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Monday, February 8, 2016

2016/010) Solve |e^{it} - 1| = 2 for -\pi<\theta<=\pi

e^ {it} = \cos t + i \sin t
so e^{it} - 1 = (\cos t-1) + i \sin t
take mod and square
(\cos t-1)^2 + \sin ^2 t = 4
or \cos^2 t - 2 \cos t + 1 + \sin ^2 t = 2
or 2 - 2\cos t = 4 or cos t = - 1 and hence t = \pi

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