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Complex Argument and ArcTan

  1. Oct 4, 2007 #1
    According to Wolfram's MathWorld, the Inverse Tangent (ArcTan) is a multivalued function.

    ...but the plot on their website (and the one I just did with my copy of Mathematica) shows it's a one-to-one function.

    Okay, so it does say the ArcTan[] function returns the Principal Value of inverse Tan, but I still don't get it...

    Moving on to complex numbers, in Mathematica the Arg[] function is apparently an alias for ArcTan[x,y] (rather than ArcTan[x], since it determines if Pi should be added to it). This is embarassing at this level to ask, but how can I do what Mathematica's ArcTan[x,y] does with my numeric calculator? So far we've been told to sketch the Argand diagram and determine which quadrant it's in to see how we should use Inverse Tan, but I'm looking for something a little faster, I'm not keen on "visual" methods.

    Thanks
     
  2. jcsd
  3. Oct 4, 2007 #2
    Your calculator doesn't have inverse trig functions? Anyway, for a complex number of the form a+bi if all you're intested in the quadrant, how about just looking at whether a and b are postive or negative? If both are positive (like 4+13i) then it must be in the first quadrant, etc....
     
  4. Oct 4, 2007 #3
    My calculator has inverse Tan, yes, but that's comparable to Mathematica's ArcTan[x] function, not MM's ArcTan[x,y] function (which determines what quadrant it's in and thus adds (or subtracts) the normal ArcTan from Pi), regular ArcTan doesn't do that.
     
  5. Oct 4, 2007 #4

    Gib Z

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    Well jackiefrost said the most non-visual method you can do with a standard calculator, to work out the quadrant look at the signs of a and b.
     
  6. Oct 4, 2007 #5

    mathwonk

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    note thast arctan is defined by the path integral of 1/[1+x^2] HENCE becomes multivalued when the path winds repeatedly around i or -i.

    similarly log is the path integral of 1/x which has bad behavior also at two points, 0 and infinity. since on the complex sphere any pair of points are equivalent under a mobius transformation, arctan has essentially the same functional behavior as log.

    i.e. just think of arctan as log, but after 0 and infinity have been moved to i and -i. you should be able to write down a formula this way expressing arctan as a log of a simple fractional linear transformation.

    in particular it is multivalued.
     
    Last edited: Oct 4, 2007
  7. Oct 5, 2007 #6

    Gib Z

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    mathwonk, may i please know what level of knowledge is required to understand perfectly what you just said? When you say path integral, what exactly do you mean :(
     
  8. Oct 6, 2007 #7

    mathwonk

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    well an integral along an interval on the real line is an integral along the straight path from one end to ther other. a path in the complex plane is a smooth functioin from an interval of the real line with values in the complex plane, i.e. it is a parametrized curve given by two complkex valued functioins of on a real interval.

    such a path looks like x(t) + iy(t), for all t in an interval. then we can integrate the differential dz/z along this path by pulling it back to the t interval (substituting x+iy for z, and dx/dt dt + idy/dt dt for dz) and doing an ordinary integral in t.

    then the log of z is just the integral of dz/z along any path from 1 to z, but since the integral makes no sense at z=0, the value will depend on how many times the path winds around zero before reaching z. hence log is multi valued.

    as usual this is explained in an elementary way in courant's calculus, i think in volume 2, at the end. that where logs first made sense to me.

    the level of knowledge, namely path integrals is the very first part of several variable calculus.
     
  9. Oct 6, 2007 #8

    Gib Z

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    Ok thank you mathwonk, I wish to learn more on the subject, so i guess courant's calculus is the way to go? I'll order a copy. Thanks again!
     
  10. Oct 6, 2007 #9

    Chris Hillman

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    Wonderful book (by one of the great mathematicians of the 20th century), but I venture to guess that many modern American students will find the textbook by the late Ralph Boas, Introduction to Complex Anaysis, a somewhat easier read. Actually, any good textbook on complex analysis will cover path integrals, branch cuts, elements of Riemann surfaces, and all that good stuff.
     
  11. Oct 7, 2007 #10

    mathwonk

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    well courant is worth owning for many reasons, but indeed it is not a text oncomplex analysis. it only has a brief intro to the topic in a final chapter as i recall. nd there are onlky a couple sentences on the definition of log as a path integral. but for me personally it was those sentences that made it become clear to me. i also wrote a lengthy discusson of the covering space aspect of logs and arctans that got erased by the browser as so often happens here to the carelss and trusting poster. maybe ill redo that.

    essentially it said that the reason log and arctan are multivalued is because their inverses, exp and tan, are periodic, hence not injective. in fact they are covering space maps, (derivative never zero), so they cannot be surjective either, since the sphere is simply connected and has no non trivial covering maps.

    thus by the somebody theiorem they must both omit exctly two points. the two points are just different, with exp omitting 0 and infinity and tan omitting i and -i.

    then if you just map z to 2iz, a mild automorphism of the sphere, and follow tan by the automorphism z-i/z+i sending i and -i to 0 and infinity, you get that this composition equals exp(2iz). i.e. recall that exp(iz) = cosz + isin(z), so that sinz = [exp(iz) - exp)-iz)]/2i and cosz = [expiz+exp(-iz)]/2, hence tanz = [exp(iz) - exp(-iz)]/i[exp(iz)+exp(-iz)].

    thus ???? oh phoo,

    start over:

    transform tanz by the automorphism above, getting [tan(z)-i]/[tan(z)+i],

    then using the formula above that exp(iz) = cosz + isinz, one gets

    [tan(z)-i]/[tan(z)+i] = -exp(2iz). so exp and tan are almost the same, hence also arctan and log are too.
     
  12. Oct 10, 2007 #11

    Gib Z

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    I always wondered why integrals of the form [tex]\int \frac{1}{1 \pm x^n} dx[/tex] could be expressed in terms of only arctans and logs and never anything else!
     
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