# Convergence of the arguments of a convergent sequence

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.

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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

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
$|arg(z)-arg(A)|<\epsilon$.
We do need to use monotonicity of $\arcsin(x)$ here at the end, and continuity, and $\arcsin(0)=0$.