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Sketching exponential curves with complex numbers

  1. Jan 5, 2012 #1
    How do you go about sketching y as a function of t for t≥0

    y= e(0.5t + i(√7/2)t) - e(0.5t-i(√7/2)t)


    I know it goes through the origin, and the gradient is positive here. But i'm unsure on how to deal with the imaginary numbers when I have a graph of y vs t.
     
  2. jcsd
  3. Jan 5, 2012 #2
    Can you try to simplify it using the definition of complex exponentiation. That is:

    [tex]e^{a+ib}=e^a(\cos(b)+i\sin(b))[/tex]
     
  4. Jan 5, 2012 #3
    Thanks for that idea, I have done that and am left with

    y=e0.5t ( i 2 sin (√7/2)t )

    But as I am meant to be plotting y as a function of t, I don't see how the imaginary part will factor in
     
  5. Jan 5, 2012 #4
    Well, it will be immediately obvious that no point on the curve will have a real part. So the curve will only move on the imaginary axis.
     
  6. Jan 5, 2012 #5

    HallsofIvy

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    In general, a function from R to C (real numbers to complex numbers) requires two dimensions for the value as well as one dimension for the argument- in other words, a three dimensional graph! However, as micromass points out, [itex]y= [e^{0.5t}sin(\sqrt{7}{2}t)]i[/itex] is always imaginary so you you really only need a single dimension for that.

    Graph [itex]y= e^{0.5t}sin((\sqrt{7}/2)t)[/itex], clearly labeling the "y" axis as imaginary numbers.
     
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