brad89 Posted June 6, 2005 Share Posted June 6, 2005 I did a google and found out that the square root of i is [sqrt 2]/2 + [sqrt 2]/2i, but the reasoning behind it is stuff way out of my league. Can anybody simplify this to a reasoning easier to understand? Link to comment Share on other sites More sharing options...

timo Posted June 6, 2005 Share Posted June 6, 2005 It´s often usefull to write complex numbers a+bi in a polar coordinate way as r*exp(ip) with r and p both being reals. If you draw the complex plane in the usual way, r will be the distance of a+bi from the origin and p will be the counter-clockwise angle of this connection line to the positive Re-axis. The multiplication rule in this representation is z1 * z2 = r1*exp(ip1) *r2*exp(ip2) = (r1*r2) * exp(i(p1+p2) or in other words: Multiply the distances, add the angles. So for the square root of a complex number z=d*exp(ip) you are looking for a number z' =d' *exp(ip') such that d'*d' = d and p'+p' = p (you have to pay attention that the addition is modulo 2pi, so p/2 is as well a solution for p' as p/2+pi is). The imaginary number can be written as i = 1*exp(i*pi/2) so d=1 and p=pi/2. From d=1 it follows directly that d'=1 (negative distances are not possible). From p=pi/2 => p' = pi/4 or pi/4+pi. Link to comment Share on other sites More sharing options...

BigMoosie Posted June 8, 2005 Share Posted June 8, 2005 A simple explaination: [math]i[/math] will be our imaginary constant. [math]\sqrt{i} = a + bi[/math] All solutions can be expressed like that where [math]a[/math] and [math]b[/math] are real numbers. [math]i = a^2 - b^2 + 2abi[/math] We can break that apart into 2 equations, with the reals in one and the imaginaries in the other, we get: (I) [math] 0 = a^{2} - b^{2}[/math] and (II) [math] i = 2abi[/math] (II) [math] 1 = 2ab[/math] Normal algebraic manipulation with (I) and (II) using quadratic formula and such will conclude that: [math]a = \frac{1}{2}[/math] and [math]b=\frac{\sqrt{2}}{2}[/math] Thus [math]\sqrt{i} = \tfrac{1}{2} + \tfrac{\sqrt{2}}{2}[/math] Link to comment Share on other sites More sharing options...

DQW Posted June 17, 2005 Share Posted June 17, 2005 A simple explaination: [math]i[/math] will be our imaginary constant. [math]\sqrt{i} = a + bi[/math] All solutions can be expressed like that where [math]a[/math] and [math]b[/math] are real numbers. [math]i = a^2 - b^2 + 2abi[/math] We can break that apart into 2 equations' date=' with the reals in one and the imaginaries in the other, we get: (I) [math'] 0 = a^{2} - b^{2}[/math] and (II) [math] i = 2abi[/math] (II) [math] 1 = 2ab[/math] Normal algebraic manipulation with (I) and (II) using quadratic formula and such will conclude that: [math]a = \frac{1}{2}[/math] and [math]b=\frac{\sqrt{2}}{2}[/math] Thus [math]\sqrt{i} = \tfrac{1}{2} + \tfrac{\sqrt{2}}{2}[/math] I think you mean [imath]a = b = 1/\sqrt{2} [/imath] and [imath]\sqrt{i} = 1/\sqrt{2} + i/\sqrt{2}[/imath] Link to comment Share on other sites More sharing options...

BigMoosie Posted June 18, 2005 Share Posted June 18, 2005 Ooops... I think my brain farted, thanks for correcting me. I think instead of [math]\sqrt{i} = \tfrac{1}{2} + \tfrac{\sqrt{2}}{2}[/math] I meant [math]\sqrt{i} = \tfrac{1}{\sqrt{2}} + \tfrac{\sqrt{2}i}{2}[/math]... it was close Link to comment Share on other sites More sharing options...

bloodhound Posted July 25, 2005 Share Posted July 25, 2005 blah. you get two solutions in this example, which are intiutively +/- blah blah. generally when you take a nth root of a complex number it will spit out n values. but its been so long i am not entirely sure ;_; maybe someone can reassure me. Link to comment Share on other sites More sharing options...

twanvitas Posted December 24, 2006 Share Posted December 24, 2006 I think you mean [imath]a = b = 1/\sqrt{2} [/imath] and [imath]\sqrt{i} = 1/\sqrt{2} + i/\sqrt{2}[/imath] Actually there are two solutions also [imath]a = b = -1/\sqrt{2} [/imath] Link to comment Share on other sites More sharing options...

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