# Base e to an imaginary exponent seeming contradiction

• B
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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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Delta2

fresh_42
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scottdave and Orodruin
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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##.

scottdave
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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.

scottdave
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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.

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 ##

Homework Helper
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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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WWGD
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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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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!

WWGD
Gold Member
Log of a complex number? Overload for this EE!
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.

FactChecker
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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.

WWGD
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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.

FactChecker
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That I've dealt with! Nyquist stability criterion etc etc.
This is at the heart of it.

WWGD
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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?

FactChecker
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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.

WWGD
WWGD
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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.

FactChecker
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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.

WWGD
WWGD
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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.

Ray Vickson
Homework Helper
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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##.

WWGD