(-|z| \leq b \leq |z|\)
04

Prove the inequality on the left side

We need to show that \(-|z| \leq b\). Since \(|z|\) is non-negative, we have \(-|z| \leq 0\). Also, as \(b\) is the imaginary part, it can be positive, negative, or zero. Therefore, \(-|z| \leq b\).
05

Prove the inequality on the right side

Now, we need to show that \(b \leq |z|\). From step 2, we know that \(|z| = \sqrt{a^2 + b^2}\). Notice that \(a^2\) is a non-negative number, so the value of \(a^2 + b^2\) is at least \(b^2\). Hence, \(\sqrt{a^2 + b^2} \geq \sqrt{b^2}\), which simplifies to \(|z| \geq |b|\). Since \(|b|\) is also non-negative, and \(b\) represents the imaginary part, \(b \leq |z|\).
06

Combine the inequalities

By combining the inequalities from Steps 4 and 5, we get \(-|z| \leq b \leq |z|\). This proves that for any complex number \(z\), its imaginary part falls within the interval \([-|z|, |z|]\).

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Most popular questions from this chapter

Explain why \(\left(\cos 1^{\circ}+i \sin 1^{\circ}\right)^{360}=1\).

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Verify that $$ (\sqrt{3}+i)^{6}=-64 $$.

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