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, Namaste Nov 29 '18 at 17:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Adrian Keister, supinf, Paul Plummer, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.


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

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

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

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

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


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

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.