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From the identities $$e^{i\pi}=-1$$$$\sin x = \frac{e^{ix} - e^{-ix}} {2i},$$ it is quite easy to get the results $\sqrt{2}=i^{1/2} - i^{3/2}$ and $\sqrt{3}=i^{1/3} - i^{5/3}$

I found these two while playing around with Euler's Identity and Euler's formula. After getting these, I searched for more, perhaps of the form $\sqrt{p} = i^{1/p} - i^{p+k/p}$ where $p$ is a prime and $p+k$ is the prime after it. But, surprisingly, I didn't find any. Can anyone prove that these are the only identities of this form, or find some more if they can?

Edit: I now see that it is not possible for $p>3$. But, there may then exist results like $\sqrt{p} = i^{1/p} - i^{a/p} - i^{b/p} - \dots$ I am now interested in finding results of this form.

If we also bring $\tan x$ in the mix, since it has a range of all the real numbers, we do not have pesky problems like the ones faced by $\sin x$. Hence, we get $$\tan x = \frac {i^{2x/\pi} - i^{-2x/\pi}} {i(i^{2x/\pi} + i^{-2x/\pi})}$$ Though simplifying expressions like these might be tougher, it gets rid of the modulus problem.

Here's another one (kind of): $$\frac{\sqrt{5} - 1}{2} = i^{4/5} - i^{6/5}$$ This comes from $\sin \frac{\pi}{10} = \frac{\sqrt{5}-1}{4}$. How can we somehow convert this one to the general form with $\sqrt{5}$ on one side and only $i^k$ terms on the other side?

Edit: Are these unique, or can a single irrational be expressed in two ways like this?

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    Note that $\lvert i\rvert=1$, so the magnitude of your expression could be at most $2$ – J. W. Tanner Nov 21 '24 at 13:15
  • See also this – J. W. Tanner Nov 21 '24 at 15:25
  • I see. My point is that, while the pattern you discovered doesn’t continue, it is related to an important result in (college-level) algebraic number theory that relates the square root of $\pm p$ to $e^{2\pi i/p}$; e.g., $\cos(2\pi i/5)=(\sqrt5-1)/4$ – J. W. Tanner Nov 21 '24 at 16:26
  • Thank you for explaining. –  Nov 21 '24 at 17:38
  • It's great that you're learning this stuff in 9th grade! Good job! – Integreek Nov 26 '24 at 16:48
  • There is a problem with the values of expressions like $i^{5/3}$. For instance, if $v^3$ equals $i^5$ and $\varepsilon$ is any of three complex cubic roots of $1$ then $(v\varepsilon)^3$ equals $i^5$ too, so the value of $i^{5/3}$ is not clear. – Alex Ravsky Jan 03 '25 at 18:59

2 Answers2

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Your formula can’t work for $p>4$, since $\lvert i\rvert=1$ so the magnitude of your expression is at most $2$, but you might be interested in this.

J. W. Tanner
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  • Please check the updated question –  Jan 03 '25 at 17:48
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    @AkshajMishra: Using the answer in the link I gave, you can see for example that $\sqrt{13}=$ $i^{4/13}+i^{16/13}+i^{36/13}+i^{12/13}+i^{48/13}+i^{40/13}-i^{8/13}-i^{20/13}-i^{24/13}-i^{28/13}-i^{32/13}-i^{44/13}$ – J. W. Tanner Jan 03 '25 at 18:32
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    @AkshajMishra: also $\sqrt7=i^{25/7}+i^{1/7}+i^{9/7}-i^{5/7}-i^{13/7}-i^{17/7}$ and $\sqrt{11}=i^{37/11}+i^{5/11}+i^{25/11}+i^{9/11}+i^{1/11}-i^{41/11}-i^{13/11}-i^{17/11}-i^{21/11}-i^{29/11}$ – J. W. Tanner Jan 03 '25 at 18:40
  • I don't understand the answer in the link you gave, but nice solutions! –  Jan 04 '25 at 14:47
  • @AkshajMishra in order to understand the linked answer, you should know what the symbol $\zeta_n$ means. It is $exp(2 \pi i/n)$. So what you call $i^{1/p}$ would in the other answer be written as $\zeta_{4p}$ and conversely $\zeta_{13} = i^{4/13}$. – Vincent Jan 04 '25 at 15:02
  • Then the answer assumes that you know what is the outcome of the Gauss sum it mentions at the end. This is quite a bold assumption, I didn't know it either. Luckily we have Wikipedia: https://en.wikipedia.org/wiki/Quadratic_Gauss_sum – Vincent Jan 04 '25 at 15:05
  • But that said: I agree with J. W. Tanner that the other answer answers your question as you can use it to find expressions like the one J. W. Tanner types above for every $\sqrt{p}$. – Vincent Jan 04 '25 at 15:06
  • @Vincent Can you explain the comment in the second answer (my comment, the last one)? Does it follow from the above answer? –  Jan 05 '25 at 03:30
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If I understand your question correctly, then the following might be what you want.

Let us use the fact that $$\sin\frac{\pi}{10}\cos\frac{\pi}{5}=\frac 14\tag1$$

Proof of $(1)$ : Using $\sin A\cos A=\frac 12\sin 2A$, we get $\sin\frac{\pi}{10}\cos\frac{\pi}{5}=\color{blue}{\sin\frac{\pi}{10}}\cos\frac{\pi}{5}\times\frac{\color{blue}{\cos\frac{\pi}{10}}}{\cos\frac{\pi}{10}}=\frac{\color{blue}{\frac 12\sin\frac{\pi}{5}}\cos\frac{\pi}{5}}{\cos\frac{\pi}{10}}=\frac{\frac 14\sin\frac{2\pi}{5}}{\cos\frac{\pi}{10}}=\frac 14$

From $(1)$, we get $$1=i^{2/5}-i^{4/5}+i^{6/5}-i^{8/5}\tag2$$

You already have $$\frac{\sqrt 5-1}{2}=i^{4/5}-i^{6/5}\tag3$$

So, it follows from $(2)(3)$ that $$\frac{\sqrt 5+1}{2}=i^{2/5}-i^{8/5}\tag4$$

Therefore, from $(3)(4)$, we finally obtain $$\color{red}{\sqrt 5=i^{2/5}+i^{4/5}-i^{6/5}-i^{8/5}}$$

mathlove
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  • Can you please explain how $(2)$ follows from $(1)$? –  Jan 03 '25 at 17:54
  • Even though I don't understand how you got $(2)$ from $(1)$, this seems to be a nice solution. Can it be proved that there exists something like this for all primes? –  Jan 03 '25 at 17:58
  • @Akshaj Mishra : "how $(2)$ follows from $(1)$?" I used a very similar way as yours with $\cos x=\frac{e^{ix}+e^{-ix}}{2}$. "Can it be proved that there exists something like this for all primes?" I don't know, but I'll try. Sorry, but I don't understand your last comment. – mathlove Jan 03 '25 at 18:30
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    I mean that if you add the numerators of the exponents of $i$, like $\sqrt 5=i^{2/5}+i^{4/5}-i^{6/5}-i^{8/5}$ has the exponents of $i$ to be $2,4,6,8$ when you add them you get a multiple of $5$. –  Jan 04 '25 at 14:51