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.