Base e to an imaginary exponent seeming contradiction

[rude man]

<Moderator's note: Moved from a homework forum.>

1. Homework Statement
Given 0 < a < 1, i = √(-1),
e^i2πa = cos 2πa + i sin 2πa
but also, e^i2πa = (e^i2π)^a = 1^a = 1

How to resolve the apparent contradiction?

Homework Equations

e^ab = (e^a)b
e^ix = cos x + i sin x

The Attempt at a Solution

No clue! This is embarrassing!

 

[fresh_42]

https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[FactChecker]

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. $1^{0.5} = \pm 1$.

 

[rude man]

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 - I am plane.
Thanks for triggering my curiosity. I think I opened a can of worms I'd rather not have.

 

[rude man]

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]

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 $z^n =1$

 

[rude man]

OK:
1^a = (e^i2π)^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.

 

[WWGD]

OK:
1^a = (e^i2π)^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 : $e^{i2\pi}=e^{i2k\pi}= cos(2\pi k)+iSin(2\pi k)$.

 

[rude man]

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]

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]

According to wolfram alpha there are an infinire number of roots of 1, laying on the unit circle in the re - I am 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.

 

[rude man]

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]

That I've dealt with! Nyquist stability criterion etc etc.

This is at the heart of it.

 

[WWGD]

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]

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]

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]

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]

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]

<Moderator's note: Moved from a homework forum.>

1. Homework Statement
Given 0 < a < 1, i = √(-1),
e^i2πa = cos 2πa + i sin 2πa
but also, e^i2πa = (e^i2π)^a = 1^a = 1

How to resolve the apparent contradiction?

Homework Equations

e^ab = (e^a)b
e^ix = 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]

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.