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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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 Gib Bogle Posts: 42 Registered: 3/28/11
Re: f(z) + abs(z) = 0, where z is complex
Posted: Jul 10, 2011 7:45 PM

On 7/11/2011 7:06 AM, slawek wrote:
> U?ytkownik "Gordon Sande" napisa? w wiadomo?ci grup
> dyskusyjnych:ivcrqr\$f5g\$1@dont-email.me...
>
> On 2011-07-10 15:24:35 -0300, slawek said:
>

>>> I am looking for a numerical solution of f(z) + abs(z) = 0, where z
>>> is a complex number and f is the analytic function.
>>>
>>> Newton method does not work, because there is no derivative of the
>>> non-analytic abs(z).
>>>
>>> TIA
>>> slawek

>
>> What happens if one considers f(x,y) insead of f(z) and
>> sqrt(x^2+y^2) instead of abs(z) and uses a method suitable
>> for two real variables?

>
> We will have got two variables and only one equation.

You have two equations, the real part and the imag part.

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