2022-01-17

How to combine complex powers?

It is not posssible to combine complex powers in the usual way:

$\left({x}^{y}\right)}^{z}={x}^{yz$

There was mention of multi-valued functions, is there some theory that makes this all work out correctly?

It is not posssible to combine complex powers in the usual way:

There was mention of multi-valued functions, is there some theory that makes this all work out correctly?

nick1337

Expert2022-01-19Added 573 answers

Step 1
For complex numbers the general way of defining ${x}^{y}$ (with x nonzero) is as $\mathrm{exp}[(ly)$ where l is one of the logarithms of x. But x does not have a unique logarithm, it has many. If l is one logarithm of x then the general one is $l+2n\pi i$ with n an integer. Hence the general value of ${x}^{y}$ is
$\mathrm{exp}(ly+2\pi iny)=\mathrm{exp}(ly)\mathrm{exp}(2\pi iny)$
Before proceeding further note that if y is an integer then $\mathrm{exp}(2\pi iny)-1$ so that ${x}^{n}$ is uniquely determined. More generally if y is a rational with denominator d then $\mathrm{exp}(\pi iny)$ (and so ${x}^{y}$ ) takes d different values. But if y is irrational, then ${x}^{y}$ takes infinitely many variables.
In the problem in hand ${x}^{y}$ takes the values $\mathrm{exp}[(ly)\mathrm{exp}(2\pi iny)$ as above, and ${x}^{yz}$ the values
$\mathrm{exp}(lyz)\mathrm{exp}(2\pi inyz)$
However in general each value of ${x}^{y}$ gives rise to infinitely many values of $({x}^{y}{)}^{z}$ , namely
$\mathrm{exp}((ly+2\pi iny)z+2\pi imz)$
so $({x}^{y}{)}^{z}$ has a doubly-infinite family of values. In general, there will be values of $({x}^{y}{)}^{z}$ which are not values of ${x}^{y}$ so the paradox in the previous posting should be expected.

Vasquez

Skilled2022-01-19Added 457 answers

Step 1
I guess the problem in general arises because ab can have multiple values. We have
$\mathrm{log}(({x}^{y}{)}^{z})=z(\mathrm{log}({x}^{y})+2\pi im)=z(y(\mathrm{log}(x)+2\pi in)+2\pi im)$
$=zy\mathrm{log}(x)+2\pi i(ny+m)z$
$=\mathrm{log}({x}^{yz}{e}^{2\pi i(ny+m)z})$
It seems a correction factor of ${e}^{2\pi i(ny+m)z}$ is necessary.
So all that can be said is that
$\mathrm{\exists}m,\text{}n\in \mathbb{Z}:({x}^{y}{)}^{z}={x}^{yz}{e}^{2\pi i(ny+m)z}$
In particular, we can say that the correction term is 1 if $(ny+m)z$ is always an integer, which happens for example if we know $y,z\in \mathbb{Z}$

alenahelenash

Skilled2022-01-24Added 366 answers

The original question asked about a theory that would make this all work out correctly. The previous answers are all correct, but they did not point out the underlying, unifying factor which makes all these computations much simpler: the Unwinding number. The unwinding number $K(z)$ is defined to be
$\mathrm{ln}(\mathrm{exp}(z))=z+2\pi iK(z)$
It is an integer which tells you how many times you need to unwind z before you get into the range where $\mathrm{ln}\circ \mathrm{exp}$ is the identity. This then allows a lot of identities to be fixed with relative ease.
I also find the paper Reasoning about the Elementary Functions of Complex Analysis by Corless, Davenport, Jeffrey, Litt and Watt quite instructive in this domain.