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Prove the image is a circle

  1. Jun 22, 2008 #1
    I stumbled across this question while self-learning topology, and I am not sure how to proceed. I need to prove that the image of [tex]f(x)=\cos x + i \sin x[/tex] is the unit circle. The statement is obvious, which makes it all the harder for me to figure out what I actually need to do!
     
    Last edited: Jun 22, 2008
  2. jcsd
  3. Jun 22, 2008 #2
    it's not
     
  4. Jun 22, 2008 #3
    That's strange. What am I missing?
     
  5. Jun 22, 2008 #4
    actually i have no idea about how complex functions work. for all i know it might well be.
     
    Last edited: Jun 22, 2008
  6. Jun 22, 2008 #5

    morphism

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    First off, you need to specify what the domain of f is for this question to make sense!
     
  7. Jun 22, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi CrumpledPaper! Welcome to PF! :smile:

    (ignore ice … it's this hot weather! :wink:)

    I supsect they want you to prove that |f(x)| = 1, and that arg(f(x)) takes on all possible values. :smile:
     
  8. Jun 22, 2008 #7
    The question examines [tex]f:\mathbb{R} \rightarrow \mathbb{C}[/tex]

    tiny-tim, thanks for your help. I understand the "unit" portion of the proof. My difficulty is precisely in understanding how to prove that the function takes on all possible values. I thought a circle could be defined as [tex]x(t) = \cos t[/tex] and [tex]y(t) = \sin t[/tex], in which case I feel like there's nothing left to prove. Could you provide me with a first step so I see what a rigorous proof of this would entail? Thanks!
     
  9. Jun 22, 2008 #8
    lol @arg(f(x)) taking on all all possible values, the domain of f(x) is all reals because the domain of cos and sin are all reals. actually that's not rigorous at all
     
  10. Jun 22, 2008 #9
    What would constitute a rigorous proof for this problem?
     
  11. Jun 22, 2008 #10

    morphism

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    That depends on what your definition of the "unit circle" is. A common definition that doesn't completely trivialize the problem is: { z in C : |z|=1 }.
     
  12. Jun 22, 2008 #11
    I don't seem to be getting anywhere still. Here's the complete problem:

    Let the function [tex]f: \mathbb{R} \rightarrow \mathbb{C}[/tex] be defined by [tex]f(x) = \cos x + i \sin x[/tex] for each real [tex]x[/tex]. If [tex]a + ib[/tex] is a complex number, then its absolute value is [tex]\left|a + ib\right| = (a^2 + b^2)^{\frac{1}{2}}[/tex]; it is clear that [tex]\left| f(x) \right| = 1[/tex] for each real [tex]x[/tex]. Hence the image of [tex]f[/tex] lies on the circle of unit radius which is centered at the origin in the plane. The function [tex]f[/tex] may be visualized as assigning to each real number the point on that circle arrived at by going counterclockwise around the circle a distance [tex]x[/tex], starting at [tex]1 = 1 + i0; x < 0[/tex] requires the distance [tex] \left| x \right| [/tex] in the clockwise direction. This makes it clear that the image of [tex]f[/tex] equals the unit circle; can you prove it rigorously?
     
  13. Jun 22, 2008 #12

    D H

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    Prove that
    1. For each real x, the mapping [tex]x\rightarrow\exp(ix)[/tex] is a point on the complex unit circle.
    2. For each point on the complex unit circle there exists a real x that maps to this point via this same mapping.
     
  14. Jun 22, 2008 #13
    Excellent. Thanks.
     
  15. Jun 22, 2008 #14
    why don't you just take the norm of the function? it'll be 1
     
  16. Jun 22, 2008 #15

    D H

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    That shows the image of [itex]\exp(ix)[/itex] is a subset of the complex unit circle. It does not show that the image is the complex unit circle.

    Edited to add:
    As a counterexample, consider the function [itex]f(x)=1[/itex]. While the norm of this function is always one, that fact alone does not suffice to show the image of this function is the complex unit circle (which obviously is not the case).
     
    Last edited: Jun 22, 2008
  17. Jun 23, 2008 #16
    Good point, although in this case I don't think [tex]f(x)=1[/tex] is a valid counterexample because the norm is not one in [tex]\mathbb{C}[/tex] for all points along the line (i.e. [tex](1 + 100i)[/tex]). A better counterexample would be a function defined by any single point on the circle.
     
  18. Jun 23, 2008 #17

    D H

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    f(x)=1 is identically one for all x. In other words, its image is a single point on the unit circle.
     
  19. Jun 23, 2008 #18
    My bad. You're right =)
     
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