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

1. Mar 19, 2009

### latentcorpse

i need to check $z^a$ is holomorphic on $\mathbb{C} \backslash \{0\}, a \in \mathbb{C}$ but am having difficulty arranging it into a form that i can use the Cauchy Riemann equations on. so far is have:

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

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

so if i let $z=u(x,y)+i v(x,y)$ we get

$u(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}$
$v(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}$

2. Mar 19, 2009

### yyat

Re: Holomorphic

Unless you allow multi-values functions, $$z^a$$ is not holomorphic on $$\mathbb{C}\setminus\{0\}$$ for $$a\notin \mathbb{Z}$$, but only on $$\mathbb{C}$$ minus a branch cut (e.g. along the negative real axis). To prove that $$z^a$$ is holomorphic on this region, use the fact that exp, log are holomorphic and that compositions of holomorphic functions are holomorphic.

3. Mar 19, 2009

### latentcorpse

Re: Holomorphic

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

also can you talk me throught the branch cut here:

we have $z=re^{i \theta}$
so $z^a=r^a e^{i a \theta$

so if we change $\theta$ by $2 \pi$ we have the same value of z corresponding to more than one value of $z^a$ (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"?

4. Mar 19, 2009

### yyat

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 $$\mathbb{C}_-=\mathbb{C}\setminus\{x\in\mathbb{R}:x\le 0\}$$. Then $$z^a=e^{a\log(z)}$$, taking the principal value of log, is single-valued holomorphic function on $$\mathbb{C}_-$$. The function "jumps" when crossing the negative real axis, so it can not be extended to a holomorphic function on $$\mathbb{C}\setminus 0$$.
The deeper topological reason is that $$\mathbb{C}_-$$ is simply connected (all loops in it can be contracted to a point), while $$\mathbb{C}\setminus 0$$ 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
5. Mar 19, 2009

### latentcorpse

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?

6. Mar 20, 2009

### yyat

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