Ytukyg

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 $i^{i^i}$.

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

Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, 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 $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$

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

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

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

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

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

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