what's wrong with this?

[kochibacha]

as far as i know (x^a)^b=x^a*b isn't it?

((-1)^(1/3))^2 = 1

(-1)^(2/3) = -0.5 + 0.866025404 i

could someone explain this weird math inequality?

 

[JohnZ]

(-1)^(2/3) = -0.5 + 0.866025404 i

how do you get this number?

 

[kochibacha]

how do you get this number?

from google and wolfram which are very reliable

https://www.google.co.th/search?q=(...7j0.232j0j7&sourceid=chrome&es_sm=93&ie=UTF-8

http://www.wolframalpha.com/input/?i=(-1)^(2/3)

 

[JohnZ]

from Wolfram, if you select Real-value root, then (-1)^2/3 is 1.
-0.5 + 0.866 * i is the solution in complex number system.
But don't know why they are not match

 

[stapel]

The different values probably stem from how the values were calculated. Different methods (different assumptions, such as "real" or "complex", etc) may lead to different solutions.

I'm just guessing, but that's what my experience suggests, based on some odd results I've encountered in graphing calculators.

 

[HallsofIvy]

as far as i know (x^a)^b=x^a*b isn't it?

No- not for all numbers. That is true for a and b real and x a positive real number.

((-1)^(1/3))^2 = 1

(-1)^(2/3) = -0.5 + 0.866025404 i.

could someone explain this weird math inequality?

When you are dealing with complex numbers, -1, like any number, has three third roots.
They are -1, \(\displaystyle \frac{1}{2}+ \frac{\sqrt{3}}{2}i\), and \(\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}\). Squaring gives 1, \(\displaystyle -\frac{1}{2}+ \frac{\sqrt{3}}{2}\) (the root you have above), and \(\displaystyle -\frac{1}{2}- \frac{\sqrt{3}}{2}\).

 

[kochibacha]

No- not for all numbers. That is true for a and b real and x a positive real number.


When you are dealing with complex numbers, -1, like any number, has three third roots.
They are -1, \(\displaystyle \frac{1}{2}+ \frac{\sqrt{3}}{2}i\), and \(\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}\). Squaring gives 1, \(\displaystyle -\frac{1}{2}+ \frac{\sqrt{3}}{2}\) (the root you have above), and \(\displaystyle -\frac{1}{2}- \frac{\sqrt{3}}{2}\).

thx a lot

so when dealing with physical phenomena only the roots which are real number would make sense right? i just wonder if the complex roots means anything in solving physical phenomena

 

[HallsofIvy]

Complex roots can have physical meaning if you have a mathematical model for your physical situation that uses complex numbers.
And there are such models- such as the Dirac operators in Quantum Mechanics.