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Chapter 6: Problem 19

Suppose \(z\) is a nonzero complex number. Show that \(\bar{z}=\frac{1}{z}\) if and only if \(|z|=1\).

Short Answer

Expert verified
Given a nonzero complex number \(z = a + bi\), we showed that \(\bar{z} = \frac{1}{z}\) if and only if \(|z| = 1\). We defined the conjugate of \(z\) as \(\bar{z} = a - bi\) and expressed the reciprocal of \(z\) in the standard form as \(\frac{1}{z} = \frac{a - bi}{a^2 + b^2}\). By showing that \(|z| = 1\) implies \(\bar{z} = \frac{1}{z}\) and vice versa, we proved the desired relationship between the conjugate, magnitude, and reciprocal of a complex number.

Step by step solution

01

Define and Express Complex Number Conjugate and Magnitude

Let \(z = a + bi\) be a nonzero complex number, where \(a\) and \(b\) are real numbers. The conjugate of \(z\) is given by \(\bar{z} = a - bi\). The magnitude of \(z\) is given by \(|z|= \sqrt{a^2 + b^2}\).
02

Express the Reciprocal of z

The reciprocal of the complex number \(z\) is given by \(\frac{1}{z}=\frac{1}{a+bi}\). To express it in standard form, multiply both numerator and denominator by the conjugate of \(z\): \[\frac{1}{z}=\frac{1}{a+bi}\cdot\frac{(a-bi)}{(a-bi)}=\frac{a-bi}{a^2+b^2}\]
03

Show \(\bar{z}=\frac{1}{z}\) when \(|z|=1\)

First, we will show that if \(|z|=1\), then \(\bar{z}=\frac{1}{z}\). If \(|z|=1\), then: \[\sqrt{a^2 + b^2} = 1\] \[a^2 + b^2 = 1\] Now, given the expression for \(\frac{1}{z}\) from Step 2, we have: \[\frac{1}{z}=\frac{a-bi}{a^2+b^2}\] Since \(a^2 + b^2 = 1\), we have: \[\frac{1}{z}=\frac{a-bi}{1}= a - bi\] This is the conjugate of \(z\), so \(\bar{z}=\frac{1}{z}\) when \(|z|=1\).
04

Show \(|z|=1\) when \(\bar{z}=\frac{1}{z}\)

Now, we will show that if \(\bar{z}=\frac{1}{z}\), then \(|z|=1\). Given that \(\bar{z}=\frac{1}{z}\), we have \(a-bi = \frac{a-bi}{a^2 + b^2}\). Equating the numerators, we have: \[a^2+2abi+b^2 = a^2+b^2\] Since \(a\) and \(b\) are real numbers, their product must be real, so \(2abi = 0\). Since \(z\) is nonzero and complex, either \(a\) or \(b\) must be non-zero. Therefore, we must have: \[a^2 + b^2 = a^2 + b^2\] As shown in Step 3, when this equation is true, \(|z|=1\). Thus, we have shown that if \(\bar{z}=\frac{1}{z}\), then \(|z|=1\). #Conclusion# In conclusion, we have shown that for a nonzero complex number \(z\), \(\bar{z}=\frac{1}{z}\) if and only if \(|z|=1\).

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Most popular questions from this chapter

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

Show that multiplication of complex numbers is associative, meaning that $$ u(w z)=(u w) z $$ for all complex numbers \(u, w,\) and \(z\).

Suppose \(a \neq 0\) and \(b^{2}<4 a c .\) Verify by direct calculation that $$ \begin{array}{l} a x^{2}+b x+c= \\ a\left(x-\frac{-b+\sqrt{4 a c-b^{2}} i}{2 a}\right)\left(x-\frac{-b-\sqrt{4 a c-b^{2}} i}{2 a}\right) \end{array} $$.

Show that if \(\mathbf{u}, \mathbf{v}\) and \(\mathbf{w}\) are vectors, then $$ \mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w} $$.

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ \frac{5+6 i}{2+3 i} $$
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