Limits at Infinity for the Argument Function in Complex Variables

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SUMMARY

The discussion centers on finding the limit at infinity of the function h(z) = Arg(z), where z ≠ 0, as presented in "Complex Variables Second Edition" by Stephen D. Fisher. Participants confirm that the principal argument Arg(z) is defined within the interval [-π, π) and that there are infinitely many values of θ corresponding to different directions in the complex plane. The consensus is that the limit is undefined because z can approach infinity along any line of arg(z) = C, leading to different values for C, thus confirming the limit does not exist.

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  • Understanding of complex variables and functions
  • Familiarity with the principal argument function Arg(z)
  • Knowledge of limits in complex analysis
  • Basic concepts of infinity in the complex plane
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Tsunoyukami
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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. h(z) = Arg z , z \neq 0
" (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 \theta_{o} in the interval [-\pi, \pi)that satisfies the equation z = |z|(cos\theta + isin\theta).

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:


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

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


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 z \neq 0 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! :)
 
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Hi Tsunoyukami! :smile:
Tsunoyukami said:
Does the restriction z \neq 0 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 …

Yes, it's simply because the function Arg is not defined for z = 0, and has nothing to do with limits at ∞.
First I would like to check my understanding of the function Arg(z). Arg(z) is the 0principal argument of z. It is the value \theta_{o} in the interval [-\pi, \pi)that satisfies the equation z = |z|(cos\theta + isin\theta).

That's correct.
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:

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

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


Is this the correct approach to this question or am I missing something crucial?

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:
 
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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