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I'm trying to understand this proof for why $\log z$ is not continuous:

\begin{align} \lim_{n \to \infty} \log \left(-1-\frac{i}{n}\right) & = \lim_{n\to\infty}\left(\log\left|-1-\frac{i}{n}\right|+i \arg\left(-1-\frac{i}{n}\right)\right) \\ & =\log|-1|+i(-\pi) \\& = -i\pi,\end{align}

whereas

$$\log(-1)=\log|-1| + i\arg(-1)=i\pi \neq -i\pi.$$

I don't understand why

$$\lim_{n\to\infty}i\arg\left(-1-\frac{i}{n}\right) = -i\pi \neq i\arg(-1) = i\pi,$$

but

$$\lim_{n\to\infty}\log\left|-1-\frac{i}{n}\right| = \log|-1| = 0.$$

The idea makes sense: since the limit is tending towards $-1$ from the negative axis, you'll get $-i\pi$, which is not equal to $i\pi$. But I don't get the notation.

Isn't

$$\lim_{n\to\infty}i\arg\left(-1-\frac{i}{n}\right) = i\arg(-1) = i\pi?$$

Kenny Wong
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Anonymous73648
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    Would you mind sharing with us which definition of $\log$ you are using? – Another User May 27 '23 at 13:10
  • $\log(z) = \log(re^{j \theta}) = \log(r) + j \theta$ But $\theta$ has a discontinuity at $\pm \pi$ or at $0,2\pi$. Sine $e^{j \theta}$ is periodic, we cannot unique assign the value of $\theta$ without discontinuity. This is same for any function periodic. From this, you can see that you can move the location of discontinuity to any value of $\theta$ –  May 27 '23 at 13:11
  • @AnotherUser Of course. I'm using the principal logarithm: Log(w) = log|w| + iArg(w) – Anonymous73648 May 27 '23 at 13:13
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    And which definition of $\operatorname{Arg}$ are you using? – Another User May 27 '23 at 13:14
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    The trouble is in the fact that $\mathrm{Arg}$ is not continuous – FShrike May 27 '23 at 13:15
  • @AnotherUser Principal argument: $-\pi < Arg(w) \leqslant \pi$ – Anonymous73648 May 27 '23 at 13:16
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    "Isn't ..." - there, you assume $\arg$ is continuous. But this is not true – FShrike May 27 '23 at 13:16
  • Basically $\log(-1-\frac{i}{n}) \rightarrow -i\pi$ and $\log(-1+\frac{i}{n}) \rightarrow i\pi$. This is the discontinuity. –  May 27 '23 at 13:17
  • Consider the assertions $\lim_{n\to\infty}i\operatorname{arg}\left(-1-\frac in\right)=-i\pi$ and $i\operatorname{arg}(-1)=\pi$. Which one of them don't you understand? – Another User May 27 '23 at 13:20
  • @AnotherUser I think my misunderstanding must come from the limit. My thought process is , $\lim_{n \to \infty}iArg(-1- \frac{i}{n})=\lim_{n \to \infty}iArg(-1)$ because $\lim_{n \to \infty}(-1 - \frac{i}{n}) = \lim_{n \to \infty}(-1)=-1$. So I imagine there's something wrong with that assumption (Sorry for the slow responses btw) – Anonymous73648 May 27 '23 at 13:32
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    The assertion $i\operatorname{Arg}\left(-1-\frac in\right)=\lim_{n\to\infty}i\operatorname{Arg}(-1)$ assumes that the $\operatorname{Arg}$ function is continuous. But it is not. – Another User May 27 '23 at 13:37
  • Think I'm getting it now. Thank you so much to all of you for your help! – Anonymous73648 May 27 '23 at 14:06

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