Limits at Infinity for the Argument Function in Complex Variables

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The discussion centers on finding the limit at infinity for the function h(z) = Arg(z) where z ≠ 0. The principal argument, Arg(z), is defined within the interval [-π, π) and varies based on the direction from which z approaches infinity in the complex plane. Participants clarify that because there are infinitely many directions (or angles) for approaching infinity in ℂ, the limit does not converge to a single value, rendering it undefined. The restriction of z ≠ 0 is due to Arg(z) being undefined at that point, not affecting the limit at infinity. Ultimately, the conclusion is that the limit at infinity for Arg(z) is indeed undefined due to the multiple values of θ possible.
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! :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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