Given that \({z_1} = 3\left( {\cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right)} \right)\) and \({z_2} = 3\left( {\cos \left( {\frac{\pi }{6}} \right) + i\sin \left( {\frac{\pi }{6}} \right)} \right)\) then \({z_1}{z_2} = \,\)?

Given that \({z_1} = 6\left( {\cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right)} \right)\) and \({\rm{ }}{z_2} = 3\left( {\cos \left( {\frac{\pi }{6}} \right) + i\sin \left( {\frac{\pi }{6}} \right)} \right)\) then \({\rm{ }}\dfrac{{{z_1}}}{{{z_2}}} = \,\,\)?

\({\left( {1 + i\sqrt 3 } \right)^4} = \)

\({\left( {1 + i} \right)^6} - {\left( {1 - i} \right)^6} = \)

Given that \({z^n} + \dfrac{1}{{{z^n}}} = 2\cos n\,\theta {\rm{ }}\) and by applying the Binomial theorem, show that: 

\({\rm{co}}{{\rm{s}}^3}\theta = \dfrac{1}{4}\left( {a\cos 3\theta + b\cos \theta } \right)\) where the values of \(a\) and \(b\) are?

By expressing \(1 + i\) and \(1 - i\) in the modulus/argument form and applying De Moivre's theorem then \({\left( {1 + i} \right)^{12}} + {\left( {1 - i} \right)^{12}} = \,\,?\)

Simplify \({(1 + i)^5} + {(1 - i)^5}\)