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nachanga
Posted: Thu Mar 08, 2007 10:42 am
Guest
Hello everyone.

I have a possibly silly question.

I know that an even root of a negative number only have complex
solutions. For instance, square root of -3 is a complex number.
I know that an odd root of a negative number has a real solution. For
instance, cubic root of -3 has a real solution.

Now this is my question: (-2)^PI has a real solution or only complex?

I know I can approximate PI as much as I want with a rational number
such as p/q where p is even. Then (-2)^(p/q) is real.

But I can also aproximate PI as much as I want with p'/q' where p' is
odd and q' is even. Then (-2)^(p'/q') is complex.

Then (-2)^PI has a real solution or only complex?

Thank you in advance for your help.
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Narcoleptic Insomniac
Posted: Thu Mar 08, 2007 10:42 am
Guest
On Mar 8, 2007 8:42 AM CT, nachanga wrote:

Quote:
Hello everyone.

I have a possibly silly question.

I know that an even root of a negative number only
have complex solutions. For instance, square root of
-3 is a complex number. I know that an odd root of a
negative number has a real solution. For instance,
cubic root of -3 has a real solution.

Now this is my question: (-2)^PI has a real solution
or only complex?

I know I can approximate PI as much as I want with a
rational number such as p/q where p is even. Then
(-2)^(p/q) is real.

But I can also aproximate PI as much as I want with
p'/q' where p' is odd and q' is even. Then
(-2)^(p'/q') is complex.

Then (-2)^PI has a real solution or only complex?

Thank you in advance for your help.

Please only post a topic once, or cross-post the topics!

If I would have known this, I wouldn't have taken the
time to post reply at:

http://mathforum.org/kb/message.jspa?messageID=5561451&tstart=0
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Robert Israel
Posted: Thu Mar 08, 2007 12:08 pm
Guest
"nachanga" <basura666@gmail.com> writes:

Quote:

Hello everyone.

I have a possibly silly question.

I know that an even root of a negative number only have complex
solutions. For instance, square root of -3 is a complex number.
I know that an odd root of a negative number has a real solution. For
instance, cubic root of -3 has a real solution.

Now this is my question: (-2)^PI has a real solution or only complex?

I know I can approximate PI as much as I want with a rational number
such as p/q where p is even. Then (-2)^(p/q) is real.

But I can also aproximate PI as much as I want with p'/q' where p' is
odd and q' is even. Then (-2)^(p'/q') is complex.

Then (-2)^PI has a real solution or only complex?

By definition, the values of (-2)^pi are those of exp(pi log(-2))
= exp(pi ln(2) + i pi^2 (2n + 1)) for integers n. For that to be
real, you'd need pi^2 (2n + 1) to be an integer multiple of pi.
Since pi is irrational, that is impossible.
--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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Olin Perry Norton
Posted: Thu Mar 08, 2007 3:59 pm
Guest
nachanga wrote:
Quote:
Hello everyone.

I have a possibly silly question.

I know that an even root of a negative number only have complex
solutions. For instance, square root of -3 is a complex number.
I know that an odd root of a negative number has a real solution. For
instance, cubic root of -3 has a real solution.

[snip]


The equation x^3 = -3 is a cubic polynomial.
It has three solutions. One of these solutions
is real. The other two are complex.

Olin Perry Norton
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nachanga
Posted: Fri Mar 09, 2007 3:17 am
Guest
On Mar 8, 6:48 pm, Narcoleptic Insomniac
<i_have_narcoleptic_insom...@yahoo.com> wrote:
Quote:
On Mar 8, 2007 8:42 AM CT, nachanga wrote:





Hello everyone.

I have a possibly silly question.

I know that an even root of a negative number only
have complex solutions. For instance, square root of
-3 is a complex number. I know that an odd root of a
negative number has a real solution. For instance,
cubic root of -3 has a real solution.

Now this is my question: (-2)^PI has a real solution
or only complex?

I know I can approximate PI as much as I want with a
rational number such as p/q where p is even. Then
(-2)^(p/q) is real.

But I can also aproximate PI as much as I want with
p'/q' where p' is odd and q' is even. Then
(-2)^(p'/q') is complex.

Then (-2)^PI has a real solution or only complex?

Thank you in advance for your help.

Please only post a topic once, or cross-post the topics!

If I would have known this, I wouldn't have taken the
time to post reply at:

http://mathforum.org/kb/message.jspa?messageID=5561451&tstart=0- Hide quoted text -

- Show quoted text -

I'm sorry for posting twice, but I wasn't sure if I posted in the
right place the first time.

Thank you very much for your answers. They totally resolved my
question.
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