- #1
Bashyboy
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b1. Homework Statement
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}##
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.
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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?
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}##
Homework Equations
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?