[PeteB570]

At least it’s not an imaginary number.

Now those gave me tons of trouble in Math 161.

[fhayek]

Well, to be fair, the term ‘imaginary’ number was a poor choice of a name. Imaginary numbers are just as real as ‘real’ numbers.

[mikrofon]

From the Euler Identity:
e^iπ + 1 = 0

it follows that-

π = 2i ln(i)