1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook