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Graphing Euler Relationship as exponential spiral

  1. Apr 19, 2007 #1
    Hi everyone,

    I am trying to get an intuitive grasp of the Euler Relationship
    e^i(theta)=cos(theta)+i sin(theta)
    and also understand how to graph the exponential spiral, as demonstrated on this web page:

    Ok, first the neuron stimulation.

    I grasp how cos corresponds to the Real axis of the unit circle and sin is the imaginary axis. I grasp how to plot a complex number on the imaginary plane.

    I don't grasp why cos(theta) + i sin(theta) gives me a point on the unit circle, when on a real plane I'd have to invoke Pythagoras: sqrt((cos(theta))^2 + (sin(theta))^2) = 1 to get a point on the circle.

    Regarding the complex exponential tool at the link above, when I try to calculate the "a" and "b" calculations on the right side using the "a" and "b" selections on the left side, I get confused. For example, if I set b=0 and adjust a on the left, on the right side I get just e^a. This is obvious even to me. If I set a=0 and adjust b on the left, then on the right side, "a" is cos(left b) and "b" is sin(left b), so left-side b really is theta. I follow so far. How do I calculate the right-side a and b when the left-side a and b are not zero?

    More fundamentally, how do you guys mentally picture what the function e^ix looks like? I want to picture a circle, but I'm not getting it. :frown:

    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Apr 21, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Well the reason why cos x + i sin x gives a point on the unit circle without using Pythagoras is because the planes are different :)

    On the complex plane, the components of a number are sort of like directions. 1 + 2i means 1 to the right, then 2 up. or cos x to the right (if cos x is positive) and then sin x up or down. Hopefully that will help you visualize why e^ix is a circle.
  4. Apr 21, 2007 #3
    Rewrite the right hand side, then it will become clearer:
    Right hand side:
    [tex]e^{(a+bi)} = e^a \cdot e^{bi} [/tex]
    Now if you look carefully at the expression you will notice that, [tex]e^a[/tex] is just a real number and [tex]e^{bi}[/tex] is a number on a circle in the complex plane.

    (i) Now, let [tex]a[/tex] be a constant, so [tex]e^a[/tex] is also constant (choose for example a=0 as you mentioned):
    [tex]e^a = K[/tex]
    Thus we can write: [tex]e^a \cdot e^{bi} = K \cdot e^{bi}[/tex]

    Let's examine the expression [tex]K \cdot e^{bi}[/tex].
    How can we draw this in the complex plane?
    If we change b only, then you will only move on the circle (with radius K) in the complex plane.

    (ii) What happens, if you simultaneously change a and b?
    Changing b results in a movement on the circle, but if you additionally
    change a, you will also change K, the radius.

    That is why you get a spiral because you also change the radius.

    For example you could change a and b simultaneously by multiplying
    both of them with a factor t:

    To make this clearer, let's use the program on your website:
    1. On the left hand side set a=0.25 and b=2.00
    2. On the right hand side, there is this yellow "t-bar". Change the value
    from 0 to 1.
    - While changing t, observe the left hand side graphics. The yellow
    dot moves from the red to the blue dot.
    - While changing t, this time observe the right hand side graphics.
    The yellow dot moves along the spiral from the red to the blue dot.
    Last edited: Apr 21, 2007
  5. Apr 22, 2007 #4
    Gib Z - Thanks for the reply.
    I did mention I understand how to plot a complex number on the plane - the way you said. And I have noticed that the complex plane is different than a real plane regarding applicability of Pythagoras' theorem. I have been stuck on the why is it different part. It's just occurred to me though, that comparing the complex plane and the real plane is not a valid comparison. A complex number is not two-dimensional like a point in an x/y plane is two dimensional. A complex number is (I think) one-dimensional; for example when a complex number has the imaginary portion = zero. Extending this thought, I should be able to plot a two dimensional point where x and y are complex numbers. Perhaps I can graph the point using one Real plot with real X and Y axes and one Imaginary plot with imaginary X and Y axes. Anybody hear of such a thing?

    Edgardo - Thanks for your reply too.
    The part where you set e^a to K was interesting. As you said, setting left a to zero (K=1) does result in left b moving a point on the circle on the graph at the right. However, setting left a to, say, 0.1 (K=1.105) results in left b moving a point on a spiral, even though left a is fixed (constant). So the radius isn't just K.

    Part of what I'm trying to figure out is how to actually calculate the right a and b values. For example, if I set left a to 0.25 and left b to 1.23, I end up with right a = 0.42 and right b = 1.21. So far my feeble calculations produce different values.

    If anybody else can add more thoughts, great. Otherwise, I'll post follow-ups if I find out any further answers.
  6. Apr 22, 2007 #5
    Keep "a" constant at a=0.1 and change "b".
    While varying "b" observe the blue dot on the right screen.
    You'll notice that the dot moves on the blue circle and
    not in a spiral.

    Let's introduce a new notation. Instead of saying "left a and b values" and
    "right a and b values", we say:
    Left hand side: a+bi
    Right hand side: x+yi

    So your question is, how will x and y look like, if you have a and b as input.
    In your example, you set a=0.25 and b=1.23,
    and you end up with x=0.42 and y=1.21

    Let us review again how the calculation is done.
    You have a+bi as input and on the right hand side you get [tex]e^{a+bi}[/tex].

    But we want to know x and y:
    You have a+bi as input and on the right hand side you get x+yi.

    So the question is, how do x and y look like. Well, let's just
    manipulate [tex]e^{a+bi}[/tex] until we get it to the form x+yi.

    [tex]e^{a+bi} = e^a e^{bi} = e^a (\mbox{cos}(b) + i \mbox{sin}(b)) = (e^a \mbox{cos} (b)) + i (e^a \mbox{sin} (b)) = x + yi[/tex]

    Thus, by comparing the last equation
    [tex](e^a \mbox{cos} (b)) + i (e^a \mbox{sin} (b)) = x + yi[/tex]
    we get

    [tex]x = e^a \mbox{cos}(b)[/tex] and

    [tex]y = e^a \mbox{sin}(b)[/tex]

    Let's check this with the values from your example:
    a = 0.25 and b = 1.23 yields
    [tex]x = e^{0.25} \mbox{cos}(1.23) = 0.43[/tex]
    [tex]y = e^{0.25} \mbox{sin}(1.23) = 1.21[/tex]
    (Don't forget to set "rad" instead of "deg" on your calculator).
    Last edited: Apr 22, 2007
  7. Apr 23, 2007 #6
    You've done it! Somehow I wasn't seeing the blue circle for what it is, namely e^a, like you said the first time. It never occurred to me to substitute the imaginary portion of e^(a+ib) with the full Euler's relationship like you did :bugeye:. Understanding the calculations that produce the graph on the right is the key I needed to understand the equation.

    Thank you very much.
    Last edited: Apr 23, 2007
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