How to write the principal argument of $-2i$ ?
I cannot just write $-\pi/2$, altough it is obvious, I have to justify it somehow. Can I say that $\lim\limits_{x\to 0}\arctan\left({\frac {-2}x}\right)$ ?
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If it is obvious why cannot you "just write that"? It indeed is obvious, in particular if you already defined the principal argument as an angle in $;[0,2\pi);$ . – Timbuc Feb 25 '15 at 09:44
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I agree with @Timbuc. I would simply write "obviously $\operatorname{Arg}(-2i) = -\pi/2$", perhaps noting that it lies on the negative imaginary axis. – Kaj Hansen Feb 25 '15 at 09:45
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@Timbuc the interval is $(-\pi, \pi]$ – OBDA Feb 25 '15 at 09:45
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@OBDA That's fine just as well: exactly the same. – Timbuc Feb 25 '15 at 09:47
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@Timbuc OK, I agree, shall I delete the question ? – OBDA Feb 25 '15 at 09:50
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@OBDA Not at all. These things can help others as well with little doubts. In fact, I forgot to upvote your question, so now I do it. – Timbuc Feb 25 '15 at 09:50
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@OBDA It depends on how you calculate the principal argument, but usually it is done by atan$2$, from which it, of course, is obvious. – Eff Feb 25 '15 at 10:11
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By definition, for any branch of the logarithm function determines a branch of the complex logarithm and vice versa, via $$\arg z = \Im \log z$$ Now, if we take the branch of $\arg$ that takes values in $(-\pi, \pi]$, then for complex numbers $w$ with imaginary part in that range, we have $\log \exp w = w$. In particular, if we can find such a $w$ such that $\exp w = -2i$, then we will have $$\arg (-2i) = \Im \log (-2i) = \Im \log \exp w = \Im w.$$ Now, certainly $$w = \log 2 - \frac{\pi}{2} i$$ satisfies these conditions, so $$\arg (-2i) = -\frac{\pi}{2}$$ as desired.
NB that the argument doesn't depend on the real part of $w$, so it is enough to determine the imaginary part of a $w$ that meets the criteria.
Travis Willse
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