Derivative of z^z where z is complex

• Raziel2701
In summary, we need to define a branch for the function f(z) = z^z, which can be done by choosing a particular determination (branch) for log(z). By choosing the principal branch, we can show that f(z) is analytic on an open set U⊆C. To find the derivative f'(z), we can use the ordinary method, which results in (1+log(z))z^z.
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?

$e^u$ 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:

$$\log(z)=\ln|z|+i\Theta,\quad -\pi<\Theta\leq \pi[/itex] Now, if $z=re^{i\theta}$ then $\log(z)=\ln(r)+i\theta=u(r,\theta)+iv(r,\theta)$ and it's easy to verify the CR equations in polar coordinates for log(z) except at the origin. Therefore, [tex]e^{z\log(z)}$$

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:

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

1. What is the formula for the derivative of z^z where z is complex?

The formula for the derivative of z^z where z is complex is dz^z = z^z(lnz + 1)dz.

2. How do you calculate the derivative of z^z where z is a complex number?

To calculate the derivative of z^z where z is a complex number, you can use the formula dz^z = z^z(lnz + 1)dz or you can use the definition of the derivative, which is the limit of (f(z+h) - f(z))/h as h approaches 0.

3. What is the significance of the derivative of z^z where z is complex?

The derivative of z^z where z is complex is significant because it allows us to find the slope of the function at any point and to determine whether the function is increasing or decreasing at that point.

4. Can the derivative of z^z where z is complex be calculated using the power rule?

No, the power rule can only be applied to functions with a constant base and variable exponent. Since z is a complex number, it cannot be treated as a constant, and therefore the power rule cannot be used.

5. Are there any special cases to consider when calculating the derivative of z^z where z is complex?

Yes, there are a few special cases to consider when calculating the derivative of z^z where z is complex. One is when z is equal to 0, in which case the derivative is also 0. Another is when z is equal to 1, in which case the derivative is 1. Additionally, when z is a negative real number, the derivative is undefined.

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