M Mumma Member Joined May 19, 2004 Messages 586 Location Sydney Gender Male HSC 2006 Aug 6, 2006 #1 Z is conjugate z z2 = iZ
S shimmerz_777 Member Joined Sep 19, 2005 Messages 130 Gender Male HSC 2006 Aug 6, 2006 #2 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
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
S shimmerz_777 Member Joined Sep 19, 2005 Messages 130 Gender Male HSC 2006 Aug 6, 2006 #3 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
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
S shimmerz_777 Member Joined Sep 19, 2005 Messages 130 Gender Male HSC 2006 Aug 6, 2006 #4 ha ive done it again with u
S suamara New Member Joined Jul 9, 2006 Messages 4 Gender Male HSC N/A Aug 6, 2006 #5 2xy=x x(2y-1) =0 x=0 , y=1/2 ie x=0 is a solution as well
S shimmerz_777 Member Joined Sep 19, 2005 Messages 130 Gender Male HSC 2006 Aug 6, 2006 #6 where does the z=i come from? EDIT-- oh i see dont worry
STx Boom Bap Joined Sep 5, 2004 Messages 473 Location Sydney Gender Male HSC 2006 Aug 6, 2006 #7 so how did u find the z=i solution?, i know it satisfies the equation though..
S shimmerz_777 Member Joined Sep 19, 2005 Messages 130 Gender Male HSC 2006 Aug 6, 2006 #8 check suamara's line