Given that \(z = \cos \theta + i\sin \theta = {e^{i\theta }}\) and \(z - \frac{1}{z} = k\sin \theta \), the value of \(k\) is?
\(k=2i\)
Convert \(4\left( { - \cos \dfrac{\pi }{6} - i\sin \dfrac{\pi }{6}} \right)\) to the form \(r{e^{i\theta }}\).
\(4{e^{-i\,\dfrac{{5\pi }}{6}}}\)
Given that \({\rm{ }}z = \cos \theta + i\sin \theta = {e^{i\,\theta }}\) then \(z + \dfrac{1}{z} = \,\)?
\(2\cos \theta \)
\(2\sin \theta \)
\( - 2\cos \theta \)
\( - 2\sin \theta \)
If \(\dfrac{{1 + 3i}}{{1 - 2i}} = r{e^{i\,\theta }}\) then the values of \(r\) and \({\rm{ }}\theta {\rm{ }}\) are ?
\(r = \sqrt 2 ,\,\,\theta = \dfrac{{3\pi }}{4}\)
Given that \(z = \cos \theta + i\sin \theta = {e^{i\,\theta }}\) then \({\left( {1 + i} \right)^8} = \,\)?
\(16\)
Solve \({e^{ - ix}} = \dfrac{{\left( {2 + i} \right)\left( {1 - 3i} \right)}}{{5\sqrt 2 }}\) for \(0 < x < \pi \)
\(x = \dfrac{\pi }{4}\)
Convert \({\rm{ }}\,2\left( { - \cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right){\rm{ }}\) to the form \(r{e^{i\theta }}\).
\(2{e^{\dfrac{{2\pi \iota }}{3}}}\)
If \(\cos \dfrac{\pi }{6} = 1 - 2{\sin ^2}\dfrac{\pi }{{12}}\) then the complex number \(\dfrac{1}{2}\left( {\sqrt {2 + \sqrt 3 } + i\sqrt {2 - \sqrt 3 } } \right){\rm{ = }}\,\)?
\({e^{\dfrac{{\pi \iota }}{{12}}}}\)
