How Many Solutions To ##z^{1/4}##

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Homework Statement


Is ##z^{1/4}## multi-valued or single-valued? How many possible values can it have in general?

Homework Equations



##z## raised to some complex power ##c## is defined as ##z^c = e^{c \log z}##.

The Attempt at a Solution



##
z^{\frac{1}{4}} = e^{\frac{1}{4} \log z} \iff
##

##
z^{\frac{1}{4}} = e^{\frac{1}{4} (\ln|z| + i \arg z)} \iff
##

##
z^{\frac{1}{4}} = e^{\ln |z|^{1/4} + \frac{i}{4} (\theta + 2 n \pi)} \iff
##

##
z^{\frac{1}{4}} = \sqrt[4]{|z|} e^{i(\frac{\theta}{4} + \frac{\pi}{2} n)}
##

I would say, "yes, it is multi-valued, as each value of ##n## gives you a distinct solution." However, in general, I am unsure of how many solutions there are. For a given ##z##, doesn't ##\sqrt[4]{|z|}## give us four distinct roots; and if we let ##n## be 0, 1, 2, 3, would we get four distinct solutions (n=4 and beyond would give us repetitive solutions)? Furthermore, how do I know the solutions I get from ##\sqrt[4]{|z|}## correspond to the ones gotten by enumerating values of ##n##?
 
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Bashyboy said:
For a given ##z##, doesn't ##\sqrt[4]{|z|}## give us four distinct roots; and if we let ##n## be 0, 1, 2, 3, would we get four distinct solutions (n=4 and beyond would give us repetitive solutions)? Furthermore, how do I know the solutions I get from ##\sqrt[4]{|z|}## correspond to the ones gotten by enumerating values of ##n##?

Since |z| is a non-negative real number, both |z|^{1/4} and \sqrt[4]{|z|} always mean the unique non-negative real number c such that c^4 = |z|. Thus there are in general four distinct values of z^{1/4}, corresponding to the four distinct values of n \mod 4.
 
Oh, I see.

I actually have another one which is also giving me trouble:

## (1- \sqrt{3})^{1/5} = e^{ \frac{1}{5} (\ln |1 - \sqrt{3}| + i (\pi + 2 \pi n))} ##

## (1- \sqrt{3})^{1/5} = e^{ \ln (1 -\sqrt{3})^{1/5} + i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

## (1- \sqrt{3})^{1/5} = (1 - \sqrt{3})^{1/5} e^{i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

If I let ##n## go through the integers from 0 to 4, I will get the angles ##\frac{\pi}{5}##, ##\frac{3 \pi}{5}##, ##\frac{5 \pi }{5}##, ##\frac{7 \pi}{5}##, and ##\frac{9 \pi}{5}##.

What would the principal argument be? I see three...I thought the principal argument was that unique angle in the range ##(-\pi,\pi]##, yet I see three.

EDIT: I don't know why I can't get the latex to render properly. I ran the code in TexStudio and it rendered perfectly. Could anyone spot the issue?

 
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Bashyboy said:
EDIT: I don't know why I can't get the latex to render properly. I ran the code in TexStudio and it rendered perfectly. Could anyone spot the issue?
You had an extra right brace in there, now removed.
 
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Bashyboy said:
Oh, I see.

I actually have another one which is also giving me trouble:

## (1- \sqrt{3})^{1/5} = e^{ \frac{1}{5} (\ln |1 - \sqrt{3}| + i (\pi + 2 \pi n))} ##

## (1- \sqrt{3})^{1/5} = e^{ \ln (1 -\sqrt{3})^{1/5} + i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

## (1- \sqrt{3})^{1/5} = (1 - \sqrt{3})^{1/5} e^{i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

If I let ##n## go through the integers from 0 to 4, I will get the angles ##\frac{\pi}{5}##, ##\frac{3 \pi}{5}##, ##\frac{5 \pi }{5}##, ##\frac{7 \pi}{5}##, and ##\frac{9 \pi}{5}##.

What would the principal argument be? I see three...I thought the principal argument was that unique angle in the range ##(-\pi,\pi]##, yet I see three.

I see immediately the problem that 1 < \sqrt{3}, so |1 - \sqrt{3}| = \sqrt{3} - 1. Thus (1 - \sqrt{3})^{1/5} = (\sqrt{3} - 1)^{1/5}e^{i\pi(1 + 2n)/5}. There are five distinct roots. Taking n = 0,1,2 yields principal arguments of \frac \pi 5, \frac {3\pi}5 and \pi respectively. This last is the negative real number -(\sqrt{3} - 1)^{1/5}. Taking n = 4, 5 gives values outside the range (-\pi,\pi] so they must be reduced mod 2\pi to obtain -\frac{3\pi}5 and -\frac{\pi}5 respectively.
 
Oh, I see. Each distinct has their own principle value. For some reason, I was thinking that only one of the five distinct roots would be the principal argument.
 
Okay, I see what the problem is. I was confusing the terms "principal argument" and "principal value."
 

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