# Finding in which domain a complex function $z^\omega$ is analytic

Given the complex function $z^\omega$ (which I rewrote as $z^\omega=e^{\omega\ln{z}}$), I need to find the domain in which it is analytic.

What I know about analytic functions:

• They must satisfy the Cauchy-Riemann equations
• They are of class C$^\infty$

But I'm not sure how these help me find the domain in which the function is analytic.

How should I get started/what should I be looking for?

Edit

Since a function is analytic in a domain $D$ if it is continuous (or differentiable?) everywhere in that domain, I assume that I just need to find where the function ceases to be continuous or differentiable?

• Complex exponentiation is not in general uniquely defined. – Wojowu Nov 4 '15 at 20:47
• Your last rewrite is not correct. In general $e^{\omega\ln z}\ne e^\omega e^{\ln z}$ – skyking Nov 4 '15 at 21:13
• yep, just came back to fix that, actually – galois Nov 4 '15 at 21:23

First of all you have to decide if you would consider it a function of $z$ orv $\omega$ (or even both).

Second you will have to realize that exponentiation is not uniquely defined for complex numbers (making either $z^\omega$ a multi-function or making you have to decide branch).

Third it's normally not analytic for $z=0$. The exceptions are (possibly) if $\omega$ is a non-negative integer and you consider it a function of $z$.

Otherwise it's analytic everywhere (in the domain of your branch if you may have selected).

Although your note about analytic functions is correct that's redundant requirements. A function of one complex variable is analytic in a region if it's (complexly) derivable there. You only have to check that, in general you get a problem if $z=0$ with the possible exception that $\omega$ is a positive integer, then we have that it becomes derivable everywhere.

If you consider it a function of both $z$ and $\omega$ you will have to do a little more, it then have to be investigated a little more. We have to see that the function is $C^\infty$, to see this we observe that they are all of the form:

$$\partial^n z^\omega = P(\omega, \ln z)z^{\omega+\alpha}$$

where $P$ is a polynomial and $\alpha$ is a constant. It's trivial for $n=0$ as the function is on this form (with $P=1$ and $\alpha=0$). The only thing we need to see is that the partial derivates of a function on that form is again on that form (using chain rule, rule for derivate of product and that the derivate of a polynomial is again a polynomial):

$$\partial_\omega P(\omega,\ln z)z^{\omega+\alpha} = P'_1(\omega, \ln z)z^{\omega+\alpha}+P(\omega,\ln z)z^{\omega+\alpha}\ln z = (P'_1(\omega, \ln z)+P(\omega\ln z)\ln z)z^{\omega+\alpha}$$

$$\partial_z P(\omega,\ln z)z^{\omega+\alpha} = P'_2(\omega,\ln z){1\over z}z^{\omega+\alpha} + P(\omega, \ln z)\omega z^{\omega+\alpha-1} = (P'_2(\omega, \ln z) + P(\omega, \ln z)\omega)z^{\omega+\alpha-1}$$

These are continuous whenever $z\ne 0$. So it will be analytical everywhere else.