solve z^2 = iZ (1 Viewer)

shimmerz_777

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i just tried it then, for some reason i got real and imaginary parts equaling 0, ill try again, mayb thats not the only answer
 

shimmerz_777

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on second thought i found a mistake,

let z = x + iy
(x + iy)^2 = i (x-iy)
x^2 - y^2 +2ixy = y + ix
equating real and imaginary

2xy = x.... 2y=1 y= 1/2

x^2 - y^2 = y
x^2 - 1/4 = 1/2

x= sqrt 3/4
 

suamara

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2xy=x

x(2y-1) =0

x=0 , y=1/2


ie x=0 is a solution as well
 

STx

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so how did u find the z=i solution?, i know it satisfies the equation though..
 

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