Question 194801
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From the positive x-axis to the positive y-axis: *[tex \LARGE \frac{\pi}{2}], then from the positive y-axis to the negative x-axis another *[tex \LARGE \frac{\pi}{2}] for a total of *[tex \LARGE \frac{2\pi}{2} = \pi].  Then one more quarter circle, *[tex \LARGE \frac{3\pi}{2}].  All the way back to the beginning: *[tex \LARGE \frac{4\pi}{2} = 2\pi].  Yep, *[tex \LARGE \frac{5\pi}{2}] is the same thing as *[tex \LARGE \frac{\pi}{2}].  In fact, all angles of the form *[tex \LARGE \theta + 2k\pi] are the same angle *[tex \LARGE \forall\ k\ \in\ \I]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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