1.4 Powers and Roots
Suppose that \(\, z = r \big (\cos \theta + i\sin \theta \big )\) and \(w = \rho \big ( \cos \phi + i \sin \phi \big )\) are polar forms of complex numbers \(z\) and \(w\).
Now consider \(\, \rho ^n \big (\cos (n\phi ) + i\sin (n\phi )\big ) = r\big (\cos \theta + i \sin \theta \big )\,\) from this, we have \(\, \rho = \sqrt [n]{r}\,\) and \(\, \cos n\phi = \cos \theta \,\) and \(\, \sin n \phi = \sin \theta \).
Hence we have the relationship \(\, n \phi = \theta + 2\pi k\) \[\therefore \quad \phi = \frac {\theta + 2\pi k}{n}\qquad k = 0,\, 1,\,,\ldots , n-1\]
These are distinct roots of \(z\). Thus the formula for the \(n^{\text {th}}\) roots of a complex number
\(\, z = r\big ( \cos \theta + i \sin \theta \big )\) are given by
\[\omega _k = \sqrt [n]{r} \Bigg \{ \cos \Big (\frac {\theta + 2\pi k}{n}\Big ) + i \sin \Big (\frac {\theta + 2\pi k}{n}\Big )\Bigg \}\]
Example 1.11. Find the cube roots of \(\, z = i\)
Solution
Here \(\, r = 1\,,\, \theta = Arg(i) = \frac {\pi }{2}\,\) and therefore the polar form of \(z\) is \(\, z = \cos \frac {\pi }{2} + i \sin \frac {\pi }{2}\).
Hence we obtain \(\omega _k = \displaystyle {\sqrt [3]{1}\Bigg \{ \cos \Big (\frac {\frac {\pi }{2} + 2\pi k}{3}\Big ) + i \sin \Big ( \frac {\frac {\pi }{2} + 2\pi k}{3}\Big )\Bigg \}}\quad k = 0, 1, 2\)
\begin {align*} k = 0\,, \quad \omega _0 & = \cos \frac {\pi }{6} + i\sin \frac {\pi }{6} = \frac {\sqrt {3}}{2} + \frac {i}{2}\\\\ k = 1\,, \quad \omega _1 & = \cos \frac {5\pi }{6} + i\sin \frac {5\pi }{6} = \frac {-\sqrt {3}}{2} + \frac {1}{2}i\\\\ k = 2\,, \quad \omega _3 & = \cos \frac {3\pi }{2} + i\sin \frac {3\pi }{2}\\\\ \end {align*}
Example 1.12. Find the fourth roots of \(\, z = 1 + i\)
Solution
Here \(\, r = \sqrt {2}\,\) and \(\, \theta = \operatorname {arg}(z) = \frac {\pi }{4}\,\) and we get
\[\omega _k = \sqrt [4]{2}\Bigg \{ \cos \Big (\frac {\pi /4 + 2\pi k}{4}\Big ) + i \sin \Big (\frac {\pi /4 + 2\pi k}{4}\Big )\Bigg \}\quad k = 0, 1, 2,3\]
\begin {align*} \omega _0 & = \sqrt [4]{2} \Big [ \cos \Big (\frac {\pi }{16}\Big ) + i\sin \Big (\frac {\pi }{16}\Big )\Big ]\\\\ \omega _1 & = \sqrt [4]{2} \Big [ \cos \Big (\frac {9\pi }{16}\Big ) + i\sin \Big (\frac {9\pi }{16}\Big )\Big ]\\\\ \omega _2 & = \sqrt [4]{2} \Big [ \cos \Big (\frac {17\pi }{16}\Big ) + i\sin \Big (\frac {17\pi }{16}\Big )\Big ]\\\\ \omega _3 & = \sqrt [4]{2} \Big [ \cos \Big (\frac {25\pi }{16}\Big ) + i\sin \Big (\frac {25\pi }{16}\Big )\Big ]\\\\ \end {align*}
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