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Mathematics Calculus Wolfram Language

How to calculate the gradient of the arg of a complex exponential?

Brian McNeil

Posted 11 months ago

I am trying to find the gradient of the phase of complex exponential functions. For example: Grad[Arg[Exp[-I (x + z)]], {x, z}] Should be {1,1} for all x,z (Reals). However, instead I get: {-I E^(-I (x + z)) Derivative[1][Arg][E^(-I (x + z))], -I E^(-I (x + z)) Derivative[1][Arg][E^(-I (x + z))]} I think it has to do with the fact that `Arg[Exp[-I (x + z)]]` doesn't return (x+z). I assumed that was because it doesn't know that x,z are Reals. I also tried `Refine[Arg[Exp[-I (x + z)]], x \[Element] Reals && z \[Element] Reals]` but that doesn't return (x+z) either. Thanks!

POSTED BY: Brian McNeil

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Michael Rogers

Michael Rogers, Emory University

Posted 11 months ago

`ComplexEpand` is useful for expanding non-differentiable complex functions into differentiable real functions: Grad[ComplexExpand@Arg[Exp[-I (x + z)]], {x, z}] (* {-(Cos[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)) - Sin[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2), -(Cos[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)) - Sin[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)} *)

ComplexEpand is useful for expanding non-differentiable complex functions into differentiable real functions:

Grad[ComplexExpand@Arg[Exp[-I (x + z)]], {x, z}]
(*
{-(Cos[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)) - Sin[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2),
 -(Cos[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)) - Sin[x + z]^2/(Cos[x + z]^2 + Sin[x + z]^2)}
*)

POSTED BY: Michael Rogers

David Moore

David Moore, University of California San Diego

Posted 11 months ago

Here's another way to get the correct result, using the mathematical property that `Log[z]==Log[Abs[z]]+I Arg[z]` In[]:= Im[Grad[Log[Exp[-I (x + z)]], {x, z}]] Out[]= {-1, -1} The problem is that `Grad` is trying to take the complex derivative of `Arg`, which is not a complex differentiable function.