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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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    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).2012-10-14

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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 $\arcsin(0)=0$.