Hi,

I am trying to figure out what $\displaystyle (-3)^\frac{2}{3}$ is. I calculated it, possibly incorrectly as $\displaystyle \approx{2.08}$ by $\displaystyle \sqrt[3]{(-3)^2}$. I would calculate $\displaystyle -\sqrt[3]{3}, as \approx{-1.4422}$ then square that to get $\displaystyle \approx{2.08}$

All the maths applications I have calculate it as -1.0400 + 1.8014i. Which is $\displaystyle (\sqrt[3]{3}[\frac{1+i\sqrt{3}}{2}])^2$. This complex fraction is courtesy of my HP.

My questions are twofold. Why if cube roots are defined for all values on the number line do the calculators, and matlab produce an answer with an imaginary part? I thought that all even rational exponents produce positive real results, and not imaginary results?

If the imaginary result above is correct, and I can trust the calculator, how on earth did it get to that complex fraction?

Thanks

Regards

Craig.