imaginary number $i$ equals $-6/3.4641$?

Question

$$-4^3 = -64$$ so the third root of $-64$ should be $-4$ than. $$\sqrt[3]{-64} = -4$$ But if you calculate the third root of -64 with WolframAlpha( http://www.wolframalpha.com/input/?i=third+root+of+-64 ) you get a complex number with an imaginary part of $$3.4641016151 i$$ and a real part of $$2$$

so if the third root of $4-64$ equals $-4$ AND $2 + 3.46410162 i$ (which i know is a bit foolish) than you could actually reform it like this $$ \sqrt[3]{-64} \approx 2 + 3.46410162 i | -2$$ $$ \sqrt[3]{-64} -2 \approx -6 \approx 3.46410162 i |/3.46410162$$ $$ \frac{\sqrt[3]{-64} -2}{3.46410162} ≈ \frac{-6}{3.46410162} ≈ i$$

and this have to be totally wrong, so my question is, where exactly is the mistake?

Answer 1

If we allow complex number solutions, there are three distinct solutions to the equation $x^{3}+64=0$. One of them is $-4$. Wolfram|Alpha is simply giving you the other root(s). They are not the same, and it is incorrect to say "$-4=2+(3.46\ldots)i$", because, in $\mathbb{C}$, we define two numbers to be equal if and only if they have equal real part and equal imaginary part; we know that $2\neq-4$, so these two complex numbers have unequal real part, and so are not equal.

What this means more generally is that, although in $\mathbb{R}$ we have theorems such as "$x^{3}=y^{3} \Rightarrow x=y$", in $\mathbb{C}$ the situation is quite different: in $\mathbb{C}$, the implication does not hold in general.

Answer 2

Note that $x^3+64=0$ implies $(x+4)(x^2-4x+16)=0$. So you have the three cubics roots of $-64$.

Answer 3

Every complex number has three cube roots: $$ (-4)^3 = -64 $$ $$ \left( 2 \pm 2i\sqrt 3 \right)^3 = -64 $$ You can find those last two cube roots by saying $-64 = 64(\cos180^\circ+i\sin180^\circ)$, so the cube roots are $$ 4(\cos60^\circ\pm i\sin60^\circ). $$ Notice that $3\times (-60^\circ) = -180^\circ$ and for present purposes $+180^\circ$ and $-180^\circ$ are the same since they both express the same thing in polar coordinates.

The fact that $(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta) = \cos(\alpha+\beta)+i\sin(\alpha+\beta))$ perhaps deserves more attention in courses where these things are taught. In this instance in place of $\alpha+\beta$ we have $60^\circ+60^\circ+60^\circ$.

You can also use algebra, as follows. $$ x^3 = -64 $$ $$ x^3 + 64 = 0 $$ Then since we know $-4$ is one solution we have $$ (x+4)(\cdots \cdots \cdots) = 0. $$ Dividing $x^3+64$ by $x+4$ (by long division if you like), we get $$ (x+4)(x^2 - 4x + 16) = 0. $$ Therefore $$ x+4 = 0 \quad\text{or}\quad x^2 - 4x+16 = 0. $$ Then solve the quadratic equation by any of the usual methods.

Your question seems to confuse $-4^3$ with $(-4)^3$. In this case they are equal, but if you confuse $-4^2$ with $(-4)^2$ it will lead you quickly into errors.

Answer 4

Cube roots of negative real numbers are not defined in a universally agreed-upon way. As other answers here explain, there are always three distinct complex cube roots to a nonzero number. So if you try to talk about the cube root, you have a problem.

If you know you are working with real numbers only, then since precisely one of the three (complex) cube roots of a negative real number is real, it would seem OK to declare the cube root of a negative real to be that one real cube root.

But Wolfram Alpha and other applications are going to consider complex cube roots. For them, if you ask for the cube root, they will go with the one that has the smallest angular argument. For a negative real number, this cube root ends up as a complex number in the first quadrant.

These two systems shouldn't be used simultaneously by say, choosing the first when your number is negative real, and the second when it is not real. Because then you have the problem that numbers like $\sqrt[3]{-1}$ and $\sqrt[3]{-1+0.0001i}$, which should be very close to each other if we want to respect continuity, end up being very far apart.