Does (-2)^(⅔) have an imaginary component?

[SafiBTA]

Some calculators say (-2)^2/3 is equal to $-\frac{1}{2^\frac{1}{3}}+i\frac{3^\frac{1}{2}}{2^\frac{1}{3}}$ while others say its equal to $4^{\frac{1}{3}}$ i.e. $|-\frac{1}{2^\frac{1}{3}}+i\frac{3^\frac{1}{2}}{2^\frac{1}{3}}|$.

I think I am right to imply from above that (-2)^2/3 does have an imaginary component. But if it is so, then why do many calculus books, while speaking about cusps, treat x^2/3 as a real-valued function by plotting it in the negative x region of the real plane?

What did I miss?

 

[Mentallic]

Actually, $(-2)^{2/3}$ has 3 roots. Two of them are complex and one is real. In fact, this is true for any positive x in $(-x)^{2/3}$.

Some calculators choose

Some calculators say (-2)^2/3 is equal to $-\frac{1}{2^\frac{1}{3}}+i\frac{3^\frac{1}{2}}{2^\frac{1}{3}}$

because this is the principal root (smallest argument/angle in the complex plane).

Other calculators choose

while others say its equal to $4^{\frac{1}{3}}$ i.e. $|-\frac{1}{2^\frac{1}{3}}+i\frac{3^\frac{1}{2}}{2^\frac{1}{3}}|$.

because this is the real valued root (second largest argument in complex plane).

and the final root is the complex conjugate of the principal root, which is

$$-\frac{1}{2^\frac{1}{3}}-i\frac{3^\frac{1}{2}}{2^\frac{1}{3}}$$

Note that all of these roots have equal magnitude.

I think I am right to imply from above that (-2)^2/3 does have an imaginary component. But if it is so, then why do many calculus books, while speaking about cusps, treat x^2/3 as a real-valued function by plotting it in the negative x region of the real plane?

What did I miss?

So you have 2 choices. You can choose the principal root or its conjugate for x<0 and be stuck with complex values, so you'll be dealing with complex values in your derivatives, or you can choose the real valued root and be able to answer the questions as they're expecting you to answer them. Your choice ;)