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I'm defining branch cut functions, and I'm using $\arg(z)$ as a building block. So I just spent an hour at the whiteboard assuming that $\arg(z)$ goes from $0$ to $2\pi$, and then I implement the code, and everything goes horribly wrong.

I just realized that the problem is that the the Arg function built into Mathematica goes from $-\pi$ to $\pi$. Is there anyway I can redefine Arg so that my code will work?

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5  
arg[z_] /; Im[z] < 0 := Arg[z] + 2 Pi; arg[z_] /; Im[z] >= 0 := Arg[z] – Artes Apr 12 '14 at 17:31
    
Thanks! a great ton! – David Apr 12 '14 at 17:50
1  
See also the arg I use here. – Jens Apr 12 '14 at 18:10
1  
RELATED – Vitaliy Kaurov Apr 12 '14 at 20:53
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arg[z_] := Pi+ArcTan[Re[z],Im[z]] – Daniel Lichtblau Apr 13 '14 at 19:15

Let me test several versions of the redefined arg:

x = RandomComplex[{-1 - I, 1 + I}, 1000000];

arg1[x_] := Mod[Arg@x, 2 π];
arg2[x_] := Arg[-x] + π;
arg3[z_] := π + ArcTan[-Re[z], -Im[z]];

Max[Abs[arg1[x] - arg2[x]], Abs[arg1[x] - arg3[x]]]
(* 8.88178*10^-16 *)

arg1[x]; // AbsoluteTiming
(* {0.16715, Null} *)

arg2[x]; // AbsoluteTiming
(* {0.154602, Null} *)

arg3[x]; // AbsoluteTiming
(* {0.090001, Null} *)

It is surprising to me that ArcTan works sufficiently faster then Arg.

However, I do not recommend to redefine the build-in Arg. Sometimes it works

Block[{Arg = arg3}, Arg[Exp[-I]]]
(* -1 + 2 π *)

and sometimes not (why?)

Block[{Arg = arg3}, Plot[Arg@Exp[I φ], {φ, -2 π, 2 π}]] 

enter image description here

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2  
The Block thing is an issue with the way Compile is used on the expression being plotted. You can fix it by setting the (undocumented?) Compiled option for Plot to False, or by moving the Block into the Plot statement. Either way, I think it's probably a bug and definitely worth a question of its own.... – Pillsy Nov 12 '15 at 18:23

From a comment by Artes, this seemed to solve the problem for the OP:

arg[z_] /; Im[z] < 0 := Arg[z] + 2 Pi; arg[z_] /; Im[z] >= 0 := Arg[z]
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