# Derivative of $z^{c}$

1. Nov 11, 2014

### Bashyboy

b1. The problem statement, all variables and given/known data
Let $c$ and $z$ denote complex numbers. Then

1. When a branch is chosen for $z^c$, then $z^c$ is analytic in the domain determined by that branch.

2. $\frac{d}{dz} z^c = c z^{c-1}$

2. Relevant equations

3. The attempt at a solution

In regards to number one, we have that $z^c$ is defined as $z^c = e^{c \log z}$ which can be written as $z^c = e^{c(\ln |z| + i \arg z}$. For $z^c$ to be analytic, it must first be a function. In the textbook I am using, the author discussing how we can remove a branch of the domain of a multi-valued function$f(z)$ so that it can be become a function.

For a given complex number, $\ln|z|$ will return a single value; however, $\arg z$ will return multiple values, whereby we get different values of $z^c$. Thus, the problem of $z^c$ being multivalued lies with the fact that $\arg z$ is multivalued. So, if I were to restrict the domain so that $\arg z$ was unique, then $z^c$ would be single-valued.

I know that if I replaced $\arg z$ with $Arg ~z$, then $z^c$ would be multivalued. So, if I restricted the domain by removing all of the complex numbers along the negative real axis, including zero, then what would happen to $\arg z$?

How could I formulate a general restriction? I was thinking that if we remove the ray which makes an angle of $\theta$ with the positive real axis, all remove all those points, then the range of $\arg z$ would become $(\theta - \pi, \pi + \theta)$. But I am unsure of how to justify this.
______________________________________________________________________________

Number two is giving me some difficulty. At my disposal, I have $\frac{d}{dz} e^z = e^z$ and the chain rule.

If I take the derivative of $z^c = e^{c(\ln |z| + i \arg z}$, I would have to take the derivative of $e^z$, which I already know, and $c(\ln |z| + i \arg z)$, which I do not know. I would venture to guess that $\frac{d}{dz} \ln |z| = \frac{1}{|z|}$, but this has not been established. What would the derivative of $\arg z$ be?

2. Nov 12, 2014

### Bashyboy

Okay, I was able to figure out how to differentiate the function $z^c$ (of course, to be able to differentiate this function, we first have to remove a branch of the function, of which I am still uncertain of how to do). I found in my textbook that $e^{\log z} = z$. Using the chain rule, I could compute the derivative and find out what the derivative of $\log z$ is, when you remove a branch from the function:

$\frac{d}{dz} [e^{\log z}] = \frac{d}{dz} z \iff$

$\frac{d}{dz}[\log z] e^{\log z} = 1 \iff$

Because $e^{\log z}$ is never zero, we can divide by this quantity to get

$\frac{d}{dz} [\log z ] = \frac{1}{e^{\log z}} \iff$

$\frac{d}{dz} [\log z ] = \frac{1}{z}$

Now we have the requisite knowledge to differentiate $z^c$:

$\frac{d}{dz} z^c = \frac{d}{dz} [e^{c \log z}] \iff$

$\frac{d}{dz} z^c = \frac{d}{dz} [e^{\log z^c}] \iff$

$\frac{d}{dz} z^c = \frac{d}{dz} [\log z^c] e^{\log z^c} \iff$

$\frac{d}{dz} z^c = c \frac{1}{z} e^{c \log z} \iff$

$\frac{d}{dz} z^c = c \frac{1}{z} z^c \iff$

$\frac{d}{dz} z^c = c z^{c-1}$

So, this much I figured out. However, I am still having difficulty with this concept of removing a branch. In general, how do you remove an arbitrary branch from the function $z^c$, so that it becomes a single-valued function, rather than a multi-valued one?

Last edited: Nov 12, 2014