Question 482564
Use *[tex \LARGE i = e^{i\frac{\pi}{2}} = \cos(\frac{\pi}{2}) + i*\sin(\frac{\pi}{2})] (I know this uses a circular definition but you can still find the square root of i this way)


Take square root of both sides:


*[tex \LARGE \sqrt{i} = \sqrt{e^{i\frac{\pi}{2}}} = e^{i\frac{\pi}{4}} = \cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i]


Similar to real numbers, there can be a negative square root (e.g. sqrt(9) being equal to -3 as well as 3), but these roots are usually ignored.