The statement $z_n \to z_0 \neq 0$ is equivalent to $$ \lim_{n \to \infty} |z_n| = |z_0| \quad\text{and}\quad \lim_{n \to \infty} \mathrm{Arg} \ z_n = \mathrm{Arg} \ z_0. $$

I want to prove the above statement , where $z$ is a complex number and $\mathrm{Arg}$ is the principal argument of $z$.

I used the definition of limit and triangle inequality than I can get the first part.

For the second part, from what I know, $\mathrm{Arg} \ z$ is depend on which quadrant that the complex number lies on and for quadrant 1,4 it is Argz=arctan(y/x),for 2 it is arctan(y/x)+pi,for 3 it is arctan(y/x) -pi . I don’t know how to do the calculation when I try to prove this by using definition of limit.

  • $\begingroup$ How is $\mathrm{Arg} \ z$ acutally defined? In partilcular, what is the domain of $\mathrm{Arg}$? If we try to define an argument on the whole of $\mathbb{C} \setminus \{0\}$ then this statement is false. $\endgroup$ – Jendrik Stelzner Jan 4 '16 at 4:53
  • $\begingroup$ for quadrant 1,4 Argz=arctan(y/x),for 2 it is arctan(y/x)+pi,for 3 it is arctan(y/x) -pi $\endgroup$ – Gillian Cheung Jan 4 '16 at 4:56
  • $\begingroup$ if $f$ is injective, then $\lim x_n = l$ is equivalent to $\lim f(x_n) = f(l)$ $\endgroup$ – reuns Jan 4 '16 at 5:36
  • $\begingroup$ math.stackexchange.com/questions/214301/… i think i can use the above question to ask this question because they are approximately the same, is it correct? $\endgroup$ – Gillian Cheung Jan 4 '16 at 6:38
  • $\begingroup$ can any one give me some hint on how to prove the argument part $\endgroup$ – Gillian Cheung Jan 4 '16 at 7:39

The claim is not true: $$z_n = -1 + \dfrac{(-1)^n}{n}\,i$$ clearly converges to $-1$, but $\operatorname{Arg}(z_n) \approx \pi$ for even $n$ and $\approx -\pi$ for odd $n$.

(The implication in the other direction is valid though.)

  • $\begingroup$ If Arg z only take from -pi/2 to pi/2, then the claim will be true? anyway thank you for your help $\endgroup$ – Gillian Cheung Jan 5 '16 at 1:44

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