1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of ##z^{c}##

  1. Nov 11, 2014 #1
    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. jcsd
  3. Nov 12, 2014 #2
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Derivative of ##z^{c}##
  1. Derivative of Im(z) (Replies: 4)

Loading...