Special Results involving Exponential Forms Q4

Special Results involving Exponential Forms Q4

Show that \({{e}^{\mathrm{i}\theta }}-1=\left( 2\sin \frac{\theta }{2} \right){{e}^{\mathrm{i}\left( \frac{\theta }{2}+\frac{\pi }{2} \right)}}\).

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Answer:\( e^{i\theta} - 1 = \left(2\sin\frac{\theta}{2}\right)e^{i\left(\frac{\theta}{2} + \frac{\pi}{2}\right)} \) (shown)\)

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