Why does:
$$(-2)^{\frac{2}{3}}$$
have a complex component?
I thought it would be equal to:
$$((-2)^2)^{\frac{1}{3}}$$
$$= 4^{\frac{1}{3}}$$
which doesn't have a complex component.
But Wolfram Alpha says it has a complex component:
Asked
Active
Viewed 70 times
1
-
It's a bit dangerous to apply the rules of exponents just like that if the base is a negative number. In essence, your solution is a solution to the equation $x^3=4$ and that has three solutions, complex included – imranfat Nov 08 '13 at 03:22
-
Related : http://math.stackexchange.com/questions/192742/how-to-solve-x3-1 – lab bhattacharjee Nov 08 '13 at 03:27
2 Answers
2
It took the cube root first. There are three cube roots of a given value. It took one with an imaginary component. Exactly why it did that I do not know. I've found with wolfram-alpha you have to be really careful and hamfisted when there are cube roots involved, otherwise it will produce non-real roots for real inputs. You have to do it like this:
http://www.wolframalpha.com/input/?i=%28%28-2%29^2%29^%281%2F3%29
You may have to go to the Mathematica SE to learn more technical reasons for why it does this.
zibadawa timmy
- 4,425
0
More generally, you can think of the question "What is $(-2)^{\frac{2}3}$?" as the statement "$x=(-2)^{\frac{2}3}$ Solve for $x$." which is another way of saying "Solve $x^{\frac{3}2}+2=0$ for x."
That last polynomial has (by the Fundamental Theorem of Algebra) 3 roots in the complex plane because it is equivalent to solving $x^3-4=0$. One of the roots is $\sqrt[3] 4$ as you point out, but the others are $\sqrt[3] 4e^{i\frac{2\pi}3}$ and $\sqrt[3] 4e^{i\frac{4\pi}3}$. (You may notice that $-2=2e^{in\pi}$ for $n\in\{\text{Odds}\}$; therefore, $(-2)^{\frac{2}3}=\sqrt[3]{(-2)^2}e^{in\frac{2\pi}3}$.)
Geoffrey
- 2,382