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Complex numbers

  1. Sep 17, 2005 #1
    Anyone got a good link to a place that explains complex numbers?
     
  2. jcsd
  3. Sep 17, 2005 #2

    arildno

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    What do you want to know about them?
     
  4. Sep 17, 2005 #3
    I'm having a hard time rewriting from one form to another, carthesian - polar and so on.
     
  5. Sep 17, 2005 #4

    Hurkyl

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    Well, the procedure is essentially identical to converting between rectangular and polar coordinates on the good ol' real plane, so if that's where you're having trouble, you can pick up one of your old textbooks and review.
     
  6. Sep 17, 2005 #5

    arildno

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    As Hurkyl said, just think of the good ole plane here.

    Examples:
    Suppose that a complex number z is given by:
    z=a+ib
    where a,b are real numbers, and i the imaginary unit.
    Then, multiply z with 1 in the following manner:
    [tex]z=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}}(a+ib)={\sqrt{a^{2}+b^{2}}}(\frac{a}{\sqrt{a^{2}+b^{2}}}+i\frac{b}{\sqrt{a^{2}+b^{2}}})[/tex]
    Find the angle [tex]\theta[/tex] that is the solution of the system of equations:
    [tex]\frac{a} {\sqrt{a^{2}+b^{2}}}=\cos\theta,\frac{b}{\sqrt{a^{2}+b^{2}}}=\sin\theta[/tex]
    Thus, defining [tex]|z|={\sqrt{a^{2}+b^{2}}}[/tex], we get:
    [tex]z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}[/tex]
    by definition of the complex exponential.
     
  7. Sep 25, 2005 #6
    buy a ti 89 or voyage 200 and your problems are forever solved
     
  8. Sep 25, 2005 #7
  9. Sep 28, 2005 #8
    Euler Identity

    Here is a cool trick for calculating pi derived from Euler Identity.

    e^(i*(pi/2)) = Cos(90) + i*Sin(90)

    ln(e^(i*(pi/2)) = ln(Cos(90) +i*Sin(90))

    i*(pi/2)*lne = ln(Cos(90) +i*Sin(90))

    pi = (1/i)*(2)*ln(Cos(90) +i*Sin(90))

    pi = (i^4/i)*(2)*ln(Cos(90) +i*Sin(90))

    pi = (-i)*(2)*ln(Cos(90) +i*Sin(90))

    pi = (-2i)*ln(Cos(90) +i*Sin(90))

    pi = ln((Cos(90) +i*Sin(90))^(-2i))

    pi = ln(1/(Cos(90) +i*Sin(90))^(2i))

    Just a cool trick!

    Best Regards,

    Edwin G. Schasteen
     
  10. Sep 29, 2005 #9
    "buy a ti 89 or voyage 200 and your problems are forever solved"

    if his problem is understanding how certain things work, then i think his problem would stay untouched if he bought one of these caluclators.
     
  11. Sep 29, 2005 #10
    I have a TI-83 plus and a TI Voyage 200, and I carry them both with me everywhere I go! They are truely amazing computation devices for those of us that are numerically challenged or just plain lazy. :smile:

    Best Regards,

    Edwin
     
  12. Oct 4, 2005 #11
     
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