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Chapter 6: Problem 20

Suppose \(z\) is a complex number whose real part has absolute value equal to \(|z| .\) Show that \(z\) is a real number.

Short Answer

Expert verified
Given a complex number \(z = a + bi\) where the absolute value of its real part is equal to the absolute value of \(z\) itself, \(|a| = |z|\). We know that \(|z| = \sqrt{a^2 + b^2}\). By substitution, we have \(a^2 = a^2 + b^2\), which leads to \(b^2 = 0\). Thus, \(b = 0\), indicating that \(z\) is a real number.

Step by step solution

01

1. Define and Express Complex Number \(z\)

We know that a complex number can be represented in the form \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part of \(z\). The given information states that \(|a| = |z|\), where \(z = a + bi\).
02

2. Calculate Absolute Value of \(z\)

Recall that the absolute value or modulus of a complex number is calculated as \(|z| = \sqrt{a^2 + b^2}\). Given that \(|a| = |z|\), it can also be expressed as \(a^2 = a^2 + b^2\).
03

3. Solve for the Imaginary Part \(b\)

Now, we'll solve for \(b\) using the equation we derived in step 2: \(a^2 = a^2 + b^2\) \(b^2 = a^2 - a^2\) Therefore, \(b^2 = 0\) Since \(b^2 = 0\), we can deduce that \(b = 0\).
04

4. Concluding the Proof

Since we have shown that the imaginary part \(b\) of the complex number \(z\) is equal to zero, we can conclude that \(z\) is a real number. Therefore, we have successfully proven that if the absolute value of the real part of a complex number is equal to its absolute value, the complex number must be a real number.

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Most popular questions from this chapter

Show that \(\overline{w-z}=\bar{w}-\bar{z}\) for all complex numbers \(\mathcal{w}\) and \(z\).

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