Can one be many?

Algebra Level 4

" Let $z$ be a complex number, if $z = 1$ then $z^n = 1$ where $n$ is a real number "

How can you describe the previous statement?

Notation: $z^n=e^{n\ln(z)}$ , where $\ln(z)$ is the multiple-valued logarithmic function.

Always false We can't know Depends on the value of

n

Always true

1 solution

Ahmad Hesham
Apr 4, 2016

If $n$ is a real integer, then obviously the statement is true.

However, taking $n$ as a non integer - and let $n = \frac{p}{q}$ where $p & q$ are coprime integers, will produce $\sqrt [ q ]{ { 1 }^{ p } }$ ,obviously $1^p$ equals $1$ , but $\sqrt [ q ]{ { 1 } }$ has a $(q-1)$ number of values beside $1$ .

How I approached this was to first note that $\large 1 = e^{i*2\pi k}$ for any integer $k$ . Then $\large (e^{i*2\pi k})^{n} = e^{i*2\pi kn} = 1$ for any integer $n$ , but if, for example, $n = \frac{1}{2}$ then $\large (e^{i*2\pi k})^{n} = e^{i*k\pi},$ which equals $1$ for even $k$ and $-1$ for odd $k$ .

Brian Charlesworth - 5 years, 2 months ago

I would say the statement is "ambiguous"; it depends on the definition of $z^n$ we are using.

$z^n$ is defined as $e^{\ln(z)n}$ . If we use the principal value of the logarithm, $\ln(1)=0$ , then $1^n=1$ for all complex $n$ . But if we consider the logarithm as a multivalued function, $\ln(1)=2\pi ik$ for integer $k$ , then $1^n=e^{2\pi i kn}$ is multivalued as well.

Otto Bretscher - 5 years, 2 months ago

Hmmm... Yes, I suppose it does come down to definitions, (as it so often does). :) We can identify $z$ such that the statement is true, and we can identify $z$ where it is not necessarily true and depends on the value of $n$ , which is why I went with "depends on the value of $n$ " rather than "we can't know", which is not always the case. Yet as you point out, with the logarithm defined as a multivalued function, even if we do know $n$ we won't be able to identify a unique value for $1^{n}$ . The two answer options I've mentioned are not mutually exclusive, but neither is completely accurate, either, and I'm not sure how best to rectify the ambiguity. Should Ahmad offer a working definition in the text of the question, or perhaps change the "we can't know" option to something along the lines of "we can't always be sure"?

Brian Charlesworth - 5 years, 2 months ago

@Brian Charlesworth – I always work with the multiple-valued logarithmic function as the general form, and all the text books I have encountered define it as a multiple-valued function, it didn't cross my mind that not everybody consider it like that. I'll add a definition to eliminate the ambiguity from the statement

Ahmad Hesham - 5 years, 2 months ago

@Ahmad Hesham – I'm afraid that the remark on "Notation" does not really settle the issue since the argument is often restriced, usually to $-\pi<\theta\leq \pi$ .

Otto Bretscher - 5 years, 2 months ago

@Otto Bretscher – $Arg(z)$ is the principle value of the argument, which is restricted to $-\pi<Arg(z)\le \pi$ . However, $arg(z)$ is multiple-valued " $arg(z)=Arg(z)+2k\pi ,k=0,\pm 1,\pm 2,...$ ", and varies from $-\infty$ to $\infty$ , It's the main reason the logarithmic function is considered multiple-valued function.

Ahmad Hesham - 5 years, 2 months ago

@Ahmad Hesham – No! This is only one possible definition, but it is not generally agreed upon. Wolfram's Math World states that $\arg(1)=0$ , as does Wolfram Alpha .

It is always a bit futile to argue over definitions. It is best to accept that many definitions in math are not "set in stone."

To make the probem meaningful, I suggest to write something like "Notation: $z^n:=e^{\ln(z)n}$ where $\ln(z)$ is the multivalued complex logarithm."

A small mistake in your solution: If $q$ is a positive integer and if we use the multivalued root, then $\sqrt[q]{1}$ has $q$ values, of course, not $q-1$ .

Otto Bretscher - 5 years, 2 months ago

@Otto Bretscher – You're right, I think your suggestion is the best way to avoid any ambiguity.

I'll edit the mistake of the number of values, I meant to write " q - 1 values other than 1" but I changed it and didn't notice the error.

Thank you for correcting my misconceptions.

Ahmad Hesham - 5 years, 2 months ago

@Ahmad Hesham – Exactly! "Avoiding any ambiguity" is a great guideline when writing math problems. (I have learned that the hard way by writing a lot of exams, with a lot of ambiguities, as it turned out.) Thanks!

Otto Bretscher - 5 years, 2 months ago

Can one be many?

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