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Suppose the sequence $z_{n}$ converges to a nonzero limit $A$ and let $\Phi_{n}$ be any sequence of values of $Arg (z_{n})$ satisfying the inequality $$|\Phi_{m}-\Phi_{n}|<\pi$$ for $m>N, n>N$. Prove that $\Phi_{n}$ converges to one of the values of $Arg(A)$.

I really couldn't find the way to start this. In my previuos question I asked about a similar proof. Maybe this should have something in commun.

My idea for the case where $A$ is not negative real number: let $Arg(z_{m})=\arg z_{m}+2k\pi$ and $Arg(z_{n})=\arg z_{n}+2t\pi$. Since $|\Phi_{m}-\Phi_{n}|<\pi$ and $\arg z_{m}\to \arg A$ and $\arg z_{n} \to \arg A$, $t=k=C$ so $\lim_{n \to \infty} \Phi_{n} = \arg A + 2C\pi$, which would be the value of some of the arguments of $A$.

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Suggestion: Split the proof into two cases, (1) $A \notin \mathbb{R}_- = \{z | \text{Im} z = 0, \ \ \text{Re} z \leq 0\}$, and (2) $A \in \mathbb{R}_-$. –  copper.hat Oct 15 '12 at 16:24
    
Maybe you mean to include that capital phi_m are values of arg(A). Once this is assumed, the rest should resemble an answer I gave to the similar question, since in approaching A the difference mod 2Pi will eventually be zero anyway. Then you can use any angle containing A to determine which branch of arg to use. –  coffeemath Oct 15 '12 at 16:53

3 Answers 3

up vote 2 down vote accepted

You can write $\displaystyle z_n=r_ne^{i \Phi_n}$. As $(z_n)$ converges to $A$, then $\displaystyle \left(\frac{z_n}{r_n} \right)=(e^{i\Phi_n})$ converges to $\displaystyle \frac{A}{r}$. So $\displaystyle (e^{i \Phi_n})$ is a Cauchy sequence: $\displaystyle |e^{i \Phi_n}- e^{i\Phi_m} | \underset{n,m \to + \infty}{\longrightarrow} 0$. But $\displaystyle |e^{i \Phi_n}- e^{i\Phi_m} |= \left| e^{i \frac{\Phi_n+ \Phi_m}{2}} \left( e^{i \frac{\Phi_n-\Phi_m}{2}}- e^{-i \frac{\Phi_n-\Phi_m}{2}} \right) \right|=2 \left| \sin \left( \frac{\Phi_n-\Phi_m}{2} \right) \right|$ and $\displaystyle \frac{\Phi_n-\Phi_m}{2} \in ]-\pi/2,\pi/2[$ so by composing by $\arcsin$ we get: $\displaystyle |\Phi_n-\Phi_m| \underset{n,m \to + \infty}{\longrightarrow} 0$.

Thus, $(\Phi_n)$ is a Cauchy sequence and converges to some $\Phi$. So $\displaystyle \left(\frac{z}{r_n} \right)$ converges to $\displaystyle e^{i \Phi}= \frac{A}{r}$, hence $\displaystyle A=r e^{i \Phi}$.

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thankyou @Seirios. Just could you clarify this for me please: $\left| e^{i \frac{\Phi_n+ \Phi_m}{2}} \left( e^{i \frac{\Phi_n-\Phi_m}{2}}- e^{-i \frac{\Phi_n-\Phi_m}{2}} \right) \right|=2 \left| \sin \left( \frac{\Phi_n-\Phi_m}{2} \right) \right|$ –  Mykolas Oct 15 '12 at 16:49
    
For $\theta \in \mathbb{R}$, $\sin(\theta)= \frac{1}{2} \left( e^{i \theta}- e^{-i \theta} \right)$. –  Seirios Oct 15 '12 at 17:20
    
Yes @Seirios, but I don't understand where $e^{i\frac{\Phi_{n}+\Phi_{m}}{2}}$ did go? –  Mykolas Oct 15 '12 at 18:27
    
It vanished because its modulus is one. –  Seirios Oct 15 '12 at 19:54
    
Ok, got it. Thnakyou @Seirios –  Mykolas Oct 15 '12 at 19:57

Supposing A is not on the negative real axis, it has argument say $\theta$ where $-\pi<\theta<\pi$. Then the arguments of the $z_{n}$ which also satisfy $-\pi<\arg(z_{n})<\pi$ will converge to $\arg(A)$ by what appears in the answer accepted for the article you reference as question.

And if it happens that A is on the negative real axis you can simply work with -A which is on the positive axis, and use that $z_{n}\to-A$ iff $(-z_{n})\to A$.

In either case, you have your sequence $z_{n}$ such that one choice of their arguments converges to the argument of $A$. If you now throw in your inequality that the randomly chosen arguments of the $z_{n}$ are all within $\pi$ of each other (for $n,m>N$ etc), then the differences of these arguments are all $0 \mod 2\pi$, so that since they are less thn $\pi$ they are all the same for sufficiently large $n$. Whatever this limiting value of $\arg(z_{n})$ is, it is certainly one of the values of $\arg(A)$.

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thankyou @coffeemath –  Mykolas Oct 15 '12 at 19:58
    
My reply was dashed off. The theta in first paragraph should be said to be -pi < theta < pi. I also should edit the LaTex into it... –  coffeemath Oct 15 '12 at 20:19

Here is a useful bound: Suppose $|x| < \pi$. Then $1-\cos x \geq \frac{1}{24} x^2$. To see this, note that $\cos x \geq 1 - \frac{x^2}{2!} + \frac{x^4}{4!}$ which gives $1-\cos \theta \geq \frac{x^2}{2} (1-\frac{x^2}{12})$. Since $\pi < \frac{22}{7}$ (see here for a simple, elementary proof), we have $x^2 < \pi^2 < (\frac{22}{7})^2 < 11$. A simple rearrangement gives $1-\frac{x^2}{12} > \frac{1}{12}$ from which the result follows. We also have $|e^{ix} -1 | \geq |\cos x -1| \geq \frac{1}{24} x^2$.

Since $z_n \to A$, we have $|z_n| \to |A|$. Letting $u_n = \frac{z_n}{|z_n|}$, we have $u_n \to \frac{A}{|A|}$. Furthermore, since $\text{Arg}\, z_n = \text{Arg}\, u_n$, we have $u_n = e^{i \Phi_n}$. Then $|u_n-u_m| = |e^{i \Phi_n}-e^{i \Phi_m}| = |e^{i (\Phi_n-\Phi_m)} -1|$. Since $|\Phi_{m}-\Phi_{n}|<\pi$, we can use the above estimate to get $|u_n-u_m| \geq \frac{1}{24} (\Phi_{m}-\Phi_{n})^2$. Since $u_n$ is Cauchy, it follows from this estimate that $\Phi_n$ is also Cauchy, hence it converges to some value $\hat{\Phi}$. Since $u_n \to \frac{A}{|A|}$ and $z \mapsto e^z$ is continuous, it follows that $\hat{\Phi} \in \text{Arg}\, \frac{A}{|A|} = \text{Arg}\, A $.

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