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Homework Help: Complex numbers problem

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Trying to write
    [tex]

    -8\pi - 8\pi\sqrt3 j
    [/tex]

    in exponential

    I got the coefficient as 16 pi

    but to get the theta in top of the exponential I have to do

    the inverse tangent of [tex]
    \frac{-8\pi} {-8\pi\sqrt3 j}
    [/tex]


    I know it is pi over 3, but what is the easiest way to find the multiplicative factor of pi? it is four I just want to know how to find it for other harder examples



    answer [tex]

    16e^{\frac{j4\pi} 3}
    [/tex]


    Input appreciated
     
  2. jcsd
  3. Sep 18, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor

    The "argument", or [itex]\theta[/itex] for x+ jy is arctan(y/x), not arctan(x/y) and certainly not arctan(x/jy)! You want
    [tex]arctan(\frac{-8\pi\sqrt{3}}{-8\pi}= arctan(\sqrt{3})[/tex]
    because, of course, the "-8" terms cancel. Now you could use a calculator or, perhaps better, imagine a right triangle with opposite side of length [itex]\sqrt{3}[/itex] and near side 1 (because tan= opposite side/near side). By the Pythagorean theorem, the hypotenuse has length [itex]\sqrt{(\sqrt{3})^2+ 1}= 2[/itex]. That is, you are looking for an angle that has sine (opposite side divided by hypotenuse) [tex]\frac{-\sqrt{3}}{2}[/itex] and cosine [tex]\frac{-1}{2}[/tex]. The negatives are because the real and imaginary parts of the number you give are both - so the point is in the fourth quadrant, not the first.
    [itex]sin(\theta)= \sqrt{3}/2[/itex] and [itex]cos(\theta)= 1/2[/itex] should be among the "special angles" you learned long ago.

    No, you do NOT know "it is pi over 3" because pi/3 is in the first quadrant and the value you want is in the fourth. 4pi/3 does happen to be in the fourth quadrant.


     
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