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Complex Number

  1. Jan 11, 2005 #1
    Hi all,

    Looking for some help on the following problem. Any replies much appreciated.

    I have the complex number

    exp(i.x)

    If x = - infinity,

    is this zero?? Is there any intuitive/straightforward value that it should be? I decomposed the expression into cos and sin and it looks like the number doesnt converge to anything!

    regards,
     
  2. jcsd
  3. Jan 11, 2005 #2

    dextercioby

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    Of course it doesn't converge.U need a complex number of different modulus.The one u've chosen has modulus one,i.e. 1.U cannot make it's argument go to infinity,coz that would mean rotating the unit vector (which is represented by e^(ix)) an ininite amount of times and the phase could not be determined.

    Daniel.
     
  4. Jan 11, 2005 #3

    matt grime

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    It is undefined, and there is no reason why it should be defined.
     
  5. Jan 11, 2005 #4
    Well, it doesn't go to infinity because of the very reason that you quoted. But don't worry : there is NO problem because this function is NOT DEFINED in the infinity. So, there is no problem and why should there be any ???


    marlon
     
  6. Jan 11, 2005 #5
    Got it.

    Many thanks for the replies.

    Regards,
    Sam
     
  7. Jan 11, 2005 #6

    Hurkyl

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    A comment: a number is a number, it doesn't make sense to say it "converges" to anything. While the distinction is often blurred, it is still important to remember that the concept of limit is different than that of a number.
     
    Last edited: Jan 11, 2005
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