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Homework Help: Limits at Infinity for the Argument Function in Complex Variables

  1. May 25, 2013 #1
    ℂI am working on an assignment and have come across a question that I'm not quite sure how to approach. Here it is, with my "solution" and reasoning:

    "[F]ind the limit at ∞ of the given function, or explain why it does not exist.

    24. [itex]h(z) = Arg z , z \neq 0[/itex]
    " (Complex Variables Second Edition by Stephen D. Fisher; pg. 41-42)

    First I would like to check my understanding of the function Arg(z). Arg(z) is the principal argument of z. It is the value [itex]\theta_{o}[/itex] in the interval [itex][-\pi, \pi)[/itex]that satisfies the equation [itex]z = |z|(cos\theta + isin\theta)[/itex].

    However, unlike the notion of positive and negative infinity in ℝ, there exist an infinite number of infinities in ℂ, each corresponding to a different value of θ. As a result, I argue that:

    [itex]\stackrel{lim}{z\rightarrow∞} h(z) = \stackrel{lim}{z\rightarrow∞} Arg z, z \neq 0 = Arg ∞ = θ [/itex]

    This is true because there are an infinite number of infinities in ℂ; so θ is any element of the set [itex]\theta_{o} = [-\pi, \pi)[/itex].

    Is this the correct approach to this question or am I missing something crucial? This question is a bit "tricky" for me. Does the restriction [itex]z \neq 0 [/itex] limit the solution (ie. does it remove a particular solution from the set described above, or is that simply because the function Arg is not defined for z = 0? (I believe its the latter, but better to be sure!)

    Thanks for any and all responses! :)
  2. jcsd
  3. May 25, 2013 #2


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    Hi Tsunoyukami! :smile:
    Yes, it's simply because the function Arg is not defined for z = 0, and has nothing to do with limits at ∞.
    That's correct.
    That's basically correct, but it would be more straightforward to say that z can approach ∞ along any line of argz = C for any constant 0 ≤ C < 2π, and the limit along that line will be that constant, C …

    since that applies for any such C, the limit is undefined. :wink:
  4. May 26, 2013 #3
    Thanks a lot!

    After reading "the limit is undefined" I thought to myself: why didn't I realize that? Thanks again! :)
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