# Find the different principal values. [closed]

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 $$i^{i^i}$$.

## closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, NamasteNov 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 $$m$$.

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