# Wish to solve exponential congruence equation

1. Nov 19, 2013

### jackmell

Hi,

Is it possible to find a value of $n$ for the following expression other than by exhaustive search?

$$e^{\frac{2n\pi i p}{q}}=e^{-\pi i p/q},\quad n=1,2,\cdots, q-1, \quad (p,q)\in \mathbb{N}\backslash 0$$

I can write a short program to search for the value of n, but that would be infeasible if p and q were millions of digits long.

2. Nov 19, 2013

### mathman

Your question is confusing. You ask for values of n, but you say n = 1,2,....,q-1. What are you looking for?

3. Nov 19, 2013

### jackmell

Hi mathman. Sorry for the confusion. I'm looking for a positive integer $n$ in the range of one to $q-1$ which satisfies the equation. There should be only one value of n in that range which does this. This problem is in regards to computing the analytically-continuous branch across a multi-valued function.

For example, consider the function $w=z^{-3/4}(1-z)^{-7/5}$ which produces the equation for unknown n:

$$e^{-14 n \pi i/5}=e^{14 \pi i/5}=e^{4\pi i/5}$$

such that n can be only 1, 2, 3, or 4.

Now:

For n=1:

$$e^{-14 \pi i/5}\neq e^{4\pi i/5}$$

For n=2:

$$e^{-28 \pi i/5}\neq e^{4\pi i/5}$$

For n=3:

$$e^{-42\pi i/5}\neq e^{4\pi i/5}$$

For n=4:

$$e^{-56\pi i/5}=e^{-6\pi i/5}=e^{4\pi i/5}$$

so n=4 is the solution. But that is my problem: is this the only way to solve these kinds of equations? Do I need to go through them one by one until I find the correct n or is there an algorithm or function that I could simply compute to find the answer?

Last edited: Nov 19, 2013
4. Nov 20, 2013

### JJacquelin

Hi jackmell !
The relationship given in your first post :
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q)
is not consistent with the exemple given in your second post :
exp(-14*n*pi*i/5) = exp(14*pi*i/5)
If the exemple is correct, then the first equation should be :
exp(n*pi*i*p/q) = exp(-pi*i*p/q)
Where is the mistake ?

Anyway, it seems rather easy to solve the problem :
if exp(2*n*pi*i*p/q) = exp(-pi*i*p/q) is correct then
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q + 2*k*pi*i)
where k is any positive or negative integer.
2*n*pi*i*p/q = -pi*i*p/q + 2*k*pi*i
2*n = -1 + 2*k*q/p
Depending on p and q, there are some conditions for (-1 + 2*k*q/p) to be even. If these conditions are fullfiled, one can find the formula leading to couples (k, n).

5. Nov 20, 2013

### jackmell

. . . Jesus, then I'm no better then those in the homework forum I criticize for not being clear. Hope they don't see this. Ok, let me try to be precisely clear and I think from your work JJacquelin, I have found a solution to the problem!

I wish to determine for a particular starting branch of the function $w=z^{r/s}(1-z)^{p/q}$, which branch I will land on when I make one complete revolution around the singular point $z=1$. If I begin on the principal branch at say $z=1/2$, then by analyzing the argument change of the function around this singular point, I arrive at the expression:

$$e^{p/q i(\pi+2n\pi)}=e^{p/qi(-\pi)}$$
or:
$$e^{2n\pi i p/q}=e^{-2\pi ip/q}$$

Now following what you said above:

$$e^{2n\pi i p/q}=e^{-2\pi ip/q+2k\pi i},\quad n=1,2,\cdots,q-1 \quad \text{and }k\in \mathbb{Z}$$

or:

$$2\pi i(np/q)=2\pi i(k-p/q)$$

$$np=kq-p$$
$$(n+1)=k(q/p)$$

and this is the critical point: Since I know beforehand that there can be only one solution to this problem, one n only, I can always let $k=p$ which means the ending branch will always be $q-1$ and this is consistent with the experimental results above as well as other specific cases I've analyzed: the ending branch was always the last one. And likely by a similar argument, characterize the branching geometry likewise around the origin thereby completely characterizing the analytic continuation of the beta function via the pochhammer contour for rational exponents in that thread I'm working on in the Topology forum. Really a beautiful problem in my opinion guys if anyone reading this is interested and just so you know, it's really a struggle for me and I appreciate the help. :)

Thanks a bunch JJacquelin!

Last edited: Nov 20, 2013
6. Nov 20, 2013

### JJacquelin

Don't mention it !
So, were was only a "2" missing in the equation. After correcting it, all's well that ends well.
Sure, it's an interresting problem, especially with the explanation of the context.