Proof that 1ln(5)/2πi = 5

[jcsd]

Is this correct (I was just a little worried that I've got a power of 1 that's not equal to 1)?:

1^ln(5)/2πi = 5

proof:

using x = π in Euelr's identity:

e^xi = cos(x) + isin(x)

e^πi = -1

square both sides:

e^2πi = 1

putting both sides to the power of ln(5)/2πi


1^ln(5)/2πi = (e^2πi)^ln(5)/2πi = e^ln(5) = 5

 

[Hurkyl]

In general, the law

(a^b)^c = a^(bc)

does not hold in the complex domain.

 

[jcsd]

Is there anyway to prove this incorrect, because it seems to me the well-known identity:

iⁱ = e^-π/2

can be derived from Euler's identity using (a^b)^c = a^(bc), at least suggesting the possibility that it may hold.

Of course, using one of the terms in my derivation, you get a patently incorrect answer:

e^2πi = 1 =>

2πi = ln(1) = 0 =>

-4π² = 0

 

[Integral]

Originally posted by jcsd
Is there anyway to prove this incorrect, because it seems to me the well-known identity:

iⁱ = e^-π/2

can be derived from Euler's identity using (a^b)^c = a^(bc), at least suggesting the possibility that it may hold.

Of course, using one of the terms in my derivation, you get a patently incorrect answer:

e^2πi = 1 =>

2πi = ln(1) = 0 =>

-4π² = 0

Why must this be incorrect? Remember you are in the complex plane now, not the Reals, much of what you know about the Reals does not apply.

 

[quantumdude]

Originally posted by Integral
Why must this be incorrect?

Because -4p² does not equal zero, silly!

Remember you are in the complex plane now, not the Reals, much of what you know about the Reals does not apply.

That's right. In this case, the thing that does not apply is the inverse relationship between ln(z) and z=re^iθ. In the complex plane, it is...

ln(z)=ln(r)+i(θ+2pk),

for any integer k. In this case, z=e^2pi and so:

ln(e^2pi)=ln(1)+i(2pi+2pk),

for any integer k. Note that ln(z) is multivalued.

edit: fixed superscript bracket

 

[Hurkyl]

You don't even have to leave the real domain to find counterexamples:

((-1)^2)^(1/2) = 1

but

(-1)^(2 * 1/2) = -1

So even in the simpler case, we see that the identity is only valid on certain domains.


In the complex plane, (the principal value of) the exponential operator is defined to be:

a^b = exp(b Log a)

Where Log is the principal value of the logarithm function (the imaginary part is constrained to be in (-π, π]

So if we inspect the identity:

(a^b)^c = (exp(b Log a))^c = exp(c Log exp(b Log a))

while

a^(bc) = exp(bc Log a)


In the case of positive real numbers, one uses the fact that ln and e^ are inverse functions to establish the identity... but in the complex domain, exp is not an invertible function and ln is a multivalued function.

 

[jcsd]

Yes, I guessed ln would be a multi-valued function which would invalid the silly result.

Back to your example: ((-1)^2)^1/2), this is also multi-valued as the answers are -1 and 1 so the relationship at least holds for one of the values.

I think to prove whether the orginal equality is correct or incorrect I would need to go back to Euler's identities and try to prove it from there.

 

[jcsd]

If anyone's interested, the orginal equality is correct (according to a Maths PhD student anyway)

 

[Hurkyl]

My gut tells me you can say something like "The set of possible values of the LHS is a subset of the set of possible values of the RHS" but I haven't worked it out yet.

(Incidentally, if you use the principal value of the exponentiation function, the original equality is not correct)

 

[jcsd]

This is his exact reply for you to see Hurkyl (I sold him a bit short actually, he's a maths lecturer at Princeton):

It is, believe it or not, a correct equality.

The reason is that exponentiation in the complex plane is defined as follows: a^b = exp(b ln(a)). But as you've noted, ln() is a multi-valued function when applied to complex numbers. Therefore, exponentiation is also multi-valued in the complex domain.

Now consider 1^ln(5)/2πi. By definition, this must equal exp( (ln(5)/2πi) ln(1)). If we let ln(1)=0, then we get exp(0) which equals 1, which we expect. But since we're considering complex numbers, ln(1) also equals 0+2πi, in which case 1^ln(5)/2πi = exp(ln(5)) = 5. In the same fashion, we can show that 1^ln(5)/2πi = exp(2 ln(5))=25, and so on.

If we restrict ourselves to the "primary branch" of the logarithm function...that is, if we arbitrarily require that the imaginary part of ln(z) always lies between -πi and πi...then some of these problems go away. But that in turn causes other problems, like breaking the rule that ln(ab) = ln(a)+ln(b). In short, complex numbers are weird