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      <text>10.3.2.3 Operations with the exponential form</text>
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      <text><![CDATA[<br>
<h4 class="HELM_example">Example</h4>
<p>If \(z = re^{i\theta}\) and \(w = te^{i\phi}\) then find expressions for</p>
<ol class="HELM_parts_inline">
    <li>&nbsp;\(z^{-1}\)&nbsp;</li>
    <li>&nbsp;\(z^*\)&nbsp;</li>
    <li>&nbsp;\(zw\)</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>If \(z = re^{i\theta}\) then \[z^{-1} = \frac{1}{re^{i\theta}} = \frac{1}{r}e^{-i\theta}\] using the normal rules for indices.</li>
    <li>Working in polar form: if \(z = re^{i\theta} = r(\cos\theta + i\sin\theta) \) then \[z^* = r(\cos\theta - i\sin\theta) = r(cos(-\theta) + i\sin(-\theta)) = re^{-i\theta}\] since \(\cos(-\theta) = \cos{\theta}\) and \(\sin(-\theta) = -\sin{\theta}\).
        In fact this reflects the general rule: to find the complex conjugate of any expression simply replace the \(i\) by \(-i\) wherever it occurs in the expression.<br></li>
    <li>\( zw = (re^{i\theta})(te^{i\phi}) = rte^{i\theta}e^{i\phi} = rte^{i\theta + i\phi} = rte^{i(\theta + \phi)} \)&nbsp;which is again the result we are familiar with: when complex numbers are multiplied their moduli multiply and their arguments add.</li>
</ol>
<hr>
<p> We see that in some circumstances the exponential form is even more convenient than the polar form since we need not worry about cumbersome trigonometric relations. </p>]]></text>
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