Does the value of $(-37)^{1/33}$ exist?

Does the value of $$(-37)^{1/33}$$ exist? I am very much confused with this. In desmos it is giving a real value whereas on the other hand in Wolfram Alpha it is giving an imaginary value. So, please give me the right answer? I want to know who is correct? Wolfram Alpha or Desmos? In desmos the value is showing $$-1.115$$.

• There are $33$ complex numbers which satisfy $z^{33}=-37$. Exactly one of them is real.
– Mark
Commented Oct 27, 2023 at 17:16
• Wolfram Alpha gives the value considering it as a complex number, with the standard definition of n-root for complex numbers. Desmos does the same but considering it as real number. Commented Oct 27, 2023 at 17:20
• Too much reliance on software! And a better statement would be "is there a real value of this" or "what are the complex values of this and are any real". As noted, there are 33 distinct complex values, one of them is (approx) $-1.115+0i$ as a complex number and at $(-1,0)$ in the plane, then the other 32 are equally spread around the unit circle with no others landing on the $x$-axis. Commented Oct 27, 2023 at 17:26
• If you want to use software it is important to know the assumptions it makes and how it answers questions like this! If Wolfram Alpha gave a single imaginary answer, what was it? Commented Oct 27, 2023 at 17:29
• Does this answer your question? cubic root of negative numbers Commented Oct 27, 2023 at 17:31

I think this is a good question. Suppose you have a number (I'm being deliberately vague) satisfying $$x = (-37)^{1/33}$$. Using your exponent rules, this means $$x^{33} = -37$$, or equivalently $$x^{33}+37 = 0$$. This is a polynomial with degree $$33$$.

The fundamental theorem of algebra says that any polynomial of degree $$d\geq 1$$ has a root that is a complex number. It turns out that $$x^{33} + 37$$ will actually have $$33$$ many complex solutions. Let's find them!

You say that Desmos gives a real solution of $$x \approx -1.115$$. Let's call this root $$\sqrt[33]{-37}$$, the real 33rd root of $$-37$$. This is a valid solution. It turns out that every real number has real odd power roots (cube roots, fifth roots, etc.) This follows from the intermediate value theorem. Let's find the other 32 roots.

Suppose there was a number $$\zeta\neq 1$$ so that $$\zeta^{33} = 1$$. Then if I took $$(\zeta\sqrt[33]{-37})^{33} = (\zeta^{33})(\sqrt[33]{-37}^{33}) = (1)(-37) = -37$$ so $$\zeta\sqrt[33]{-37}$$ is another 33rd root of $$-37$$. Cool! In fact, we can take any power of $$\zeta$$ and the same calculation will hold (exercise for the OP!). Now let's unravel what's happening with $$\zeta$$.

We have $$\zeta^{33} = 1$$, so $$\zeta^{33} - 1 = 0$$. This is another polynomial (of degree $$33$$) having $$\zeta$$ as a root. It also has $$1$$ as a root, and it has $$31$$ more roots (all of the form $$\zeta^k$$ for $$k = 2,\dots,32$$; check that these are in fact also 33rd roots of $$1$$). These are the so-called $$33$$rd roots of unity (unity just means $$1$$).

Finally, if you are familiar with trigonometry or complex exponentials, you can show that one possible value for $$\zeta$$ is $$e^{2\pi i/33}$$ which is the same as $$\cos(2\pi/33) + i\sin(2\pi/33)$$. If you accept that $$e^{2\pi i} = 1$$ (Euler's identity), then it should be clear that $$e^{2\pi i/33}$$ is a $$33$$rd root of $$1$$. If you know De Moivre's formula, then you should show that $$\cos(2\pi/33) + i\sin(2\pi/33)$$ is also a $$33$$rd root of unity.

To wrap up my answer, this means that there are $$33$$ possible answers to $$(-37)^{1/33}$$ (or better $$x^{33} + 37 = 0$$), namely \begin{align*} x &= \zeta^k\sqrt[33]{-37} \\ &= (e^{2\pi i(k)}/33)\sqrt[33]{-37} \\ &= (\cos(2\pi j/33) + i\sin(2\pi k)/33)\sqrt[33]{-37},\;\;\;\text{for}\;\;\;k = 0,\dots,32. \end{align*}For some practice, see if you can explicitly write (without De Moivre or Euler's identity) the second, third, and fourth roots of $$1$$. If this is easy, try the sixth and eights roots as well.

Ultimately, this means that Desmos and Wolfram|Alpha are giving different correct answers (because there are more than one). As the other comments say, it is important to understand the limitations of your software, but I do not disparage the software at all. Using it and finding its limitations can be an excellent learning tool.

• Thank you very much @Nico and to all for helping me out with this question. Commented Oct 27, 2023 at 18:49
• There are $33$ roots counting multiplicity. It might be worth adding a note to the effect that in this case there are no multiple roots. Commented Oct 27, 2023 at 21:06
• Thanks hardmath! I was going to add that somewhere but couldn't find a nice place without making it too confusing. But yes, for future readers, beware of the "up to multiplicity" part of the fundamental theorem of algebra.
– Nico
Commented Oct 28, 2023 at 1:39