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Author Topic: Newton's Method on transformed equations  (Read 2533 times)
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Timeroot
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« Reply #15 on: March 21, 2010, 11:48:56 PM »

Thanks, reesej, it seems you can explain better than I. Also, I'd like to point out that - even if the direction didn't matter, and you always looked "east" or "west" - the derivative isn't just sgn(z). z=x+iy, and abs(z)=sqrt(x^2+y^2). So, if y is constant, d(abs)/dz=d(sqrt(y^2+x^2))/dx, which equals x/sqrt(y^2+x^2), or sqrt(x^2/(y^2+x^2)). At x=0, this is zero; in other words, for a purely imaginary value, the derivative in the real direction is zero, not sign(z).  cheesy
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