Derivative of z^z where z is complex

[Raziel2701]

Homework Statement

Define a single-valued branch of the function $$f(z) =z^z$$ on an open set $$U\subseteq C$$, show f is analytic on U, and find f'(z)

The Attempt at a Solution

$$z^z = e^{zlog(z)}$$ So because of the log I have to define or pick a branch where f(z) is defined. Since f(z) is more or less an increasing exponential function that starts from the y axis, could I just pick x>0 and y>0 for my branch?

I tried substituting x+iy for z, so that I may break the function apart into its real and imaginary parts to see if they satisfy the Cauchy-Riemann equations to test f(z) if it's analytic, but then I get a mess:

$$e^{(x+iy)(log(\sqrt{x^2 + y^2}) +i arg(x+iy)}$$

And I didn't pursue it because I don't even know if this is the right approach.

Also, to find f'(z), I get z^z(log(z) +1)). Is this right?

 

[jackmell]

$e^u$ is analytic if u is analytic and a particular determination (branch) of $z\log(z)$ is analytic if $\log(z)$ is analytic. So for example, choose the principal branch of log:

[tex]\log(z)=\ln|z|+i\Theta,\quad -\pi<\Theta\leq \pi[/itex]

Now, if $z=re^{i\theta}$ then $\log(z)=\ln(r)+i\theta=u(r,\theta)+iv(r,\theta)$

and it's easy to verify the CR equations in polar coordinates for log(z) except at the origin. Therefore,

$$e^{z\log(z)}$$

must then be analytic everywhere except along the negative real axis. If it's analytic, then the derivative exists and you can compute it the ordinary way:

$$\frac{d}{dz}z^z=(1+\log(z))z^z$$