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Check that z^a is holomorphic on C\{0}

  1. Mar 19, 2009 #1
    i need to check [itex]z^a[/itex] is holomorphic on [itex]\mathbb{C} \backslash \{0\}, a \in \mathbb{C}[/itex] but am having difficulty arranging it into a form that i can use the Cauchy Riemann equations on. so far is have:

    [itex]z^a=e^{a \ln{z}}=e^{a \ln{|z|}}e^{i arg(z)}[/itex]
    which when i break up the second exponential gives

    [itex]z^a=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}+i e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]

    so if i let [itex]z=u(x,y)+i v(x,y)[/itex] we get

    [itex]u(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}[/itex]
    [itex]v(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]
     
  2. jcsd
  3. Mar 19, 2009 #2
    Re: Holomorphic

    Unless you allow multi-values functions, [tex]z^a[/tex] is not holomorphic on [tex]\mathbb{C}\setminus\{0\}[/tex] for [tex]a\notin \mathbb{Z}[/tex], but only on [tex]\mathbb{C}[/tex] minus a branch cut (e.g. along the negative real axis). To prove that [tex]z^a[/tex] is holomorphic on this region, use the fact that exp, log are holomorphic and that compositions of holomorphic functions are holomorphic.
     
  4. Mar 19, 2009 #3
    Re: Holomorphic

    ok. is this possible using the Cauchy Riemann equations though?

    also can you talk me throught the branch cut here:

    we have [itex]z=re^{i \theta}[/itex]
    so [itex]z^a=r^a e^{i a \theta[/itex]

    so if we change [itex]\theta[/itex] by [itex]2 \pi[/itex] we have the same value of z corresponding to more than one value of [itex]z^a[/itex] (i.e. a multifunction). This means there are different branches corresponding to the different values the function can take for the same z.
    what do you mean by "take a branch cut along the negative real axis"?
     
  5. Mar 19, 2009 #4
    Re: Holomorphic

    It should be. Compute the partial derivates of exp, log (inverse function theorem) and apply the chain rule to exp(a log(z)).

    Consider the set [tex]\mathbb{C}_-=\mathbb{C}\setminus\{x\in\mathbb{R}:x\le 0\}[/tex]. Then [tex]z^a=e^{a\log(z)}[/tex], taking the principal value of log, is single-valued holomorphic function on [tex]\mathbb{C}_-[/tex]. The function "jumps" when crossing the negative real axis, so it can not be extended to a holomorphic function on [tex]\mathbb{C}\setminus 0[/tex].
    The deeper topological reason is that [tex]\mathbb{C}_-[/tex] is simply connected (all loops in it can be contracted to a point), while [tex]\mathbb{C}\setminus 0[/tex] is not. If you want to know more about this, check out http://en.wikipedia.org/wiki/Monodromy_theorem" [Broken].
     
    Last edited by a moderator: May 4, 2017
  6. Mar 19, 2009 #5
    Re: Holomorphic

    i see that if we take the principal argument of z and hence the proncipal value of log it does indeed jump as you cross the negative real axis - why does this mean it can't be extended to a holomorphic function on C\0?

    also did i set up the cauchy riemann bit ok in my first post?
     
  7. Mar 20, 2009 #6
    Re: Holomorphic

    Holomorphic means in particular continuous. No matter how the values on the negative real axis are defined, there will always be a discontinuity. See also the discussion on the http://en.wikipedia.org/wiki/Complex_logarithm" [Broken].

    Looks correct.
     
    Last edited by a moderator: May 4, 2017
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