**Sidebar**: What is ln(-1)?

Euler's Equation states that

e^{iπ} + 1 = 0

Thus,

e^{iπ} = -1

So

iπ = ln(-1)

### Question

What is i^{i}, expressed as a power of e?

### Answer

Since we want to express it as a power of e, we can start by saying that

e^{x} = i^{i}

for some `x`. So

`x` = ln(i^{i})

`x` = i ln(i)

Now, i is the square root of -1, so

`x` = i ln(-1^{1/2})

`x` = (i/2) ln(-1)

But as demonstrated in the sidebar,

ln(-1) = iπ

So

`x` = (i/2) · iπ

`x` = (i · i · π) /2

but i · i is -1, so

`x` = -π/2

so

i^{i} = e^{-π/2}

(Note that this is a real number! It's about 0.208.)

Thanks to Peter Hartman for showing me this elegant derivation.