Thread: prove i^2 =-1
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Old January 7th, 2009, 09:38 PM
Math Major Math Major is offline
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I'm not sure I entirely understand the question, since it's asking you to prove a definition, but I'll give it a shot.

i is defined as the square root of -1. i^2 = (-1)^(1/2)^(2) = -1 ^(2/2) = -1 ^1 = -1.

So, given (u+iv)(x+iy), we can use the FOIL method to get ux + xvi + uyi + yvi^2 = ux + i(xv + uy) + yv * -1 = (ux -yv) + (xv+uy)
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