Wolfram answer for cubed root of -1

[daviddoria]

At http://www.wolframalpha.com/ , if you type:

1) (-1)^(1/3)

It gives a complex approximation. Isn't it exactly -1?

David

 

[Pengwuino]

The solution Wolfram gave you is $${1 \over {2}} + i\sqrt{3}/2$$. Cube it and see what you get

 

[stevenb]

At http://www.wolframalpha.com/ , if you type:

1) (-1)^(1/3)

It gives a complex approximation. Isn't it exactly -1?

David

That's one answer. There is also $${\rm e^i}^{{{\pi}\over{3}}}$$ and $${\rm e^-^i}^{{{\pi}\over{3}}}$$

Generally, the n'th root has n values if you include complex numbers. The solutions are on a circle in the complex plane at equal radius from the origin and separated from each other by an angle of 2pi/n.

 

[daviddoria]

So they are just giving the first solution? I'm confused why it doesn't say, -1, e^{i\pi / 3}, e^{-i\pi / 3} ?

 

[jackmell]

Mathematica always returns the principal value for a multi-valued function. So if we define:

$$z^{1/3}=r^{1/3} e^{\frac{i}{3}(\theta+2n\pi)},\quad n=0,1,2$$

then the principal value is:

$$r^{1/3}e^{i/3\theta},\quad -\pi<\theta\leq \pi$$

so that Mathematica returns for:

$$(-1)^{1/3}=e^{i/3(\pi)}$$

if you wanted all three, enter:

Solve[z^3==-1,z]