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Challenge IV: Complex Square Roots, solved by jgens

  1. Aug 6, 2013 #1

    micromass

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    This is a well-known result in complex analysis. But let's see what people come up with anyway:

    Challenge:
    Prove that there is no continuous function ##f:\mathbb{C}\rightarrow \mathbb{C}## such that ##(f(x))^2 = x## for each ##x\in \mathbb{C}##.
     
    Last edited by a moderator: Aug 8, 2013
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  3. Aug 6, 2013 #2

    jgens

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    Suppose such a function existed and consider the map g:S1→S1 given by restriction. Then the degree of g2 is a multiple of two but the degree of idS1 is one. This contradiction completes the proof.
     
  4. Aug 6, 2013 #3
    Suppose a function [itex] f [/itex] did exist. For a complex number [itex] z=Re^{i\theta} [/itex], the only two square roots are [itex] \pm \sqrt{R}e^{\frac{i\theta}{2}} [/itex]. Call the function with the plus sign [itex] f_1 [/itex] and the function with the minus sign [itex] f_2 [/itex]

    First let's show that a function that takes on the value [itex] f(z)=\sqrt{R}e^{\frac{i\theta}{2}} [/itex] for some z's and [itex] f(z)=-\sqrt{R}e^{\frac{i\theta}{2}} [/itex] for others is discontinuous.

    For nonzero [itex] z_1=R_1e^{i\theta_1} [/itex] and [itex] z_2=R_2e^{i\theta_2} [/itex] assume that [itex] f(z_1)=\sqrt{R_1}e^{\frac{i\theta_1}{2}} [/itex] and [itex] f(z_2)=-\sqrt{R_2}e^{\frac{i\theta_2}{2}} [/itex].

    Let [itex] \gamma:[0,1] \to ℂ [/itex] such that [itex] \gamma(0)=z_1 [/itex] and [itex] \gamma(1)=z_2 [/itex] be an injection and a parameterization of a path from [itex] z_1 [/itex] to [itex] z_2 [/itex] that does not pass through the origin . Let [itex] t' [/itex] be the infimum over all [itex] t \in [0,1] [/itex] such that [itex] f(\gamma(t))= f_2(\gamma(t)) [/itex]. It is easy to show that [itex] f [/itex] is discontinuous at [itex] \gamma(t') [/itex]

    Now, we just have to deal with the case [itex] f=f_1 [/itex] or [itex] f=f_2 [/itex].

    If [itex] f(z)=\sqrt{R}e^{\frac{i\theta}{2}} [/itex], where [itex] z=Re^{i\theta} [/itex], then we discover that f isn't really a function because [itex] Re^{i\theta}=Re^{i\theta+2\pi} [/itex] but [itex] f(Re^{i\theta+2\pi})=\sqrt{R}e^{\frac{i\theta}{2}+\pi} \neq \sqrt{R}e^{\frac{i\theta}{2}}=f(Re^{i\theta}) [/itex] The case for [itex] f_2 [/itex] is similar.
     
    Last edited: Aug 6, 2013
  5. Aug 6, 2013 #4

    micromass

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    OK, so that was definitely too easy for the smart crowd of people here :smile:

    A big congratulations for jgens for his topological solution and for beating HS-Scientist in just a minute. But the latter did give a nice elementary solution. There are other solutions though, so if somebody finds them, please do post!
     
    Last edited: Aug 8, 2013
  6. Aug 6, 2013 #5
    what does a degree of a function g:S^1->S^1 mean?
     
    Last edited: Aug 6, 2013
  7. Aug 6, 2013 #6

    micromass

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  8. Aug 6, 2013 #7
    Anyway, my choice of words would be this: We define an operation [itex]f\mapsto g[/itex] by setting

    [tex]
    g:[0,2\pi[\to\mathbb{C},\quad g(\theta)=f(e^{i\theta})
    [/tex]

    The property [itex]f^2=\textrm{id}[/itex] implies

    [tex]
    g(\theta)=\epsilon(\theta)e^{i\theta/2}
    [/tex]

    with some function [itex]\epsilon:[0,2\pi[\to \{-1,1\}[/itex].

    If [itex]f[/itex] is continuous (as anti-thesis), [itex]g[/itex] must be continuous too, which implies [itex]\epsilon(\theta)[/itex] is a constant. Then we find that [itex]\lim_{z\to 1}f(z)[/itex] does not exist.

    In other words, I would prefer forcing the discontinuity somewhere with the assumptions.
     
    Last edited: Aug 6, 2013
  9. Aug 6, 2013 #8

    micromass

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    I might as well post my own solution. Fix ##z\neq 0##, consider the function

    [tex]G(w) = \frac{f(z)f(w)}{f(zw)}[/tex]

    This function is well defined on ##\mathbb{C}\setminus \{0\}##. By squaring, we see that

    [tex]G(w) = \pm 1[/tex]

    By continuity of ##G## and since ##\mathbb{C}\setminus \{0\}## is connected, we see that ##G## is constant, so we have

    [tex]f(z)f(w) = f(zw)~\text{or}~f(z)f(w) = f(zw)[/tex]

    By (if necessary) multiplying ##f## with ##-1##, we can assume that the first identity is satisfied. But then

    [tex]f(1) = f(-1)f(-1) = (f(-1))^2 = -1[/tex]

    but also

    [tex]f(1) = f(1)f(1) = (f(1))^2 = 1[/tex]

    Contradiction.
     
  10. Aug 6, 2013 #9

    jgens

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    Another solution: With a little voodoo we can actually show that f is holomorphic. This means that differentiating at zero tells us that 1 = 2f(0)f'(0) = 0 and then we are done.
     
  11. Aug 6, 2013 #10
    I've done this proof before. I did it using the definition of continuity of complex functions by the continuity of their real and imaginary parts, id est, given ##x=a+bi## and ##f(x)=u(a,b)+iv(a,b)## for real numbers ##a## and ##b## and real valued functions ##u## and ##v##, ##f## is continuous at ##x_0=a_0+ib_0## if and only if $$\lim_{(a,b)\rightarrow(a_0,b_0)}[u+iv]=u(a_0,b_0)+iv(a_0,b_0)=f(x_0).$$ I can see if I can find my proof later if anyone is interested, but it essentially came down to a statement like "if ##u## is continuous, ##v## is not."

    Suppose ##n=1##. :biggrin:
     
    Last edited: Aug 6, 2013
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