Both are in $ (0, \pi) $, but since the maximum imaginary part is positive and corresponds to $ \frac{\sqrt2}2 $, and the angle is often taken as the acute one in such contexts, but the problem asks for $ \theta $ such that $ \sin \theta = \textmax imag part $, and $ \theta $ is to be found. However, $ \frac{\sqrt2}2 = \sin \frac\pi4 = \sin \frac3\pi4 $, but $ \frac3\pi4 $ is larger and still in range. But since the root has argument $ \frac3\pi4 $, and the imaginary part is $ \sin \frac3\pi4 = \frac{\sqrt2}2 $, the value $ \theta $ such that $ \sin \theta = \frac{\sqrt2}2 $ and corresponds to the phase is naturally $ \frac\pi4 $, but the problem says "expressed as $ \sin \theta $", and to find $ \theta $. However, since $ \sin \theta = \frac{\sqrt2}2 $, - Silent Sales Machine