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Prove that $f: \{z \in \mathbb C : |z|=1\} \setminus \{-1\} \to \mathbb R$, $f(z)=\arg(z)$, is continuous.

I am lost with this exercise, I don't know which identity to use in order to show that if $z \to z_0$ in this set, then $\arg(z) \to \arg(z_0)$.

Sorry if I am asking something trivial, but I don't see why the point $-1$ is excluded from this set. And, considering the general case of the complex plane as the domain: if a sequence of complex numbers $\{z_n\}_{n \in \mathbb N}$ is converging to some number $z$, wouldn't this mean that the argument of $z_n$ is getting closer and closer to the argument of $z$? I can't see why this is not true.

Could someone please enlighten me?

dfeuer
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user100106
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  • $-1$ is excluded because the argument you're looking at is (apparently) the one that approaches $\pi$ from one side and $-\pi$ from the other. – dfeuer Jan 02 '14 at 01:26
  • Hint: try using the fact that for all $|z|=1$,$z\neq -1$, there exists some disc $D_\epsilon(z)$ which doesn't intersect $(-\infty,0]$. – Jonathan Y. Jan 02 '14 at 01:49

3 Answers3

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The function $\arg\colon \mathbb C^* \to \mathbb R/(2\pi)$ is one of the most important functions in analysis and gives the polar angle of a point $z \in \mathbb C^*\,$ "up to a multiple of $2\pi$".

Its principal value $\DeclareMathOperator{\Arg}{Arg}\Arg(z)$ is real-valued and is defined on the set $$\mathbb C^{-\cdot}:=\mathbb C\setminus \{z = x + iy \mid x\leq 0,\, y=0 \}.$$ The principal value $\Arg(z)$ is the representive of $\arg(z)$ lying in the interval $\, ]-\pi,\pi[\,$, and can be expressed in terms of the $\arctan$ function familiar from calculus as follows: $$\Arg(z)=\cases{ -{\pi\over2}+\arctan{x\over -y} & $(y<0)$ \cr \arctan{y\over x} & $(x>0)$ \cr {\pi\over2}-\arctan{x\over y} & $(y>0)$\cr} \tag{1}$$ It is easy to check that $\Arg$ is well defined on $\mathbb C^{-\cdot}$ by $(1)$. As any $z\in \mathbb C^{-\cdot}$ has a neighborhood in which a single alternative of $(1)$ applies it follows that $\Arg$ is continuous (even $C^\infty$) in all of $\mathbb C^{-\cdot}$, and a fortiori $\Arg$ is continuous on $\mathbb C^{-\cdot}\cap S^1$.

Christian Blatter
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To answer your second question, consider the sequence $x_n=\left(-\frac{1} 2\right)^n i$. This clearly converges to $0$, but the argument oscillates (does not converge).

copper.hat
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dfeuer
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  • This is wrong, $|x_n| = {1 \over 2^n}$, not one. – copper.hat Jul 02 '23 at 20:37
  • @copper.hat I was answering the general question of whether the arguments of the terms of a converging sequence have to converge to the argument of the limit. The answer is yes, except when the limit of the sequence is 0. – dfeuer Jul 03 '23 at 21:07
  • My apologies, I did not read carefully. – copper.hat Jul 03 '23 at 21:09
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See any point on the set $A = \{z \in \mathbb{C}: |z| = 1\}$ as $z = e^{i \theta}$ where $-\pi \le \theta < \pi$. $e^{- i \pi} = -1$

$arg (z) = \theta $ and $\log(z) = -i arg(z)$ where $z \in A$

On the set $A - \{-1\}$ the complex logarithm $\log(z)$ is single valued and continuous.

Take any point $z \in A$ $z \neq -1$. Consider the sequence $\{z_n\}$ in $A$ converges to $z$. We shall see the sequence $\{\log(z_n)\}$ will converge to $Arg(z)$ because of the continuity of the function $\log(z)$

For $z = -1$ the function $\log(z)$ is multivalued and not continuous. So $arg(z)$ will not also be continuous at $z = -1$.

Supriyo
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  • Isn't the argument something more like $-i$ times the logarithm? I don't know too much about complex numbers but that would seem to make more sense. – dfeuer Jan 02 '14 at 03:50
  • Thank you for pointing out the mistake – Supriyo Jan 02 '14 at 10:46
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    Aren't we fundamentally shifting the problem to the logarithm function, this way? When you say it's continuous and single valued on $S^1\setminus{-1}$, can this be substantiated independently from the problem we're dealing with here? – Jonathan Y. Jan 02 '14 at 19:12
  • It is not independent from the given problem. We know $z = r e^{i \theta}$. In $A$, $r = |z| = 1$ and so $z = e^{i \theta}$. As $A - {-1}$ we have $-\pi < \theta < \pi$ and so $\theta = Arg(z)$. Now I am taking inverse to get the $Arg(z)$ and it is from the relation $\log(z) = -i Arg(z)$. Now using the continuity of $\log(z)$ I am trying to show the continuity of $Arg(z)$. – Supriyo Jan 03 '14 at 01:59
  • @HopelessFool but without continuity of $\mathrm{Arg}$ we wouldn't have continuity of $\log$. To put it another way, the two are equivalent; a proof of either shouldn't rely on the other unless an independent proof is given--in turn--for that. – Jonathan Y. Jan 03 '14 at 03:06
  • Here I have assumed the continuity of $\log(z)$ which gives answer for continuity of Arg(z). For getting a proof of continuity of $\log(z)$ independently of Arg(z) we may consider power series representation of $\log(z)$. Also we may consider $f(z) = \log(z)$ as a solution of $z = \exp(f(z))$. – Supriyo Jan 03 '14 at 04:59