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?
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}]] 
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.... $\endgroup$ ArcTan is definitely slower than Arg in Mathematica 10.4, 10.3, 9.0 and 8.0.4. On 10.3 they timings are {0.016644, 0.025684, 0.090886} on my machine. $\endgroup$ {0.10717, 0.087667, 0.10717} in v10.3. The ratios between timings can be machine specific. I tested on my old laptop with Core 2 Duo under Linux. $\endgroup$ EDIT: See comment below by Daniel, his answer is better. I shied away from using ArcTan[] because the previous comments mentioned problems with it, but it seems to work fine... Also I have since noticed a problem with my version where pure reals also get a 2π shift.
So I stumbled upon this thead looking for the same answer as OP. While the answers here work, they are orders of magnitude slower than using the built in Arg[]. This is what I could come up with to solve my problem:
argVec[zVec_] := Module[{imShift}, imShift = Unitize[(Sign /@ Im[zVec] - 1)]*(2π//N); Arg[zVec] + imShift]; This is only about 2x slower than the Arg[] function and it was the best I could come up with. I built it specifically to work with lists, just like Arg[], but it works just as well with scalars. The //N is needed to maintain the speed since my input zVec is machine precision, and I want machine precision output, however you can remove the //N and you will also get integer output for integer input, at the cost of speed.
Cheers!
Mod: Mod[ArcTan[Re[z], Im[z]], 2.*Pi]. $\endgroup$ 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]
arg[z_] /; Im[z] < 0 := Arg[z] + 2 Pi; arg[z_] /; Im[z] >= 0 := Arg[z]$\endgroup$argI use here. $\endgroup$arg[z_] := Pi+ArcTan[Re[z],Im[z]]$\endgroup$