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Show that the principal value of $i^{i^i}$ differs from the principal value of $i^{i*i}$.

And find the set of all values of the expression $i^{i^i}$.

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closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, Namaste Nov 29 '18 at 17:00

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First, $i^{i*i}=i^{-1}=\frac{1}{i}=-i$.

Then, $i^{i^i}=i^{(e^{\frac{\pi}{2}i+2\pi mi})^i}=i^{e^{-\frac{\pi}{2}-2\pi m}}=(cis(\frac{\pi}{2}))^{(e^{-\frac{\pi}{2}-2\pi m})}=cis(\frac{\pi}{2}e^{-\frac{\pi}{2}-2\pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(\frac{\pi}{2}e^{-\frac{\pi}{2}})$.

In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(\frac{\pi}{2}e^{-\frac{\pi}{2}-2\pi m})=-i$

$$-i=cis(\frac{3\pi}{2})=cis(\frac{\pi}{2}e^{-\frac{\pi}{2}-2\pi n})$$

$$\frac{3\pi}{2}=\frac{\pi}{2} e^{-\frac{\pi}{2}-2\pi n} $$

$$3=e^{-\frac{\pi}{2}-2\pi n}$$

$$\frac{1}{3}=e^{\frac{\pi}{2}+2\pi n}$$

$$n=-\frac{\frac{\pi}{2}+ln(3)}{2\pi}=-\frac{1}{4}-\frac{ln(3)}{2\pi}$$

As $\frac{ln(3)}{2\pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.

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  • $\begingroup$ Thankyou so much! This will help me $\endgroup$ – Peter van de Berg Nov 22 '18 at 12:21

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