# Limits at Infinity for the Argument Function in Complex Variables

1. May 25, 2013

### Tsunoyukami

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! :)

2. May 25, 2013

### tiny-tim

Hi Tsunoyukami!
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.

3. May 26, 2013

### Tsunoyukami

Thanks a lot!

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