Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Prove that if the sequence $z_{n}$ converges to a nonzero limit $A$ which is not a negative real number, then $\arg z_{n}\to \arg A$, where finite number of terms of $z_{n}$ which may wanish are ignored and $\arg z $ is $-\pi< Arg z \le \pi$

My solution: by definition, there is $n_{0}=N(\epsilon)$ for which $|z_{n_{0}}-A|\le\epsilon$. So there is a neighborhood $N(A,\epsilon)$ which intersection with real negative axis is empty. So all $z_{n}$ with $n\ge n_{0}$ will be in this neighborhood. Let $\theta=|\arg z_{n}-\arg A|$ and $\epsilon<|A|$ . So $\epsilon=|A|\sin\theta$ or $\theta=\arcsin\frac{\epsilon}{|A|}$. And it goes to $0$ togehter with $\epsilon$.

Not sure if this is ok.

share|cite|improve this question
It seems off somehow; you should be using arg(z_n-A) which is not the same as arg(z_n)-arg(A). I think your formula for theta is off because of this. But: really all you need is that, with arg(z) defined as the angle of z when the origin and negative real axis have been removed, then arg(z) is continuous at points other than 0 or on the negative real axis. – coffeemath Oct 14 '12 at 12:50
@coffeemath I don't think I should be using $\arg(z_{n}-A)$, since it would give me nothing, because it could diverge even with convergent sequance. Maybe I should clarify , that I am using $\epsilon$ like radius of a circle centered at $A$ and express it like function of $\theta$. So this way I get decreasing and bounded sequence of arguments. – Mykolas Oct 14 '12 at 19:41
You start with A and then choose a small epsilon, small enough that the disc of radius epsilon around A doesn't meet the negative real axis (or the origin). That's fine. But I don't see how you get epsilon = |A|sin(theta). – coffeemath Oct 14 '12 at 20:14
up vote 1 down vote accepted

You have the assumption that $z_n$ converges to $A$ and $A$ is not the origin or on the negative real axis. You need to show that given $\epsilon>0$ there is $N$ such that $n>N$ implies $\arg(a_n)$ is within epsilon of $\arg(A)$. It's equivalent to show there is a $\delta$ such that if $z$ is within $\delta$ of $A$ then $\arg(z)$ is within $\epsilon$ of $\arg(A)$.

To do this the actual delta you use might need to be the min of some things. So first we can draw a circle around $A$ of radius say $\epsilon_1 <= \epsilon$, but such that the entire disk inside the circle is also away from the negative x axis. Next consider all the angles formed by points in this disk; a diagram reveals they will be within $\arcsin(\epsilon_1/|A|)$ of arg(A).

So if z is within $\arcsin(\epsilon_1/|A|)$ of A, we have the desired closeness


We do need to use monotonicity of $\arcsin(x)$ here at the end, and continuity, and $\arcsin(0)=0$.

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.