Question 523689
You can't claim that *[tex \LARGE i^{\frac{1}{2}} = \frac{1}{2}i]. Instead, what you could do is rewrite i using Euler's formula (which is derived from Taylor series):


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


Then,


*[tex \LARGE i^{\frac{1}{2}} = e^{i \frac{\pi}{4}}]


*[tex \LARGE = \cos {\frac{\pi}{4}} + i*\sin{ \frac{\pi}{4}}]


*[tex \LARGE = \frac{\sqrt{2}}{2} + \frac{sqrt{2}}{2}i]

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Another way to do it is to say that


*[tex \LARGE \sqrt{i} = a+bi]


*[tex \LARGE i = (a+bi)^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + 2abi]


We can equate real and imaginary coefficients. The real part is a^2 - b^2 which is equal to 0 (since the real part of i is 0), so a^2 = b^2. Also, by equating the imaginary parts, 2ab = 1, so ab = 1/2. It follows that a = b = sqrt(2)/2 works (you can check as well), so


*[tex \LARGE \sqrt{i} = \frac{2}{2} + \frac{2}{2}i]


Update: Since a and b are real numbers, then the real part of the expression we are interested in is a^2 - b^2. If this is equal to i (think of it as 0+i), then a^2 - b^2 = 0 --> a^2 = b^2. The imaginary part of the expression is 2ab, which is equal to 1 (think, i = 2abi, 2ab = 1). Then solve for a and b.