Ask Your Question

Imaginary result for cube root of -1

asked 2020-09-17 16:44:47 -0600

cybervigilante gravatar image

I get -1 as the cube root of -1,which is correct. But if I put -1 in parentheses I get an imaginary result. I was hoping the imaginary result, if cubed, would give me -1, but it got worse.


0.500000000000000 + 0.866025403784439*I
edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted

answered 2020-09-17 19:08:57 -0600

slelievre gravatar image

The power operator has higher priority than the unary negation.

So in the first case you are really doing "minus (1 to the 1/3)", and not "(minus 1) to the 1/3".

For the real n-th root, use real_nth_root.

sage: real_nth_root(-1, 3)

Regarding cubing the cube root:

sage: a = (-1)^(1/3)
sage: a
sage: a^3

Using the numerical approximation of the cube root:

sage: aa = a.n()
sage: aa
0.500000000000000 + 0.866025403784439*I
sage: aa^3
-1.00000000000000 + 3.88578058618805e-16*I

So, very close to -1, with tiny imaginary part from rounding errors.

edit flag offensive delete link more



$x^3+1=0$ has three roots in $\mathbb{C}$. Therefore it could be argued that (-1)^(1/3) is in fact a set of three numbers ; our habit of using this notation to denote one of these numbers

  • is not consistent, and

  • does not explicitly specify which of these numbers it denotes

But this is far to be the sole inconsistency in our mathematical notational conventions... Equating $\sqrt [3]{-1}$ to $-1$ is just a habit...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2020-09-18 01:23:51 -0600 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower


Asked: 2020-09-17 16:44:47 -0600

Seen: 44 times

Last updated: Sep 17