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Suppose my z = x + iy Now, if y=0, then z = x i.e it becomes a real number So, logically Arg[z] should be equal to zero for this case.

If I assign the numeric value for x, then Arg[z] produces zero in Mathematica easily. But if I do not assign the numeric value for x which is of course a real number always, then how to produce Arg[z]=0 for this case ? Because assigning or not assigning numeric value to x should not prevent Argument of z to be zero i.e Arg[z]=0.

Actually, my computation involves variables only i.e algebraic expressions, no numeric calculations. The final Mathematica output has a term Arg[r]. The parameter r , radius, is always real. So, Mathematica should understand that as r is always real number, so Arg[r] should be zero. Please guide me how to achieve this. (Pranjol Paul, Department of Mechanical Engineering, Indian Institute of Technology Guwahati, Assam, India)

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Note: x is (undefined) understood to be complex in Mathematica. Also be aware that the Arg of a real is not always 0, e.g. Arg[-1] is Pi...; you might want to look at $Assumptions etc, e.g. as in Refine[Arg[y], y >= 0] –  Pinguin Dirk Aug 21 '13 at 17:00
    
@PinguinDirk:Arg is only for positive side of x-axis and has significance only in complexes. –  Rorschach Aug 21 '13 at 18:19
    
@Blackbird, I am afraid I don't understand your comment. As reals are a subset of complexes, Arg is defined there too, see e.g. en.wikipedia.org/wiki/Argument_(complex_analysis). Thus, multiplying $i*i=-1$ you directly see that the real number $-1$ must have argument $\pi$ (as $i$ has $\pi/2$) –  Pinguin Dirk Aug 21 '13 at 19:14
    
@PinguinDirk:Arg[-n] will be Pi because the y-component sits on -x and in case of -x it makes 180 Degrees from origin starting from +ive x-axis.Tan[0]=Tan[Pi]...that's why Im is needed. –  Rorschach Aug 21 '13 at 19:33

2 Answers 2

To get rid of this unanswered question (and I do not really agree with @Blackbird's answer, see comments above), I shall put my comment in an answer:

Note: x is (undefined) understood to be complex in Mathematica. Also be aware that the Arg of a real is not always $0$, e.g. Arg[-1] is Pi...; you might want to look at $Assumptions etc, e.g. as in Refine[Arg[y], y >= 0]

Please comment if it doesn't help - if there are further questions!

(now let's hope someone will upvote this (or the other answer) so there is one less question on the list)

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+1 from me too.. –  Rorschach Oct 11 '13 at 9:38

In M all symbolics are considered to support Complex numbers and obviously Real are special cases of complexes.

z = x + y I

x + I y

Head[%]      (*Plus*)
Head[z /. {x -> 3, y -> 0}]

Integer (Not anymore complex)

Head[z /. {x -> 3, y -> 0.}]

Complex (*It considers only approx to zero as complex *)

Arg[3. + 0. I]

0 . (It return approx to Zero not zero)

And what is the meaning of calculating Arg if the expression itself is not Complex anymore if you substitute your Imaginary part as 0

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