Question: Is it true and can we quickly show that $$\Bigg|(-1/2)^{(-1/2)^{(-1/2)}}\Bigg|=e^{\pi\sqrt{2}}$$

Her was my solution. Abusively I write $a/b$ for the fraction ${a \above 1.5pt b} $. I write $i=\sqrt{-1}$. If $z$ is any complex number I denote the absolute value of $z$ by $|z|$.

Solution: We start by computing the tower top down. Observe $$(-1/2)^{(-1/2)}= {1\above 1.5pt \sqrt{{-1\above 1.5pt 2}}}={1 \above 1.5pt i \sqrt{1 \above 1.5pt 2}}={1 \above 1.5pt i {\sqrt{1} \above 1.5pt \sqrt{2}}}={\sqrt{2}\above 1.5pt i\sqrt{1}}={\sqrt{2} \above 1.5pt i}$$ I multiply the numerator and denominator of the the RHS of the above equation by the conjugate of the denominator of ${\sqrt{2}\above 1.5pt i}$ and simplify; $${\sqrt{2}(-i) \above 1.5pt i(-i)}={-i\sqrt{2}\above 1.5pt -(-1)}=-i\sqrt{2}$$ Now I can compute the base of the tower $$\left(-{1 \above 1.5pt 2}\right)^{-i\sqrt{2}}=(-1)^{-i\sqrt{2}}(2)^{-i\sqrt{2}}$$ I bring Euler's identity into play and rewrite $(-1)^{-i}$ as $e^{\pi}$. I also recall that absolute value is multiplicative for real and complex numbers. In particular. $$\Bigg|(-1)^{-i\sqrt{2}}(2)^{-i\sqrt{2}}\Bigg|=\Bigg|(-1^{-i})^{\sqrt{2}}(2^{-i})^{\sqrt{2}}\Bigg|=\Bigg|e^{\pi\sqrt{2}} \Bigg| \times\Bigg|{\overline{2^{-i}} \above 1.5pt 2^i}\Bigg|=e^{\pi\sqrt{2}}$$ This completes the solution.

Is there a quicker way to show this ?

The motivation here were cheap, possibly trivial, ways to rewrite $e^{\pi\sqrt{2}}$, using only finitely many combinations of rational numbers or integers in the rewrite. Here are a few more examples:

  • $\Bigg|(-1/4)^{(-1/4)^{(-1/4)}}\Bigg| = {e^{\pi} \above 1.5pt 4}$ in particular $\Bigg|(-1)^{(-1/4)^{(-1/4)}}\Bigg| = e^{\pi}.$ See Gelfond's constant
  • $\Bigg|(-2)^{(1/4)^{(-1/4)}}\Bigg| = 2^{\sqrt{2}}.$ See Gelfond Schneider Constant.

The rewrites are not unique.


Since you are multiplying/exponentiating, it makes sense to keep the numbers in polar coordinates. so let $x = -\frac{1}{2} = \frac{1}{2}e^{i\pi} = e^{-\log 2 + i\pi}$. Then

\begin{align} x^{x^x} = \exp(x^x\log x) &=\exp( e^{x\log x}\log x) \\ &= \exp(e^{-\frac{1}{2}(-\log 2 + i\pi)}(-\log 2 +i\pi)) \\ &= \exp(-i\sqrt{2}(-\log 2 + i\pi))\\ &=\exp(\sqrt{2}\pi +i\sqrt{2}\log 2) \end{align}

Hence $\big|x^{x^x}\big| = e^{\sqrt{2}\pi}$


You need to be clear about how you define $a^b$ when $a$ is a negative real number and $b\in\mathbb{C}$. One standard way is to stipulate



$$e^z=\sum_{n=0}^\infty{z^n\over n!}\quad\text{for } z\in\mathbb{C}\qquad\text{and}\qquad\ln x=\int_1^x{du\over u}\quad\text{for }x\in\mathbb{R^+}$$

With that convention,


and thus


If all you want is the absolute value, then


  • $\begingroup$ Do we know if the real and imaginary parts in the complex number $(-1/2)^(1/2)^(-1/2)$ are necessarily irrational ? Tranascendental ? $\endgroup$ – Antonio Hernandez Maquivar Dec 14 '17 at 20:45
  • $\begingroup$ ...and why do I need to be carefull /clear with defining $a^b$ when $a$ is negative and real with $b$ complex ? $\endgroup$ – Antonio Hernandez Maquivar Dec 14 '17 at 20:49
  • 1
    $\begingroup$ @AnthonyHernandez, you could ask about irrationality/transcendentality as a separate, stand-alone question (with a link to your question here to help provide context). As for being careful/clear, it's always a good idea to be careful and clear. But it's especially important when dealing with expressions of the form $a^b$. This is because certain "obvious" identities, such as $(ab)^c=a^cb^c$ or $(a^b)^c=a^{bc}$, which are true when $a$, $b$ and $c$ are positive reals, do not hold in general; glossing over the fine points leads people to derive nonsense like $-1=(-1)^{2/2}=((-1)^2)^{1/2}=1$. $\endgroup$ – Barry Cipra Dec 14 '17 at 21:13

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