Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to get Mathematica to evaluate the logarithm of a negative real number using the lower branch instead of the upper branch, so that while

In[1]:=   Log[3.2]
Out[1]:=  1.16315

I need

In[2]:=   Log[-3.2]
Out[2]:=  1.16315 - 3.14159 I

and not

Out[2]:=  1.16315 + 3.14159 I

I have already defined my own function loopLog that does this:

loopLog[x_: NumericQ] = If[Element[x,Reals], Conjugate[Log[x]]];

But I am not able to get it to perform any of the usual simplifications or manipulations using this function. For example, when I want to differentiate loopLog, I get

In[3]:=   D[loopLog[x],x]
Out[3]:=  If[x \[Element] Reals, Derivative[1][Conjugate][Log[x]]/x]

Instead of the much needed 1/x. What is the cleanest way to define such a logarithm function in Mathematica?

share|improve this question
x_: NumericQ this is a pattern that matches anything, but defaults to the symbol NumericQ if no argument is given. You surely meant x_?NumericQ – Rojolalalalalalalalalalalalala Dec 26 '12 at 9:46
Would be nice with some $BranchCut global variable that affects ArcTan and the rest as well, even nicer if it could be set to an arbitrary curve. – ssch Dec 26 '12 at 13:14
@Rojo Yes, bad programing on my part. It should have been x_?NumericQ – QuantumDot Dec 26 '12 at 17:14
@ssch: Well, inspired by whuber's answer, here's a Log with a branch cut along any curve of the form $z = re^{i\theta(r)}$: branchLog[z_, \[Theta]_] := With[{r = Abs[z]}, Log[z/Exp[I \[Theta][r]]] + I \[Theta][r]]. Then for ArcTan you can do ExpToTrig[TrigToExp@ArcTan[z] /. Log[z_] -> branchLog[z, \[Theta]]]... – Rahul Dec 27 '12 at 5:30
Sorry, the branch is actually along $z=-re^{i\theta(r)}$. – Rahul Dec 27 '12 at 11:06
up vote 12 down vote accepted

The question asks for a "cleanest way." Arguably, any solution that reduces the calculation to various cases is not very clean, and is probably inefficient too. These considerations suggest a simple, direct approach:

log0[x_] := -Log[1/x]

Checking a few cases (or a plot) easily confirms that log0 does what is intended with negative real numbers; e.g.,


1.16315 - 3.14159 I

Moreover, it does not change the behavior for any other complex numbers, as a contour plot of its imaginary part demonstrates:

ContourPlot[Chop[Im[Log[x + I y]] -  Im[log0[x + I y]]], {x, -2, 2}, {y, -2, 2},
  PlotPoints -> 100] 

Contour plot

Operators therefore produce the expected results:

D[log0[x], x]


Integrate[x log0[x], {x, -1, 1}]

$\frac{i \pi }{2}$

share|improve this answer

I never did this before (i.e. modify system function), but I just tried it, and it seems to work. But I just wanted to first say, that Log[z] as it works in Mathematica is the correct way. i.e. Log[z]=Log[Abs[z]]+k where k=0 when z>0 and k=I*Pi when z<0 so what you are asking to do is not the correct math way. But this is what I tried to change the definition

(*1.1631508098056809 + 3.141592653589793*I*)

Log[x_?(Element[#, Reals] && # < 0 &)] := Log[Abs[x]] - I*Pi
(*  1.1631508098056809 - 3.141592653589793*I  *)


(*  1.1631508098056809  *)

btw, the reason your D[loopLog[x],x] command did not work, is because you defined your function to accept only numerical x, but you are now taking derivative w.r.t x ? I do not think this will work.

share|improve this answer
"What you are asking to do is not the correct math way." They're just taking a slightly different branch cut from the usual one; it's hardly incorrect. There can be more than one right way in mathematics! :) – Rahul Dec 26 '12 at 11:46
_?(Element[#, Reals] && # < 0 &) can simply be written as _?Negative :) – R. M. Dec 26 '12 at 17:33
@NasserM.Abbasi Negative and Positive were introduced in version 1, when such conventions were probably not codified. At this point, it is probably left that way for backwards compatibility. – R. M. Dec 26 '12 at 20:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.