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The Arg function

  1. Dec 17, 2012 #1
    We are doing complex numbers, polar coordinates, Euler's formula etc in class and there is this thing called the Argument (Arg) function. I am having trouble understanding what this is. This is a new concept and sadly my notes aren't much help. So could someone kindly answer the following questions.
    what is this function?
    How do I evaluate it?
    Where is it used?
    Are there any properties of this function I should know about?
    Anything for further reading is also appreciated but keep in mind that this is the first I am seeing all this stuff. No wikipedia please because math on wikipedia doesn't make sense to me with all those complicated equations and symbols that I have never seen before.

    Here is a sample question from my assignment just to make sure I am getting my message across properly. Feel free to use it as an example if you want.
    z1 = [itex]\pi[/itex]/8
    z2 = 3[itex]\pi[/itex]/4
    find Arg (z1z2)

    Answer is 7[itex]\pi[/itex]/8
    which just seems like simple addition so why go through all this trouble with Arg?
  2. jcsd
  3. Dec 17, 2012 #2


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    [tex]\text{Arg}(z)=-\imath \log \left( \dfrac{z}{|{z}|} \right)=2 \arctan \left(\dfrac{\Im{(z)}}{|{z}|+\Re{(x)}}\right)[/tex]

    The reason for introducing it is that complex numbers in geometric form are easier to multiply. This is because arg is a logarithm of the unit scaled numer. In particular

    Arg(a b)=Arg(a)+Arg(b)
    Last edited: Dec 17, 2012
  4. Dec 17, 2012 #3
    in the third expression what are those fancy symbols which look like J and R
  5. Dec 17, 2012 #4


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    Those are the real and imaginary parts of z.

    In your example, I believe it should read
    Arg(z1) = [itex]\pi[/itex]/8
    Arg(z2) = 3[itex]\pi[/itex]/4
    find Arg (z1z2)

    There are different ways to represent complex numbers. If we will be adding them the algebraic form is best
    z1=a1+i b1
    z2=a2+i b2
    z1+z2=(a1+a2)+i (b1+b2)

    If we are multiplying, the trigonometric (which is related to exponential) form is best
    z1=r1(cos θ1+sin θ1)
    z2=r2(cos θ2+sin θ2)
    z1 z2=(r1 r2)((cos (θ1+θ2)+sin (θ1+θ2)))

    notice that if
    z=r(cos θ+sin θ)
  6. Dec 17, 2012 #5
    My bad about the question. Thanks for help I understand all this now.
  7. Dec 17, 2012 #6


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    You missed out a factor of ##-i## there, I think.
  8. Dec 17, 2012 #7
    There is no need to memorize all those logarithm functions if you don't want to.

    The way I define Arg is just as the angle between a point and the x-axis. So in the following picture:


    the number [itex]\varphi[/itex] would be the argument.

    Now, using trigonometric formulas, you can find easy formulas for the argument. For example, let's say that you have the point 1+10i. This corresponds to (1,10). So if we work in the triangle (0,0), (1,0), (1,10), then we can find the argument by

    [tex]\tan \varphi= \frac{10}{1}=10[/tex]

    So then you can find [itex]\varphi[/itex] easily.

    It might be also good to remark that the principal argument is always a number in [itex](-\pi,\pi][/itex].
    So if you ever get an argument equal to [itex]3\pi[/itex], then the principal argument is [itex]\pi[/itex] (note that [itex]3\pi[/itex] and [itex]\pi[/itex] correspond to the same angle!).
  9. Dec 17, 2012 #8


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    ^yes thank you.
  10. Dec 17, 2012 #9
    This is not true for general a and b. At least, not if you take the principal argument. If you define argument as a multivalued function, then it is true.
  11. Dec 17, 2012 #10
    So basically Arg is just the angle [itex]\varphi[/itex] in rei[itex]\varphi[/itex] given by tan-1 b/a. a and b come from a+bi
  12. Dec 17, 2012 #11
    Yes, but you need to be careful with that. For example, when calculating the argument of i that way. You would get [itex]\tan^{-1}(\frac{1}{0})[/itex] which is meaningless. On the other hand, [itex]Arg(i)=\frac{pi}{2}[/itex] is easy to see.

    Also, you need to take care that [itex]Arg(z)[/itex] is always in the interval [itex](-\pi,\pi][/itex] (or whatever standard you have defined in class). The [itex]\tan^{-1}[/itex] function might not always do the right job for that (depending on how things are defined).

    So don't just memorize [itex]\tan^{-1}(\frac{b}{a})[/itex], but rather try to think if your answer makes sense.
  13. Dec 17, 2012 #12
    Alright. Thank you all for the help
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