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Topic: f(z) + abs(z) = 0, where z is complex
Replies: 18   Last Post: Jul 11, 2011 9:23 AM

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 slawek Posts: 25 Registered: 12/12/08
Re: f(z) + abs(z) = 0, where z is complex
Posted: Jul 10, 2011 3:35 PM

U?ytkownik "Thomas Koenig" napisa? w wiadomo?ci grup

>Take z = x + I*y, split f(x) into its real part f1(x,y) and its
>imaginary part I*f2(x,y), then calculate the zeros of the
>functions

>f1(x,y) - sqrt(x^2+y^2)
>f2(x,y)

>using a two-dimensional Newton method.

It should work, but I think that the substitiution z = abs(z) exp(i Arg(z))
would help a little.

Date Subject Author
7/10/11 slawek
7/10/11 Gordon Sande
7/10/11 slawek
7/10/11 Gordon Sande
7/10/11 slawek
7/11/11 Gib Bogle
7/11/11 slawek
7/11/11 Gordon Sande
7/11/11 slawek
7/10/11 Gib Bogle
7/11/11 slawek
7/11/11 Gib Bogle
7/11/11 slawek
7/11/11 slawek
7/10/11 Thomas Koenig
7/10/11 slawek
7/10/11 Gib Bogle
7/11/11 Eli Osherovich
7/11/11 Eli Osherovich