Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A$.

share|cite|improve this question
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
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 $\displaystyle A=r e^{i \Phi}$.

share|cite|improve this answer
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 $\arg(A)$.

share|cite|improve this answer
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 $.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.