I was messing with the identity $e^{i\pi}=-1$ and I got that $i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}$ and on. I plugged it in to a calculator and it was infinite. It grew very fast. Does that make $i$ solvable?

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    $\begingroup$ How did you derive that expression for $i$? $\endgroup$
    – DanZimm
    Commented Nov 6, 2014 at 2:25
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    $\begingroup$ e^ipi = -1. If you do the square root of that you get i = sqrt(e^ipi), but i has just been defined to sqrt(e^ipi), and so it goes on. $\endgroup$ Commented Nov 6, 2014 at 2:29
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    $\begingroup$ People should not downvote this guy, he's simply making a question. $\endgroup$
    – DLV
    Commented Nov 6, 2014 at 2:33
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    $\begingroup$ Link. $\endgroup$
    – Lucian
    Commented Nov 6, 2014 at 3:16
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    $\begingroup$ Where did you get a calculator that allows plugging in an infinite expression? $\endgroup$ Commented Nov 6, 2014 at 22:13

3 Answers 3


I'm assuming to reach that conclusion, you took the square root of both sides and got: $$\sqrt{e^{i\pi}}=i$$ which is correct. Then, on the left hand side, you substituted for $i$ using this equality to get $$\sqrt{e^{\pi\sqrt{e^{i\pi}}}}=i$$ which is still okay - and then you tried doing this infinitely often, and lo and behold, it broke!

The main issue is that, suppose we let $$f(x)=\sqrt{e^{\pi x}}.$$ What's you've shown amount to the fact that $f$ has what is called a fixed-point at $i$ - that is $$f(i)=i.$$ And this implies things like $$f(f(i))=i$$ $$f(f(f(i)))=i$$ or, where we let $f^n(i)$ be $f$ applied to $i$ repeatedly $n$ times, we get $$f^n(i)=i.$$ So far so good. However, if we wanted to make sense of an expression which is basically an infinite nest of the function $f$ with no starting value, it would be that, no matter what $x$ we started with, the sequence $$(x,f(x),f^2(x),f^3(x),\ldots)$$ - called the orbit of $x$ under $f$ - would always converge to $i$ - and as you see, this is not the case.

A simple example of the difference between "has a fixed point" and "converges to something" would be if we took $$f(x)=2x$$ which has a fixed point at $x=0$. If we wished to think of the expression $$2\cdot (2\cdot (2\cdots)))$$ we would hope that $f$ would have that $(x,f(x),f^2(x),\ldots)$ converged for any $x$ - but it only converges for $x=0$ - otherwise, the sequence has $f^n(x)=2^nx$ which diverges. Your function acts similarly, but it is harder to visualize, as it takes place on the complex plane. So, in fact, you cannot "compute" $i$ using this method - the series will simply diverge. However, for any finite number of applications, your identity is correct.

  • $\begingroup$ For getting to the converging place is it possible to make an equation that represents the sequence and then go negatively because then it will approach a certain number instead of spreading. $\endgroup$ Commented Nov 6, 2014 at 3:07
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    $\begingroup$ Like, if we instead of going $(x,f(x),f^2(x),\ldots)$ went backwards? Yes, it is possible; we'd use the inverse of $f$, which happens to be: $$f^{-1}(x)=\frac{2}{\pi}\log(x).$$ Though this is not true in general, it does seem that iterating this function on a given argument always yields $i$ - and that sort of makes sense, given that $\log(x)$ is going to take big things and make them close to zero and - though you have to be careful about logs on the complex plane - notice that $e^{2\pi i n}=1$ for any $n\in\mathbb{Z}$, which makes defining $\log(1)$ (or anything else) nontrivial. $\endgroup$ Commented Nov 6, 2014 at 3:17
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    $\begingroup$ I disagree with your "which is correct": the square root operator is defined for positive numbers. For negative numbers, or more generally for complex numbers, it's ambiguous (thus not defined): they have two square roots, none of which is more natural than the other. That's a mistake that leads sometimes to prove that $1=-1$, for example. $\endgroup$ Commented Nov 6, 2014 at 16:51
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    $\begingroup$ @Jean In this case the principal value works - but even better is that, if we let $\sqrt{-1}=x$ (i.e. choose $i$ or $-i$ arbitrarily but be consistent), then the identity $e^{\pi x}=-1$ holds and $\sqrt{e^{\pi x}}=\sqrt{-1}=x$ holds. Moreover, defining $f[X]=\{k:k^2=e^{\pi x},x\in X\}$ - that is, taking the set of both roots to the equation - yields that $\{i\}$ and $\{-i\}$ iterate to $\{i,-i\}$, which is a fixed point. So there's some nice symmetry built into the problem which lets us avoid any nastiness with square roots. $\endgroup$ Commented Nov 6, 2014 at 22:04
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    $\begingroup$ I VERY much like the $f(x)=2x$ example at the end. That is a wonderful example of how orbits can fail to converge which is simple and concise. Chalk up one more answer for "infinite recursion of functions do awkward things when you least expect them to!" $\endgroup$
    – Cort Ammon
    Commented Nov 9, 2014 at 19:57

If you take $$\exp(i\pi)=-1=i^2\implies i=\sqrt{\exp(i\pi)}=\sqrt{\exp(\pi\sqrt{\exp(i\pi)})}$$ So the correct identity should be: $$i=\sqrt{\exp(\pi\sqrt{\exp(\pi\sqrt{\exp(\pi\ldots)})})}$$ You can't evaluate from left to right, you need to evaluate it from right to left. Note that last term in an case would be $\exp(i\pi)$ which normals calculators can't evaluate.

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    $\begingroup$ @MikeMiller that is what OP wrote $\endgroup$
    – RE60K
    Commented Nov 6, 2014 at 2:29
  • $\begingroup$ What's the exp supposed to be? Am I overlooking something that I should know $\endgroup$ Commented Nov 6, 2014 at 2:32
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    $\begingroup$ @Fmonkey2001 $\exp x\equiv e^x$ $\endgroup$
    – RE60K
    Commented Nov 6, 2014 at 6:58


Replace $i$ with $x$.


where $x=\pm i$


Here $i$ satisfies $x=+\sqrt{e^{x\pi}}$.


Hence proved.


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