Comments on: A Visual, Intuitive Guide to Imaginary Numbers http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ Learn Right, Not Rote. Fri, 03 Feb 2012 19:38:49 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: kalid http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-36887 kalid Mon, 09 Jan 2012 21:29:33 +0000 http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-36887 @sqlguy: Yep, you got it :). Raising i to a fractional exponent (like the square root, 0.5) will give a complex number halfway between real and imaginary, i.e. at a 45 degree angle. Basically, you need to get to i (90 degrees) in 2 steps, each of 45 degrees. @sqlguy: Yep, you got it :) . Raising i to a fractional exponent (like the square root, 0.5) will give a complex number halfway between real and imaginary, i.e. at a 45 degree angle. Basically, you need to get to i (90 degrees) in 2 steps, each of 45 degrees.

]]> By: sqlguy http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-35064 sqlguy Thu, 05 Jan 2012 15:08:45 +0000 http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-35064 But I guess I'm still confused. If raising i to a whole exponent gets you either a completely real or completely imaginary number, does raising i to a fractional exponent give you a complex number? But I guess I’m still confused. If raising i to a whole exponent gets you either a completely real or completely imaginary number, does raising i to a fractional exponent give you a complex number?

]]> By: sqlguy http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-35057 sqlguy Thu, 05 Jan 2012 14:49:25 +0000 http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-35057 @kalid, @Rich Thanks guys! I guess I thought that since the square root of the negative unit in the real dimension was in a new dimension orthogonal to the real, that the square root of the negative of THAT unit might be in a new dimension mutually orthogonal to the complex plane. But actually raising i to any power has rotational effects in the complex plane, is that right? @kalid, @Rich
Thanks guys! I guess I thought that since the square root of the negative unit in the real dimension was in a new dimension orthogonal to the real, that the square root of the negative of THAT unit might be in a new dimension mutually orthogonal to the complex plane. But actually raising i to any power has rotational effects in the complex plane, is that right?

]]> By: Rich http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-34638 Rich Wed, 04 Jan 2012 19:40:56 +0000 http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/#comment-34638 (actually, just realised this is the same answer as above (225): 1/sqrt(2) - i/sqrt(2) = (1-i)/sqrt(2) = ((1-i)^2) / 2 = (1-2i+i^2) / 2 = -2i/2 = - i !! (you wouldn't have thought it!) (actually, just realised this is the same answer as above (225):

1/sqrt(2) – i/sqrt(2) = (1-i)/sqrt(2) = ((1-i)^2) / 2

= (1-2i+i^2) / 2 = -2i/2 = – i !!

(you wouldn’t have thought it!)

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