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Euler's Formula - Need help!

  1. Dec 24, 2003 #1
    I just started studying complex numbers. It says complex numbers in polar form can be expressed as a power of e:

    e^ix = cosx + isinx

    I don't quite understand how this equation works.
    How do i evaluate e^ix? And how does taking e to the power of ix get me a complex number a + bi or even in polar form?
     
  2. jcsd
  3. Dec 24, 2003 #2

    NateTG

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    To get all complex numbers, you need the somewhat more general form:
    [tex]re^{i\theta}=r\sin{\theta}+ir\cos{\theta}[/tex]

    Now, the other direction is also not bad
    [tex]z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}[/tex]

    If you think of [tex]r[/tex] as the radius, and [tex]\theta[/tex] as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.
     
  4. Dec 24, 2003 #3
    So [tex]re^{i\theta}[/tex] is merely a convenience in writing?
     
  5. Dec 24, 2003 #4

    mathman

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    It is more than a convenience. One way to look at it is by power series of eix and compare to power series for sinx and cos x. You will see that eix=cosx+isinx.
     
  6. Dec 24, 2003 #5
    Ok, yes i think i'm beginning to understand. How do one evaluate xi? For example, what's 32i?
     
  7. Dec 24, 2003 #6

    krab

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    If you are thinking of x^n being the product of n x's, and want to apply that here, you are out of luck. When the exponent is imaginary, it doesn't really carry over. Think of it as a generalization.

    Are you familiar with Taylor series?

    [tex]e^x=1+x+{x^2\over 2}+{x^3\over 3!}+{x^4\over 4!}+...[/tex]

    [tex]\sin x=x-{x^3\over 3!}+{x^5\over 5!}-...[/tex]

    [tex]\cos x=1-{x^2\over 2}+{x^4\over 4!}-...[/tex]

    The result follows from these.
     
  8. Dec 24, 2003 #7

    NateTG

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    Well, you can do the following:

    [tex]3^{2i}=e^{\ln{3} *2i}=\cos{2\ln{3}}+i\sin{2\ln{3}}[/tex]

    If you're feeling a bit more adventerous, there is a power series for [tex]3^x[/tex] which is going to give you the same result.
     
  9. Dec 24, 2003 #8
    Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.
     
  10. Dec 24, 2003 #9

    Hurkyl

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    For the record, x^y is usually a multivalued function.
     
  11. Dec 24, 2003 #10

    NateTG

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    Think of it as a way to calculate [tex]x^{iy}[/tex].
     
  12. Dec 25, 2003 #11
    should that be arctan(y/x)?
     
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