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As we all know, square rooting -1 (a real number) opens up the "imaginary" dimension (defined by the presence of iota).

We can return from the imaginary dimension back to the real dimension by reversing this process (i.e by taking a square of iota).

My question is, can we return to the real dimension by using the other path? That is, shall we ever return to the real dimension if we keep taking square root of the negative value obtained by the previous square root?

That is, let i, j, k, l ... be variables such that.

i = √-1
j = √-i
k = √-j
l = √-k
...
...

Shall any of these variables return to the real dimension again? If yes, after how many square roots?

OR

In mathematical terms:

For what real value of n, so that n is a power of 2, (-1)^1/n belongs to the set of real numbers?

Sorry if this last statement appears in a bad format. I do not know how to present equations in mathematical form using this site's encoding system.

Youstay Igo
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    Taking square roots halves the angle the point makes with the real axis in the complex plane. Therefore, if you start with a non real number, square rooting can never get you back to the reals. – Asvin Jan 27 '15 at 18:08
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    I'm not the best at complex analysis so I may be breaking some rules, but could you argue $$i^4 = 1$$ and $$1=(1)^{1/16} = (i^4)^{1/16} = i^{1/4} = \sqrt{\sqrt{i}}$$ – graydad Jan 27 '15 at 18:10
  • That's right, graydad. But I want to know the answer for 1/n power of iota, not the n power. Asvin, I had the sad feeling that this would turn out to be the case. But I am still waiting if somebody can come up with a solution to it. – Youstay Igo Jan 27 '15 at 18:14

2 Answers2

2

If it were possible, there would be a real number that, when squared, gave an imaginary result (squaring is single-valued in real and complex domains). But every real when squared gives a real number. Therefore there is no such process for square root or any other roots.

Joffan
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0

For all $n=2,4,8,16,\dots$, we have that $(-1)^{\frac{1}{n}}\in\mathbb{C}$ is $\rho e^{i\theta}$, with $\rho(n)=|-1|^{\frac{1}{n}}=1$ and $$\theta(n)=\frac{\pi+2k\pi}{n},\ k\in\{0,1,\dots,n-1\}.$$ None of these values for the argument $\theta$ equals $0$ or $\pi$, hence none of these $(\rho(n),\theta(n))$ is a point of the real axis.

By the way, notice that this is not true in general for all $n\in\mathbb{N}$.

MattAllegro
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