Talk:Arctangent Logarithmic Formulation

Good as a first stab, but notice needs to be made of the fact that arctan is a multifunction. I call attention to Definition:Arctangent, for instance. Similar applies to other inverse trig functions. --prime mover (talk) 06:57, 14 December 2012 (UTC)


 * The logarithm is also a multifunction; as it happens, the identity holds as an equality of multifunctions. I have the impression though that insufficient caution with the peculiarities of complex logarithm is taken... It really cannot be well-defined on all of $\C$. --Lord_Farin (talk) 12:01, 14 December 2012 (UTC)


 * Are we using arctan to mean inverse tan now? --GFauxPas (talk) 14:33, 14 December 2012 (UTC)


 * We ought to. $\tan^{-1}$, as has been pointed out on the Tangent page, is misleading notation. arctan is de rigueur. --prime mover (talk) 14:42, 14 December 2012 (UTC)


 * Having said that, $\tan^{-1} (x)$ is a set: $\{y \in \C: \tan y = x\}$ which is well-defined. It's more complicated than it looks on the surface, and is tractable, but requires considerable fundamental underpinning. --prime mover (talk) 14:47, 14 December 2012 (UTC)


 * If we're using $y = \tan^{-1} x$ to mean $(x,y) \in \tan$, I object to using "$\iff$". (Actually, I object to even using $"="$, but that's a losing battle) --GFauxPas (talk) 14:53, 14 December 2012 (UTC)

I changed it to $ \arctan x $ (and the others too). — Timwi (talk) 19:33, 14 December 2012 (UTC)


 * I'm actually unhappy about the entire page until we have covered the multifunctional nature of logarithm in the complex plane. These grade-school definitions are all very well but they are a bit limiting and need to be expanded. This is one reason why this area has not been done; I was nerving myself up to posting up the fundamentals but got bogged down in the topology which I failed at badly. --prime mover (talk) 19:42, 14 December 2012 (UTC)


 * Well, since this is a wiki, you are free to edit and improve the pages. However, for this one, I’m not sure I see why you need a multifunctional logarithm. This theorem and its proof assume a single-valued logarithm of a complex number, namely:
 * $ \displaystyle \ln(a+bi) = \ln\left({\sqrt{a^2 + b^2} }\right) + i \begin{cases} \frac \pi 2 & : & a = 0 \\ \arctan\left({\frac b a}\right) & : & \mbox{otherwise} \end{cases} $
 * — Timwi (talk) 11:32, 15 December 2012 (UTC)


 * My books must all be wrong then. Sorry. I will amend them all to stub out all that rubbish that $\ln$ is a multifunction, e.g. "If $e^z = w$, then we write $w = \ln z$, called the natural logarithm of $z$. Thus the natural logarithmic function is the inverse of the exponential function and can be defined by
 * $w = \ln z = \ln r + i(\theta+2k\pi)\quad k=0,\pm1,\pm2,\ldots$
 * where $z = re^{i\theta}=re^{i\theta+2k\pi}$. Note that $l\ln z$ is a multiple-valued (in this case infinitely-many-valued) function. The principal-value or principal branch of $\ln z$ is sometimes defined as $\ln r + i\theta$ where $0\le\theta\le2\pi$. However, any other interval of length $2\pi$ can be used, e.g. $-\pi < \theta \le \pi$, etc."
 * Bet that Murray Spiegel is feeling a bit silly now ... --prime mover (talk) 21:43, 15 December 2012 (UTC)