" 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.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
How I approached this was to first note that 1 = e i ∗ 2 π k for any integer k . Then ( e i ∗ 2 π k ) n = e i ∗ 2 π k n = 1 for any integer n , but if, for example, n = 2 1 then ( e i ∗ 2 π k ) n = e i ∗ k π , which equals 1 for even k and − 1 for odd k .
Log in to reply
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 π i k for integer k , then 1 n = e 2 π i k n is multivalued as well.
Log in to reply
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"?
Log in to reply
@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
Log in to reply
@Ahmad Hesham – I'm afraid that the remark on "Notation" does not really settle the issue since the argument is often restriced, usually to − π < θ ≤ π .
Log in to reply
@Otto Bretscher – A r g ( z ) is the principle value of the argument, which is restricted to − π < A r g ( z ) ≤ π . However, a r g ( z ) is multiple-valued " a r g ( z ) = A r g ( z ) + 2 k π , k = 0 , ± 1 , ± 2 , . . . ", and varies from − ∞ to ∞ , It's the main reason the logarithmic function is considered multiple-valued function.
Log in to reply
@Ahmad Hesham – No! This is only one possible definition, but it is not generally agreed upon. Wolfram's Math World states that ar g ( 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 q 1 has q values, of course, not q − 1 .
Log in to reply
@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.
Log in to reply
@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!
Problem Loading...
Note Loading...
Set Loading...
If n is a real integer, then obviously the statement is true.
However, taking n as a non integer - and let n = q p where p & q are coprime integers, will produce q 1 p ,obviously 1 p equals 1 , but q 1 has a ( q − 1 ) number of values beside 1 .