Negative number raised to the two-thirds power

• Oct 2nd 2011, 11:28 AM
cristumagnus
Negative number raised to the two-thirds power
Hello. I'm a calculus 1 student in college, and I recently came across a function where x was raised to the (2/3) power. Does anyone in this forum know why a negative number raised to the 2/3 becomes an imaginary number? Type the problem into any calculator, including Google. You'll get an error or an imaginary number.

(-1)^(2/3)

But why does this give you an imaginary number? It's the same as ((-1)^2)^(1/3) or ((-1)^(1/3))^(2). It seems like when you square (-1) you get (1), and the cube root of (-1) is just (-1). Can anyone on this forum explain why this happens?
• Oct 2nd 2011, 11:33 AM
CaptainBlack
Re: Negative number raised to the two-thirds power
Quote:

Originally Posted by cristumagnus
Hello. I'm a calculus 1 student in college, and I recently came across a function where x was raised to the (2/3) power. Does anyone in this forum know why a negative number raised to the 2/3 becomes an imaginary number? Type the problem into any calculator, including Google. You'll get an error or an imaginary number.

(-1)^(2/3)

But why does this give you an imaginary number? It's the same as ((-1)^2)^(1/3) or ((-1)^(1/3))^(2). It seems like when you square (-1) you get (1), and the cube root of (-1) is just (-1). Can anyone on this forum explain why this happens?

It is because of the way the calculator compute non-integer powers (essentially taking logs, which requires that the number being raised to the power be positive, also if you calculator is smart enough to use complex logarithms it won't return the answer you expect since the branch used is not the real branch).

CB
• Oct 2nd 2011, 11:37 AM
alexmahone
Re: Negative number raised to the two-thirds power
Quote:

Originally Posted by cristumagnus
Hello. I'm a calculus 1 student in college, and I recently came across a function where x was raised to the (2/3) power. Does anyone in this forum know why a negative number raised to the 2/3 becomes an imaginary number? Type the problem into any calculator, including Google. You'll get an error or an imaginary number.

(-1)^(2/3)

But why does this give you an imaginary number? It's the same as ((-1)^2)^(1/3) or ((-1)^(1/3))^(2). It seems like when you square (-1) you get (1), and the cube root of (-1) is just (-1). Can anyone on this forum explain why this happens?

$\displaystyle (-1)^{2/3}$

$\displaystyle =(\cos \pi+i\sin \pi)^{2/3}$

$\displaystyle =\cos \frac{2}{3}\pi+i\sin \frac{2}{3}\pi$

$\displaystyle =-0.5 + 0.866025404 i$
• Oct 2nd 2011, 11:40 AM
Plato
Re: Negative number raised to the two-thirds power
• Oct 2nd 2011, 11:53 AM
CaptainBlack
Re: Negative number raised to the two-thirds power
Quote:

Originally Posted by alexmahone
$\displaystyle (-1)^{2/3}$

$\displaystyle =(\cos \pi+i\sin \pi)^{2/3}$

$\displaystyle =\cos \frac{2}{3}\pi+i\sin \frac{2}{3}\pi$

$\displaystyle =-0.5 + 0.866025404 i$

And how did you determine which branch to use?

CB
• Oct 2nd 2011, 11:56 AM
skeeter
Re: Negative number raised to the two-thirds power
Ti-84 ...
• Oct 2nd 2011, 12:02 PM
alexmahone
Re: Negative number raised to the two-thirds power
Quote:

Originally Posted by CaptainBlack
And how did you determine which branch to use?

CB

I suppose there are many ways of calculating $\displaystyle (-1)^{2/3}$.

Another would be:

$\displaystyle (-1)^{2/3}$

$\displaystyle =(\cos 3\pi+i\sin 3\pi)^{2/3}$

$\displaystyle =\cos \left(\frac{2}{3}\times 3\pi\right)+i\sin \left(\frac{2}{3}\times 3\pi\right)$

$\displaystyle =\cos 2\pi+i\sin 2\pi$

$\displaystyle =1$
• Oct 2nd 2011, 12:02 PM
cristumagnus
Re: Negative number raised to the two-thirds power
Thank you. This is the answer I got from Google. Can you tell me what math subject I can study to learn about this? I want to build my math skills. I will find books and videos if you tell me the specific subject. For example, in college algebra and trigonometry, we never studied any of this. I have become very interested in math and I want to study math.
• Oct 2nd 2011, 12:04 PM
alexmahone
Re: Negative number raised to the two-thirds power
Quote:

Originally Posted by cristumagnus
Thank you. This is the answer I got from Google. Can you tell me what math subject I can study to learn about this? I want to build my math skills. I will find books and videos if you tell me the specific subject. For example, in college algebra and trigonometry, we never studied any of this. I have become very interested in math and I want to study math.

Complex numbers.
• Oct 2nd 2011, 12:08 PM
alexmahone
Re: Negative number raised to the two-thirds power
To sum up, all 3 cube roots of unity are valid answers for $\displaystyle (-1)^{2/3}$.