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 $t = \pi$