Can Anyone Tell Me Where this "Proof" Breaks Down?

[jinydu]

Today, I was playing around with the exponential expression for the sine function, when I stumbled onto a seeming paradox that I just can't figure out. If anyone can point out where I went wrong in this "proof", please let me know.

(False) Theorem: The sine of any complex number equals 0.

Proof:

Use the formula:

sin x = ((e^(ix)) - (e^(-ix)))/2i

We know that sin (pi) = 0. Therefore:

((e^(i*pi)) - (e^(-i*pi)))/2i = 0

Multiply both sides by 2i:

(e^(i*pi)) - (e^(-i*pi)) = 0

e^(i*pi) = e^(-i*pi)

Now, let's compute the sin of (z*pi), where z is an arbitrary complex number:

sin (z*pi) = ((e^(i*z*pi)) - (e^(-i*z*pi)))/2i

Applying the property: e^(a*b) = (e^a)^b:

sin (z*pi) = ((e^(i*pi))^z - (e^(-i*pi))^z)/2i

Previously, we established that e^(i*pi) = e^(-i*pi), so substituting in:

sin (z*pi) = ((e^(i*pi))^z - (e^(i*pi))^z)/2i

But ((e^(i*pi))^z - (e^(i*pi))^z) = 0, so

sin (z*pi) = 0/2i

sin (z*pi) = 0

Note that any complex number can be expressed as z*pi, so the sine of any complex number be equal 0.

Q.E.D

I know this proof can't possibly be right, but I just can't find where it goes wrong .

[Orgasmic Troll]

Originally Posted by jinydu

Today, I was playing around with the exponential expression for the sine function, when I stumbled onto a seeming paradox that I just can't figure out. If anyone can point out where I went wrong in this "proof", please let me know.

(False) Theorem: The sine of any complex number equals 0.

Proof:

Use the formula:

sin x = ((e^(ix)) - (e^(-ix)))/2i

We know that sin (pi) = 0. Therefore:

((e^(i*pi)) - (e^(-i*pi)))/2i = 0

Multiply both sides by 2i:

(e^(i*pi)) - (e^(-i*pi)) = 0

e^(i*pi) = e^(-i*pi)

Now, let's compute the sin of (z*pi), where z is an arbitrary complex number:

sin (z*pi) = ((e^(i*z*pi)) - (e^(-i*z*pi)))/2i

Applying the property: e^(a*b) = (e^a)^b:

sin (z*pi) = ((e^(i*pi))^z - (e^(-i*pi))^z)/2i

Previously, we established that e^(i*pi) = e^(-i*pi), so substituting in:

sin (z*pi) = ((e^(i*pi))^z - (e^(i*pi))^z)/2i

But ((e^(i*pi))^z - (e^(i*pi))^z) = 0, so

sin (z*pi) = 0/2i

sin (z*pi) = 0

Note that any complex number can be expressed as z*pi, so the sine of any complex number be equal 0.

Q.E.D

I know this proof can't possibly be right, but I just can't find where it goes wrong .

e^(a*b) = (e^a)^b only holds for real variables

http://mathworld.wolfram.com/ExponentLaws.html

[jinydu]

Thanks TravisT

But if that is the problem, I may be in trouble. I routinely use that property on my homework assignments, so far without any problems.

Is it necessary for both a and b to be real for the formula to work?

[JuanTutors]

The statement "a = b" does not imply "sqrt(a) = sqrt(b)".

sqrt(-1) = +i or -i, but -1 isn't the same thing as e^(i*pi).

sqrt(e^(i*p)) = e^(i*pi/2) = i
sqrt(e^(-i*pi)) = e^(-i*pi/2) = -i

It'll come up again later, but basically the point is, f(z) = a^z, where a is any complex number, is not a function because it's not single valued. However, if you DEFINE a^z := exp(i*z*Ln(a)), then it is single valued.

Look up branches or branch cuts in your book or in Mathworld.

[Orgasmic Troll]

Originally Posted by jinydu

Thanks TravisT

But if that is the problem, I may be in trouble. I routinely use that property on my homework assignments, so far without any problems.

Is it necessary for both a and b to be real for the formula to work?

I think it would depend on the class you're taking. If it's assumed that you're working with real variables, you should be okay

[jinydu]

I'm taking a Calculus class. The professor introduced complex exponentials to allow the solution of the differential equation:

mx''(t) + cx'(t) + kx(t) = 0

Its worrying if the property e^(a*b) = (e^a)^b fails to hold because I used it on some of the homework assignments. For example, some of the questions are proofs of identities like:

(sin x)^2 = (1 - cos(2x))/2
(cos x)^2 = (1 + cos(2x))/2
(sin x)^3 = (3sin x - sin(3x))/4
(cos x)^3 = (3cosx + cos(3x))/4

Furthermore, in these questions, I was asked specifically to use the exponential formulas for sine and cosine, which I did. However, my proofs (and the answer proofs) relied on the property e^(a*b) = (e^a)^b. Does that mean that the proofs are in fact invalid?

[jinydu]

Originally Posted by TravisT

e^(a*b) = (e^a)^b only holds for real variables

http://mathworld.wolfram.com/ExponentLaws.html

I've checked with my math professor, and it turns out that this sometimes doesn't even hold in real variables. For instance, when b = 1/2, the left-hand side has one possible value while the right-hand side has two possible values. He says the rule can only be expected to work reliably when b is a positive integer.

Anyway, my original motivation for this was to find a way to construct functions with periodic roots using the complex exponential. I knew that the sine function was periodic, so I thought that if I could prove this using the exponential formula for sine, I may get insight on how to build new functions.