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

[Bashyboy]

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 $n$?

 

[pasmith]

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 $n \mod 4$.

 

[Bashyboy]

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?

 

[Mark44]

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.

 

[pasmith]

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.

 

[Bashyboy]

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.

 

[Bashyboy]

Okay, I see what the problem is. I was confusing the terms "principal argument" and "principal value."