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Re: f(z) + abs(z) = 0, where z is complex
Posted:
Jul 11, 2011 2:00 AM


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 >> nonanalytic 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 ....



