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(-2)^some decimal = a + bi? why?

  1. Feb 24, 2004 #1
    (-2)^some decimal = a + bi? why?
     
  2. jcsd
  3. Feb 24, 2004 #2

    matt grime

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    What does 3^2 mean? It's 3*3. We then manipulate things algebraically and see it makes sense to define, for positive x, x^r for r 1/n for some integer n as the n'th root. We also see that x^{-r} = 1/x^r for r any integer, moreover we see we can define any rational power in this way - the power p/q is the qth root of the pth power. all well and good, but that doesn't allow us to *define* powers of negative numbers always, and have the power be not a rational number. So we add on the symbol i, so that (-1)^{1/2}:=i (that := means defined to be equal to). Adding in imaginary numbers allows us to define powers of negative numbers. Is that a sufficient start, I mean, is it puzzling to you why (-1)^{0.5} is of the form a+ib?


    Do you need to know how to define the powers for irrationnal exponent?

    properly, x^r := exp(r*log(x))

    so it boils down to knowing exp of anything is just a power series in the anything, and defining log for arbitrary complex numbers, which can be done.
     
    Last edited: Feb 24, 2004
  4. Feb 24, 2004 #3
    Egads! I cant believe I didn't see that..blame it on my lack of sleep. Now lets pretend I never asked it.

    Thanks for the great explanation though!
     
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