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

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Homework Help Overview

The discussion revolves around the multi-valued nature of the expression ##z^{1/4}##, particularly in the context of complex numbers. Participants explore how many distinct values this expression can take and the implications of using different values of ##n## in the context of complex exponentiation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the multi-valued nature of ##z^{1/4}## and question how the distinct roots relate to the values of ##n##. There is also exploration of the principal argument in relation to the expression ##(1 - \sqrt{3})^{1/5}## and the confusion surrounding the definition of principal argument versus principal value.

Discussion Status

The discussion is ongoing, with participants providing insights into the number of distinct solutions and the nature of the principal argument. Some participants have offered clarifications regarding the relationship between distinct roots and their corresponding arguments, while others are still grappling with the definitions and implications of these concepts.

Contextual Notes

There are mentions of specific constraints related to the values of ##z## and the interpretation of complex logarithms, as well as issues with LaTeX rendering that have been noted by participants.

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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?

 
Last edited by a moderator:
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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