# Why ArcTan[1, 0. I ] yields -1.5708+0. I?

Bug introduced in 5.2 or earlier and persisting through 12.0.0

Why ArcTan[1, 0. I ] yields -1.5708+0. I ? The result should equal to 0.

Is this a bug?

• it does look like a bug. For a work around use exact zero ArcTan[1, 0*I] instead of non-exact zero ArcTan[1, 0.*I] – Nasser Aug 24 at 5:30
• @Nasser The original expression I calculated is like ArcTan[1, p q], where p is a MachinePrecision complex number and q is a real number. Let q be 0, it yeilds -1.5708+0. I, but it should be 0. And then I found this is because the result of ArcTan[1, 0. I ] is -1.5708+0. I during the calculation. – Hallow_Juan Aug 24 at 10:20
• @xzczd I have tried to plot Re[ArcTan[1, (x + I y)]] and Im[ArcTan[1, (x + I y)]] in xy plane, and there is no branch cut near x=0 and y=0. – Hallow_Juan Aug 24 at 10:24
• (1) The docs "define" ArcTan[x,y] in terms of quadrants, implying real arguments, but also give a formula valid for complex numbers, -I Log[(x + I y)/Sqrt[x^2 + y^2]]. (2) The MKL seems to define atan2 only for reals. So maybe WRI botched the implementation. The second argument 0. I seems to be a special case, since any small nonzero number yields a correct value. – Michael E2 Aug 24 at 10:40
• I added the bugs tag. There seems to be consensus (at least no dissent). Please report it to Wolfram Research. – Michael E2 Aug 25 at 14:53

It looks like a bug.

Perhaps

ArcTan[1, p q + $MinMachineNumber * I]  will be acceptable. The little noise is unlikely to be a numerical problem. The main pitfall is when p q == -$MinMachineNumber * I, which drops you back into the buggy case.

ArcTan[1. + 0. I, $MinMachineNumber*I] (* 0. + 0. I *)  Update: Runtime environments for existing code You can overload ArcTan with code like this: InternalInheritedBlock[{ArcTan}, Unprotect[ArcTan]; ArcTan[x_, y_] := func[x, y]; Protect[ArcTan]; ArcTan[1., 0. I] ]  There are several issues with doing this, and the simplest thing to do is to understand your particular use-case and choose an appropriate compromise. For instance, here is one possibility: ClearAll[runWithNewArcTan]; SetAttributes[runWithNewArcTan, HoldAll]; runWithNewArcTan[code_] := InternalInheritedBlock[{ArcTan}, Unprotect[ArcTan]; (* keeps packed arrays from being unpacked *) ClearAttributes[ArcTan, Listable]; (* makes all ArcTan[x,y] results Complex; * vectorized formula means it still works on lists * however, the formula can have rounding errors when y == 0 * of around$MachineEpsilon in magnitude *)
ArcTan[x_, y_] /; ! FreeQ[{x, y}, _Complex] :=
-I Log[(x + I y)/Sqrt[x^2 + y^2]];

Protect[ArcTan];

(* run code *)
code
];


Here is another:

ClearAll[runWithNewArcTan];
SetAttributes[runWithNewArcTan, HoldAll];
runWithNewArcTan[code_] :=
InternalInheritedBlock[{ArcTan},
Unprotect[ArcTan];
(* keeping Listable attribute means packed arrays will be unpacked *)

(* fix just the buggy values; the patterns tests are not
* vectorized, so the Listable attribute will unpack packed arrays
* even when neither definition below is used. *)
ArcTan[x_, y_] /;
Precision[{x, y}] === MachinePrecision &&
! FreeQ[{x, y}, _Complex] && Positive[x] && y == 0 := 0. + 0. I;
ArcTan[x_, y_] /;
Precision[{x, y}] === MachinePrecision &&
! FreeQ[{x, y}, _Complex] && Negative[x] && y == 0 := Pi + 0. I;
Protect[ArcTan];

(* run code *)
code
];


The last one would work well in code that does not use packed arrays. I haven't thought of a simple way that would work exactly like ArcTan[x, y]` but fix the bug.