Base e to an imaginary exponent seeming contradiction

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  • #1
rude man
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<Moderator's note: Moved from a homework forum.>

1. Homework Statement

Given 0 < a < 1, i = √(-1),
ei2πa = cos 2πa + i sin 2πa
but also, ei2πa = (ei2π)a = 1a = 1

How to resolve the apparent contradiction?


Homework Equations


eab = (ea)b
eix = cos x + i sin x


The Attempt at a Solution


No clue! This is embarrassing!
 
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Answers and Replies

  • #3
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There are multiple values for roots. None are wrong or contradictory. You are saying that ##1^a## is always and only 1. That is wrong. ##1^{0.5} = \pm 1##.
 
  • #4
rude man
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There are multiple values for roots. None are wrong or contradictory. You are saying that ##1^a## is always and only 1. That is wrong. ##1^{0.5} = \pm 1##.
According to wolfram alpha there are an infinire number of roots of 1, laying on the unit circle in the re - im plane.
Thanks for triggering my curiosity. I think I opened a can of worms I'd rather not have.
 
  • #5
rude man
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There are multiple values for roots. None are wrong or contradictory. You are saying that ##1^a## is always and only 1. That is wrong. ##1^{0.5} = \pm 1##.
Ignore all my posts except for the last one please.
 
  • #6
WWGD
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There is also the fact that the Complex Exponential is infinite-valued ( periodic with period ##2\pi##, so we need to work with branches, and standard properties of Real exponential and roots do not always extend. Exponentiation is defined in terms of complex powers: ##z^{a}: = e^{alogz}##, with ## log ## being a branch ( local inverse) of the log. But, I think from the FT Algebra, there are only n roots for ##z^n =1 ##
 
  • #7
rude man
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OK:
1a = (ei2π)a = cos 2πa + i sin 2πa.
But the 2π can be n2π, n any integer. So there are an infinite number of roots of 1. The only real root would be for n=0.
 
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  • #8
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OK:
1a = (ei2π)a= cos 2πa + i sin 2πa.
But the 2π can be n2π, n any integer. So there are an infinite number of roots of 1. The only real root would be for n=0.
Precisely : ##e^{i2\pi}=e^{i2k\pi}= cos(2\pi k)+iSin(2\pi k) ##.
 
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  • #9
rude man
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There is also the fact that the Complex Exponential is infinite-valued ( periodic with period ##2\pi##, so we need to work with branches, and standard properties of Real exponential and roots do not always extend. Exponentiation is defined in terms of complex powers: ##z^{a}: = e^{alogz}##, with ## log ## being a branch ( local inverse) of the log. But, I think from the FT Algebra, there are only n roots for ##z^n =1 ##
Log of a complex number? Overload for this EE! :H
 
  • #10
WWGD
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Log of a complex number? Overload for this EE! :H
EDIT: It is somewhat a way of describing a number in Polar coordinates. It is really not that counter intuitive. The log assigns to a Complex number z(the log of) its length plus ( one of its)its argument(s) . The argument(s) part is what makes it multivalued. For example :## log(i):=ln|i|+ i(\pi/2+ 2k \pi) ##; sort of giving all the possible ways of locating a point in the Complex plane: The number i is located at length 1 , with argument ##\pi/2 + k2\pi ##. Basically assigns to a Complex number its Polar forms with a ln scaling of the norm/length.
 
  • #11
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According to wolfram alpha there are an infinire number of roots of 1, laying on the unit circle in the re - im plane.
Thanks for triggering my curiosity. I think I opened a can of worms I'd rather not have.
As a EE, you may be interested in how this allows study of feedback systems and which feedback frequencies would accumulate to unstable behavior.
 
  • #12
rude man
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As a EE, you may be interested in how this allows study of feedback systems and which feedback frequencies would accumulate to unstable behavior.
That I've dealt with! Nyquist stability criterion etc etc.
 
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That I've dealt with! Nyquist stability criterion etc etc.
This is at the heart of it.
 
  • #14
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As a EE, you may be interested in how this allows study of feedback systems and which feedback frequencies would accumulate to unstable behavior.
By unstable you mean Chaotic, i.e., Attractor is Fractal?
 
  • #15
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By unstable you mean Chaotic, i.e., Attractor is Fractal?
That is not what I had in mind. I meant simple feedback systems and Laplace transforms.
 
  • #16
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That is not what I had in mind. I meant simple feedback systems and Laplace transforms.
I am just using big words here, I don't have that good of an understanding of feedback loops, dynamical systems.
 
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I am just using big words here, I don't have that good of an understanding of feedback loops, dynamical systems.
It's delightful mathematics.
 
  • #18
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It's delightful mathematics.
It does seem interesting. I have just an undergrad class in Chaos theory and a bit of reading here-and-there.
 
  • #19
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<Moderator's note: Moved from a homework forum.>

1. Homework Statement

Given 0 < a < 1, i = √(-1),
ei2πa = cos 2πa + i sin 2πa
but also, ei2πa = (ei2π)a = 1a = 1

How to resolve the apparent contradiction?


Homework Equations


eab = (ea)b
eix = cos x + i sin x


The Attempt at a Solution


No clue! This is embarrassing!

You are being fooled by notation. If you use the alternative notation "##\exp(z)##" instead of "##e^z##", you would not automatically assume that $$\exp(a b) = (\exp(a))^b$$
In fact, that equation would not be apparent at all, although it is true and provable if ##a## and ##b## are real or if ##a## is complex and ##b ## is an integer. You have demonstrated that it is sometimes false for complex ##a## and non-integer ##b##.
 
  • #20
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Sorry if this scares you even more but not even ##1^a=1## may hold. We have ##1^a=e^{alog1}## and the value of log1 will depend on the choice of branch of log. The branch just means a restriction of the exponential a many-valued function to a simple function. (e^a)^b=e^ab doe not always hold because you may skip or jump branches.
 
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