# An imaginary problem

• I
• Gear300
Gear300
TL;DR Summary
imaginary algebra
I saw a proof in which they came up with the ith root of i through the typical algebra.
$$i^{1/i} = i^{-i} = e^{i\frac{\pi}{2} \cdot -i} = e^{\frac{\pi}{2}} ~.$$
But it seems the proof is entirely algebraic, so we have no grounds for thinking it works anywhere. The only exception might be an analytic connection with power series, like a Laurent series. Is there such a connection, or is this as ad hoc as it seems?

## Answers and Replies

Mentor
TL;DR Summary: imaginary algebra

I saw a proof in which they came up with the ith root of i through the typical algebra.
$$i^{1/i} = i^{-i} = e^{i\frac{\pi}{2} \cdot -i} = e^{\frac{\pi}{2}} ~.$$
But it seems the proof is entirely algebraic, so we have no grounds for thinking it works anywhere. The only exception might be an analytic connection with power series, like a Laurent series. Is there such a connection, or is this as ad hoc as it seems?
It's pretty straightforward.
##i = e^{i\pi/2} \Rightarrow i^{-i} = (e^{i\pi/2})^{-i} = e^{(i\pi/2) \cdot (-i)} = e^{\pi/2}##

topsquark
Gear300
It's pretty straightforward.
##i = e^{i\pi/2} \Rightarrow i^{-i} = (e^{i\pi/2})^{-i} = e^{(i\pi/2) \cdot (-i)} = e^{\pi/2}##
So it's fine doing this sort of thing in a quantum mechanics equation?

Gold Member
MHB
TL;DR Summary: imaginary algebra

I saw a proof in which they came up with the ith root of i through the typical algebra.
$$i^{1/i} = i^{-i} = e^{i\frac{\pi}{2} \cdot -i} = e^{\frac{\pi}{2}} ~.$$
But it seems the proof is entirely algebraic, so we have no grounds for thinking it works anywhere. The only exception might be an analytic connection with power series, like a Laurent series. Is there such a connection, or is this as ad hoc as it seems?
Of course, this isn't unique.
##i^{1/i} = \left ( e^{i \pi / 2} \right )^{1/i} = \left ( e^{i \pi / 2 + 2 k \pi i } \right )^{1/i} = e^{\pi /2 + 2 k \pi}##
where k is any integer, so

##i^{1/i} = e^{5 \pi / 2}##
just as well.

-Dan

hutchphd
Mentor
So it's fine doing this sort of thing in a quantum mechanics equation?
Why not? This is mathematics.

Gear300
Of course, this isn't unique.
##i^{1/i} = \left ( e^{i \pi / 2} \right )^{1/i} = \left ( e^{i \pi / 2 + 2 k \pi i } \right )^{1/i} = e^{\pi /2 + 2 k \pi}##
where k is any integer, so

##i^{1/i} = e^{5 \pi / 2}##
just as well.

-Dan

Why not? This is mathematics.
True enough. I guess it doesn't work unless it works.

Homework Helper
Gold Member
Far from "ad hoc"; these are basic properties of exponents and polar coordinates are the best way to deal with powers in the complex plane.
I admit that I have no geometric image of numbers to complex powers, but I have to accept it because so much of it works out perfectly for real powers.