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Graphing an Equation with a Complex Term

  1. Jan 10, 2009 #1
    1. If you`re given an equation such as

    w(t) = (1+t)e^jt

    how would I go about graphing it.

    For different values of t i get a different complex number

    but how do I represent that complex number on a

    w(t), (t) axis??????




     
  2. jcsd
  3. Jan 10, 2009 #2

    Mark44

    Staff: Mentor

    What you need are three axes: one for values of t, and two axes (the complex plane) for w(t). Also, remember that e^(it) = cos(t) + i*sin(t). (Mathematicians normally use i for the imaginary unit.)
     
  4. Jan 10, 2009 #3
    thanks, can you tell me what should i do in order to draw it i got three points

    t = -10 i get 7.55 - j4.90
    t = 0 i get 1
    t = 10 i get -9.22 - 5.98

    the question said to graph for t (-10 to 10).

    but even with theses threee points im having trouble on how to draw the graph(i guess its a plane??) on a paper....i think i need more then just the three points to draw it :S any advice??
     
  5. Jan 10, 2009 #4

    Mark44

    Staff: Mentor

    Do you know how to graph points in the complex plane? The complex plane is the same as the real plane, except that the Real axis is the horizontal axis and the Imaginary axis is the vertical axis. For t = -10, you will have a point that is 7.55 units to the right of the origin, and 4.9 units down from the real axis.
     
  6. Jan 10, 2009 #5
    Yes even after I have plotted the three points i am not able to imagine the plane that this creates :S, i dont know it just seems to abstract to draw on a paper..
     
  7. Jan 11, 2009 #6

    Mark44

    Staff: Mentor

    You'll need to plot more points to get an idea of the shape of the graph.
     
  8. Jan 11, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Three points determine a plane. If this is not a plane (and you should have seen immediately that it is not) you need a lot more than 3 points to graph it.
     
  9. Jan 11, 2009 #8
    Do you think there are particular points I should try and get? like before if I had some plane

    2x + y +z = 1

    I find the three intercepts (0,0,1), (1/2,0,0) ,(0,1,0)

    and then connect the points which gives me an idea of the shape of the plane easily.

    but in this one i dont really know what points I should be looking for to give me the best picture of the plane..

    any ideas?
     
  10. Jan 11, 2009 #9
    Ok, what else do i need in order to graph it? See the thing is our curriculum is sort of weird since i'm taking vector calculus right now. If I had already completed that course probably I would have a better idea. In another course I should know how to graph this equation. Anyways can someone hint what other things I should know before i'm able to graph this this type of equation?
     
  11. Jan 11, 2009 #10

    Mark44

    Staff: Mentor

    As I said before, you should plot more points. The three you've chosen (for t = -10, t = 0, t = 10) aren't enough to get an understanding of the graph of w(t) = (1 + t)e^(it). If you plot the points for t = -10, -9, -8, ..., 0, 1, 2, 3, ..., 10, you'll get a better idea.

    It would also be helpful to label each point you plot in the complex plane, e.g., t = 2, or whatever the value of t is.
     
  12. Jan 11, 2009 #11
    so to graph the surface this function represents the only way is to plot points? Is there nothing else I can do other then just plotting various points to try imagine the plane?
     
  13. Jan 11, 2009 #12

    Mark44

    Staff: Mentor

    In the complex plane, the graph of this function is a curve, not a surface, and also not a plane, either. The most obvious thing to do would be to plot a few points on the curve, but for some reason you seem very reluctant to do that.
     
  14. Jan 11, 2009 #13
    but how is it not some type of surface because shouldn't there be THREE axis?

    t (real axis), w(t) = imaginary + real axis....


    and so wouldnt that function represent some type of surface.....

    Anyways, two more points:

    t = -5
    => -1.1 - j3.84
    t = 5
    => 1.7 - j5.755

    is one these graphs correct?
    [​IMG]
     
  15. Jan 12, 2009 #14

    Mark44

    Staff: Mentor

    No, it's not a surface. If you graph it it three dimensions, the points will define a curve in space. If you graph the points in just the complex plane, it's still a curve, but projected onto the complex plane.

    I think that the curve will look like a sort of spiral, but you still don't have enough points to be able to see it. Bite the bullet and plot some more points.
     
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