Angles of certain complex numbers

[nomadreid]

Homework Statement

prove or disprove: for every real nonzero a & positive integers n,m: n^ia does not end up at the same angle (modulo full rotations) as m^ia.

Relevant Equations

obviously the relations between the exponential function and the natural log, and the periodicity, and that e^iA gives A in radians, and so forth. What else, I'm not sure -- maybe transcendental number properties; this is not really a homework problem, but since it has the same form as one and is probably something easy , I thought this might be the most appropriate place to post.

reducing it to various forms: for example, the one in the title, or 2*pi*k(ln m) = a(ln(n/m)), and so forth. My gut feeling is that it is true (that no such foursome exists), but manipulations have not got me anywhere. Anyone push me in the right direction? I am probably overlooking something trivial, no?

 

[Mark44]

Homework Statement:: prove or disprove: for every real nonzero a & positive integers n,m: n^ia does not end up at the same angle (modulo full rotations) as m^ia.
Relevant Equations:: obviously the relations between the exponential function and the natural log, and the periodicity, and that e^iA gives A in radians, and so forth. What else, I'm not sure -- maybe transcendental number properties; this is not really a homework problem, but since it has the same form as one and is probably something easy , I thought this might be the most appropriate place to post.

reducing it to various forms: for example, the one in the title, or 2*pi*k(ln m) = a(ln(n/m)), and so forth. My gut feeling is that it is true (that no such foursome exists), but manipulations have not got me anywhere. Anyone push me in the right direction? I am probably overlooking something trivial, no?

It took me a while to decipher what you were trying to say in your thread title.

and that e^iA gives A in radians

No, $e^{iA}$ is a complex number where A is a real number; i.e., in radians.

Start by assuming that $n^{ia} = m^{ia}$, and look for a contradiction. From this assumption, it follows that $e^{ia \ln n} = e^{ia \ln m}$.
These can be equal if, as you noted, they differ by an integer multiple of $2\pi$. See what you can find by exploring this idea.

 

[BvU]

Also: how do you exclude $n=m$ in your (pseudo?) language ?

 

[nomadreid]

Mark44, thanks: oops, I didn't mean e^iA = an angle in radians, I meant that e^iA is a rotation transformation by an angle of A radians. Trying to be too cryptic.

Assuming that n^ia=m^ia would of course give me an automatic contradiction, but if we throw in n^ia=m^ia+k*2π for integer k, it is not so clear that "These can be equal", as you stated, given the restrictions that n and m are positive integers and n≠m. (Sorry for the <> instead of the ≠; this is a throwback notation.)

 

[BvU]

the <>

No need to be sorry if it doesn't occur anywhere

 

[nomadreid]

No need to be sorry if it doesn't occur anywhere

Oops: it occurred in the title that I had originally written (but this was changed, I guess by a moderator, although the new title does not really convey what I am asking -- I should figure out my titles better), and I did not think to include this detail (without which the problem is trivial) also in the problem. Mea culpa.

Now that we have gotten over my sloppiness, can we look at the actual problem? Thanks. (I still think that there does not exist such a {n,m.a, k}). Should I try to clarify by re-stating the problem?

 

[SammyS]

Mark44, thanks: oops, I didn't mean e^iA = an angle in radians, I meant that e^iA is a rotation transformation by an angle of A radians. Trying to be too cryptic.

Assuming that n^ia=m^ia would of course give me an automatic contradiction, but if we throw in n^ia=m^ia+k*2π for integer k, it is not so clear that "These can be equal", as you stated, given the restrictions that n and m are positive integers and n≠m. (Sorry for the <> instead of the ≠; this is a throwback notation.)

$m^{ia+k\cdot2\pi}$ is not what you're looking for. That's equivalent to $m^{ia}\cdot m^{2k\pi}$

Perhaps you intended to write: $m^{i(a+2k\pi)}$. That is not quite right either.

$m^{2k\pi i}$ is not $1 + 0i$ .

 

[nomadreid]

Ah, I should hide my face!

So, trying that again: I wish to see whether, given a positive real number a, there exist positive integers k,n.m such that n^ia=m^ib and ln(b) = ln(a) + 2kπ.

Better?

Thanks for your patience.

 

[SammyS]

Ah, I should hide my face!

So, trying that again: I wish to see whether, given a positive real number a, there exist positive integers k,n.m such that n^ia=m^ib and ln(b) = ln(a) + 2kπ.

Better?

Thanks for your patience.

Well, that's not correct, so it's hard to say whether it's better or not.

When I said in my previous post that: $m^{2k\pi i}$ is not $1 + 0i$, perhaps I should have asked the following question instead.

For what value of $m$ is $m^{2k\pi i}$ equal to $1 + 0i$ ?

 

[nomadreid]

Thanks for staying in there, SammyS.

Starting with the question, e^2kπ*ln(m)i=1 iff k=0 or m=1. So I understand your point upon addressing my previous faulty formulation, in that that equality would have been broken up with this as a factor. But this is not equivalent to my new formulation, which you say is also faulty, so let me first correct that. This time I will spell it out so that any faults will be more clearly visible (which I should have done at the beginning).

The idea was to find out whether m^ai and n^ai every ended up at the same angle from the positive x-axis for unequal positive integers m, n. My reasoning in formulating this was that the angles of the two expressions would be equivalent modulo 2π. The first one is e^a*ln(m)*imeaning its angle is a*ln(m); the angle of the second one is similarly a*ln(n). So a*ln(m)≡a*ln(n) mod 2π, or a*ln(m)=a*ln(n) +2kπ, or assuming that m≠1, ln(m/n)=2kπ/a which is possible for all such a such that a=2kπ/ln(q) for rational positive q and positive integers k.

In other words, if I have worked this through correctly, my intuition is wrong, and there are an uncountably infinite number of cases.

Does this work? I am grateful for any further pointers.

 

[SammyS]

What I was pushing toward was the following.

(By the way: I'm looking for an example in which n ≠ m, but n^ia and m^ia are at the same angle from the real axis in the complex plane.)​

Starting with this snippet from Post #4:

... if we throw in n^ia=m^ia+k*2π for integer k, it is not so clear that "These can be equal", as you stated, given the restrictions that n and m are positive integers and n≠m.

Suppose that you intended that to be something like:
Consider $\displaystyle n^{i a } = m^{i(a+2k\pi)}$. (That would imply that $n = m^{(1+2k\pi/a)}$.)

Well, $\displaystyle m^{i(a+2k\pi)}=m^{i a}\cdot m^{2k\pi i}$. However, $\displaystyle m^{2k\pi i}$ itself is not at an angle of 0 in the complex plane, unless $m = e$, or some number closely related to e, such as $\displaystyle e^{1/k}$.

Well now, that pretty much should get you going in the right direction.

Using this scheme, I came up with the following example.

If $m=5$ and $a=\sqrt{2\ }$, let $n = 5\cdot 85.0197$. Then each of m^ia and n^ia are at an angle of approximately 130.41° .

 

[nomadreid]

Thanks, SammyS. Three comments:
(1) your example would be fine except for the stipulation that m and n each be positive integers.
(2) In addressing my remark that "...if we throw in..." from post #4, you apparently overlooked the fact that I admitted in post #10 that this was a false formulation, and then put the correct formulation in that post (except that the typo where m≠1 should have been m≠n )
(3) So, to come up with my own example from the correct formulation : I can find an example for any two positive integers n, m, where m≠n.
Let a = 2kπ/ln(n/m) for all integers k.
For example, let n=4, m=2.
Then a ∈{[2π/ln(2)]k : k∈ℤ}, or all multiples of approximately 9.065 , so if for example a =2π/ln(2) then 2^ai and 4^ai both end up at the x-axis.

So, basically, in pointing out the mistakes in my two original sloppy () formulations of the problem, you and the other posters have prompted me to come up with the answer, so I thank all concerned.

 

[SammyS]

...
So, basically, in pointing out the mistakes in my two original sloppy () formulations of the problem, you and the other posters have prompted me to come up with the answer, so I thank all concerned.

@nomadreid ,
The fact that you have been prompted to come up with the solution is pretty much our goal here at PF.

As far as your "sloppy formulations" go, I find that I am sometimes guilty of "sloppy reading" of details in a problem statement, as was the case here. While I apologize (USA spelling) for that, it appears that my overlooking the requirement that m, n ∈ ℤ⁺ was what kept the two of us engaged in communicating in this thread.

I applaud your efforts and the outcome.

 

[nomadreid]

 

[SammyS]

It's been a while, but let's revisit this problem.

I had been puzzling over your example in Post #12, especially the case where m=2, n=4. With your formulation for α, both m^αi and n^αi end up on the x-axis (i.e.: the Real axis).

It seemed to me that having both end up on the x-axis could indicate that there might be some underlying error in the procedure yielding a trivial result.

After reviewing your posts #10 and #12, I agree with your formulation for finding α, so that m^αi = n^αi .

Another Look At the Problem:

Given two distinct positive integers, m and n, each greater than 1:
Find a real number, α, such that both m^αi and n^αi end up at the same angle with respect to the Real axis in the complex plane.

Note: Both m^αi and n^αi are on the unit circle in the complex plane, so we are to find a real number, α, such that m^αi = n^αi .​

Let's use your formulation for α.

$\displaystyle \quad \quad \alpha =\dfrac{2k\pi}{\ln(n/m)}$

Let $\displaystyle p = \log_m(n) = \dfrac{\ln(n)}{\ln(m)}$. That is to say, $\displaystyle n=m^{p}$ .
I find it helpful to consider that if n > m, then p > 1, but it is not a necessity to have n > m.

Notice that $\displaystyle \ln(n/m) = \ln(m^p/m) = (p-1)\ln(m)$ , so that $\displaystyle \alpha =\dfrac{2k\pi}{(p-1)\ln(m)}$.

Now, let's compare m^αi and n^αi .

$\displaystyle m^{\alpha i} =e^{\ln(m) \dfrac{2k\pi}{(p-1)\ln(m)}i }$

$\displaystyle \quad \quad \quad \quad =e^{i \dfrac{2k\pi}{(p-1)} }$

$\displaystyle n^{\alpha i} =m^{p\alpha i} = e^{i 2k\pi \cdot \dfrac{ p}{(p-1)} }$

$\displaystyle \quad \quad \quad \quad =e^{i 2k\pi \cdot \left(1+\frac{1}{(p-1)}\right) }$

Back to your example: For m=2, n=4, we have p=2.

That puts m^αi at an angle of 2kπ , so m^αi is on the x-axis. n^αi is also on the x-axis.

Consider other integer values for p, For p=4, m^αi may be at 120°, or 240°, 360°, depending on the value of k.

You may want to consider the case of p being a non-integer rational number as well as the case of p being irrational, still keeping m and n as positive integers.

 

[nomadreid]

Thank you, SammyS, for these interesting manipulations. Your formulation would perhaps speed up the production of values by first classifying all pairs (n,m) with the same ratio into equivalence classes, so that one could find at one stroke the value of α for an infinite number of pairs of (m,n). (Offhand, I think you would still have a countably infinite number of cases, though.)

For a single (m,n), of course, introducing p is borrowing from Peter to pay Paul; it would only complicate the calculations.

The result of ending up on the x-axis is an artifact of my selection of n and m, but for most pairs (n,m), this would not be the case.

I am, of course, putting aside the fact that the original question merely needed a single counterexample , and it morphed into the question of finding all counterexamples. Ah, cornflakes were invented while trying to cook wheat, so...