# i-th root of i

In my previous post we talked about i^i and we evaluated it. We found that the result is a real number, although you would expect it to be complex since i is an imaginary number. Is this the case for the i-th root of i?

$\dpi{100}&space;\large&space;\sqrt[i]{i}=?$

First of all, let’s write this as a power:

$\dpi{100}&space;\large&space;\sqrt[i]{i}=i^{\frac{1}{i}}=i^{-i}$

Now, this is can also be written as

$\dpi{100}&space;\large&space;i^{-i}={\left&space;(i^i&space;\right&space;)}^{(-1)}$

We already know what i^i equals

$\dpi{100}&space;\large&space;i^i=e^{-\frac{\pi}{2}}$

therefore,

$\dpi{100}&space;\large&space;i^{-i}={\left&space;(i^i&space;\right&space;)}^{(-1)}=\left&space;({e^{-\frac{\pi}{2}}}&space;\right&space;)^{\left&space;(-1&space;\right&space;)}=e^{\frac{\pi}{2}}$

This means it’s simply the inverse of i^i, and so it’s a real number as well! If you use a calculator you will see that the result is 4.810477…

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### 1 Comment »

1. mike

ok i was searching this to see if i discovered something. but i made it (i)root(i)=square root of e to the pi. damn and i wanted to name 4.810477…..

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