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The principle argument when no imaginary part

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Its not really a problem I was just wondering if the argument for any negative real number but no imaginary part was always = to pi?

    ie -1, -2,-3, -0.65.... is the arg(z)=pi for all these cases

    if so im guessing for positive real numbers with no imaginary part then Arg(z)=0

    and likewise for imaginary numbers with no real parts then Arg(z)= pi/2 and -pi/2 respectively.

    Thanks



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2012 #2
    You are using Arg(z) and arg(z) too loosely. Arg(z) for negative real numbers is [itex]\pi[/itex], and for positive real numbers, 0. arg(z) for negative real numbers is [itex]\pi + 2\pi k,\ k \in \mathbb{Z}[/itex], and for positive real numbers, [itex]0 + 2 \pi k,\ k \in \mathbb{Z}[/itex]. Remember that [itex]\text{Arg}(z) \in (-\pi,\pi][/itex]. You are correct for the imaginary numbers.

    But, yes, the principal argument is always those four values you listed. The argument has an infinite number of values.
     
  4. Apr 11, 2012 #3
    ok thanks just wanted to clarify. thankyou!!
     
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