I Proof of i-th Root of i: Analytic Connection?

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The discussion centers on the proof of the ith root of i, expressed as i^{1/i} = e^{\pi/2}, derived through algebraic manipulation. Participants question the validity of this purely algebraic proof, suggesting it lacks a broader analytical foundation, except possibly through power series like Laurent series. There is acknowledgment that the expression can yield multiple results depending on the integer k, indicating non-uniqueness in the solution. The conversation touches on the application of these mathematical principles in quantum mechanics, affirming that such manipulations are acceptable within mathematics. Overall, the discussion highlights the complexities and nuances of working with complex numbers and their powers.
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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?
 
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Gear300 said:
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}##
 
Mark44 said:
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?
 
Gear300 said:
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
 
Gear300 said:
So it's fine doing this sort of thing in a quantum mechanics equation?
Why not? This is mathematics.
 
topsquark said:
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

Mark44 said:
Why not? This is mathematics.
True enough. I guess it doesn't work unless it works.
 
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.
 
Gear300 said:
True enough. I guess it doesn't work unless it works.
I know that ##i ^i \approx 0.2##, which is quite funny.
 
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