If \((1+i)^n - 16 = 0\), where \(i=\sqrt{-1}\) and \(n\) is a positive integer, then what is the least value of \(n\)?
8
Writing \(1+i=\sqrt{2}\left(\cos45^\circ+i\sin45^\circ\right)\), we get \((1+i)^n = 2^{n/2}(\cos(45n)^\circ + i\sin(45n)^\circ)\). For this to equal 16 (real and positive), we need \(2^{n/2}=16\), i.e. \(n=8\), and at \(n=8\) the angle is \(360^\circ\), giving exactly 16. So the least value is \(n=8\).
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What is \({\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3}\) where \({\rm{i}} = \sqrt { - 1} ,\) equal to?
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What is \({\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3}\) where \({\rm{i}} = \sqrt { - 1} ,\) equal to?
