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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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Eli Osherovich

Posts: 37
Registered: 6/5/05
Re: f(z) + abs(z) = 0, where z is complex
Posted: Jul 11, 2011 2:00 AM
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On 07/11/2011 08:59 AM, Eli Osherovich wrote:
> On 07/10/2011 09:24 PM, slawek wrote:
>> 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

> This is something very similar to it.
> Google "brandwood complex gradient". However, this approach is
> essentially equivalent to treating a complex number as a pair of two
> real numbers. The only gain here that you can express all operations in
> terms of complex arithmetics (it does not buy you some extra FLOPs).

Well, the first line should be

There is ....

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