Imaginary power?

[sail3005]

Why is x^i undefined?

[Guv]

It might be messy, but I do not think it is undefined.

x = e^ln(x)

x^i = [ e^ln(x) ]^i

Thinking about [ e^2 ]^3 = e^6, suggests the following.

x^i = e^i*ln(x)

While I would not bet a big sum on the above, I think it is correct.

Even if not correct, I think that x^i is some analytical function.

[thinktank2]

Simple logical proof that x^i is a defined quantity.

let y = x^i

y^i = (x^i)^i = x^(i^2) = x^(-1) = 1/x (is a defined quantity if x is defined)

or (x^i)^i = defined

Now...

(undefined)^i = defined , is not possible.

For the right hand side to be a "defined" quantity only possibility is (x^i) is a defined quantity.

[plenderj]

Um well the power being used must be a real number.
Because you're using say a taylor expansion.

So imagine it as though you're using a for loop.
You need a real number to count with. You can't count with say ... a picturebox can you ?

i is a complex number, not a real number

[sail3005]

Originally posted by thinktank2
Simple logical proof that x^i is a defined quantity.

let y = x^i

y^i = (x^i)^i = x^(i^2) = x^(-1) = 1/x (is a defined quantity if x is defined)

or (x^i)^i = defined

Now...

(undefined)^i = defined , is not possible.

For the right hand side to be a "defined" quantity only possibility is (x^i) is a defined quantity.

But in this instance aren't you basically using i as -1 since you are squaring it?

[noble]

Guv is correct. x^i, or more commonly, functions like e^a*i are
legitimate when dealing with complex analysis.

All e^ix says is that it defines the unit circle in the complex plane.
'x' here is a real number and i is any complex quantity.

There are also proofs available for the talyor series expansion
of e^z where z = a + i*x.

These complex functions are extremely useful when trying
to compute an integral which would otherwise seem impossible
such as e^x^2.
This is where the residue theorem comes to play.

[sail3005]

I think i get it...

So, if x^i is defined, is also log(i) or ln(i) defined?

[noble]

ln(i) = (Pi/2)*i

[thinktank2]

Originally posted by sail3005


But in this instance aren't you basically using i as -1 since you are squaring it?

LOL Sail , It seems you haven't fully understood what I said..


1) (x^i)^i is defined

Proof:

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

(undefined)^i = defined is not possible , so

(x^i)^i is defined.


2) What the above thing means to your question.

Let p be a number then p/i is defined.

If (x^i)^i is defined

[(x^i)^i)]^(p/i) should also be a defined quantity

(Since it is of the form (defined)^(defined) )

[(x^i)^i)]^(p/i) = (x^i)^(i*p/i) = (x^i)^p is defined quantity

If p=1 then

(x^i)^p = (x^i)^1 = x^i

Which proves that x^i is a defined quantity.

[thinktank2]

Originally posted by noble

x^i = e^(ln(x^i))
      = e^(i*ln(x))
      = cos(ln(x)) + i*sin(ln(x))

this is the principal solution, there are infinitely many solutions
for this problem.

There is only one value for ln(x)

So there is only one value for Cos(ln(x))
and only one value for Sin(ln(x))

How can cos(ln(x)) + i*sin(ln(x)) have infinitely many Solutions ??

[noble]

the logarithm is a multivalued function.

I incorrectly specified how ln(x) has infinitely many solutions.

I should have wrote ln(z) has infinitely many solutions where
z is any complex expression.

[thinktank2]

Originally posted by noble
the logarithm is a multivalued function.

I incorrectly specified how ln(x) has infinitely many solutions.

I should have wrote ln(z) has infinitely many solutions where
z is any complex expression.

anyway, sail did not mention that x is a complex number.

So in conclusion:
When x is a real number there is only one solution.
When x is a complex number it has more than one solution.

[sail3005]

Ok, i am understanding it now. thanks for the help guys. I have two more questions.

1) Why does it say "Domain error" when i put something like 2^i in my calculator?

2) Would 2^i simplify or be evaluated to anything?

[thinktank2]

Originally posted by sail3005
Ok, i am understanding it now. thanks for the help guys. I have two more questions.

1) Why does it say "Domain error" when i put something like 2^i in my calculator?

2) Would 2^i simplify or be evaluated to anything?

1) Are you sure you are using the radian mode. If you use degree mode while doing some complex operation you may get that error.

2)I used mathematica and I got the value for 2^i to be

0.769239 + 0.638961 i


Note: Since Cos and Sin have values in the range -1 to +1
The real and imaginary parts of x^i will always be in the range -1 to +1

[sail3005]

yeah, i had it in degree mode. thanks again for all the help!

[Alphanos]

On a slightly different topic, what is the square root of i?

[thinktank2]

i can be represented as

i = cos(pi/2) + i*sin(pi/2) {since cos(pi/2) = 0 , Sin(pi/2) =1)

So, Square root of i = i^(1/2) = [ cos(pi/2) + i*sin(pi/2)]^(1/2)

By De moivre's theorem

[ cos(pi/2) + i*sin(pi/2)]^(1/2) = [ cos(1/2*pi/2) + i*sin(1/2*pi/2)]

Therefore

i^(1/2) = cos(pi/4) + i*sin(pi/4) (or)

i^(1/2) = 1/Sqrt(2) + i * 1/Sqrt(2) = (1+i)/Sqrt(2)

you can find any power of i for that matter.