Solving (-2)^(1/2) and (-2)^(1/sqrt(pi)): A Complex Analysis Approach

[davedave]

let's consider the following simple example with no ambiguity.

To evaluate (-2)^(1/2), we can use simple concepts in applied complex analysis.

it is equal to 2^(1/2) * e^((pi/2) i (2k+1)).

if k=0, i square root (2)
if k=1, -i square root (2)

This one is straightforward since we take k=0, 1 for the square root.

What if we consider the following example with negative two raised to an irrational number.

(-2)^(1/square root (pi))

it is equal to 2^(1/square root (pi)) * e^(square root (pi) i (2k+1))

What could be k? I mean k goes from zero to what value? Is there another way to evaluate this one?

Does anyone know the answer?

 

[HallsofIvy]

Yes, you are correct. In working with real numbers, we define $\sqrt{a}$ to be "the positive number whose square is a" in order to have a single valued function.

But in complex numbers, things get so complicated we basically have to abandon the "single valued" idea- and wind up talking about things like Riemann surfaces, and "cuts".

In general the "nth" root of a complex number, a, or $a^{1/n}$ has n values. Similarly, a rational root, say $a^{m/n}$ has n values. But any complex number to an irrational power has an infinite number of values.

 

[AmberKennyDou]

I'm coming! Let's see what happens