- 2bornot2b

\[\huge{a^b=e^{b~Loga}}\]
Here a and b are complex numbers.
This is the definition of complex exponent. Why is it defined so?

- schrodinger

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- anonymous

Can't we write \( \LARGE e^{b \log{a}}\) as \(\LARGE e^{\log{a} ^ b} = a^b \).
Log here is the natural log.

- 2bornot2b

@bmp are you sure that you made the above statement keeping in mind that a and b are complex numbers?

- anonymous

Isn't it because you are at the same time defining the complex log?
In analogy with the usual exp and log functions.

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- anonymous

Ah, yeah. That's true. My bad. Hmm.

- anonymous

I think @estudier is correct. The complex log will be of the form ln(z) = ln|z| + iArg(z). Because \( \LARGE e^{zw} = e ^{(z + 2\pi i)w}\) and \(\LARGE e ^{(z + 2\pi i)} = ln(z) \)And then I think the equality holds. But I am not sure.

- anonymous

Typo: \( 2 \pi i n \)

- anonymous

If you are looking to prove that
\[\huge{e^{b~Loga}=a^b}\]
Then consider LHS and then
\[\huge{e^{Loga^b}}\]
Now apply this rule
\[\huge a^{\log_b (c)} = c^{\log_b(a)}\]
Therefore we get
\[\huge{a^b}\]
LHS = RHS

- 2bornot2b

Is this true ?\[e^{Log~a^b}=e^{b~Log~a}\]
Please note we are talking about complex a and b

- 2bornot2b

Even if b were an integer, I couldn't have said that, I think.

- 2bornot2b

However I could have said that without any hesitation if b was some 1/m where m is an integer

- 2bornot2b

Please correct me if I am wrong

- anonymous

If b were an integer , then this rule would blindly apply. But in the case of complex numbers, I think you are right. We can't say that

- 2bornot2b

If a were a complex number and b were an integer, even then we couldn't have said that. Isn't it?

- anonymous

Have a look at this http://math.fullerton.edu/mathews/c2003/ComplexFunComplexPowerMod.html

- 2bornot2b

@cshalvey If you are here can you clarify one doubt, which is the correct statement?
1. "However I could have said that without any hesitation if b \(was\) some 1/m where m is an integer"
2. "However I could have said that without any hesitation if b \(were\) some 1/m where m is an integer"

- anonymous

@2bornot2b , the b statement is better :D and have a look at the link which I gave above

- 2bornot2b

Yes, I am looking at that @shivam_bhalla
Thanks for that!

- anonymous

An example to prove that the statement you gave above in your question is valid is
\[\large i^i = e^{i\ln i} = e^{i\ln\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)} = e^{i\ln e^{\frac{i\pi}{2}}} = e^{\frac{-\pi}{2}}\]

- 2bornot2b

@shivam_bhalla In case you feel that my question is not clear enough let me give you an analogy to explain what exactly I am searching for.
If someone asks you why is the complex exponential function defined in the way it's defined you can say that it is done so that the function maintains its property of returning the function on differentiation. Now after knowing this there is some meaning to the definition of complex exponential function.
I am searching for a similar thing here. My point is I could have defined it any way, but why in that particular way?

- experimentX

take log on both sides, it is equivalent.

- 2bornot2b

log or Log? a and b are complex by the way

- 2bornot2b

Can you tell me one thing, is the product \(w~log a\) commutative. If so, then I believe the problem is solved.

- anonymous

@2bornot2b , The answer to your question lies here
->http://www.milefoot.com/math/complex/exponentofi.htm
It is too long for me to post here

- 2bornot2b

And also what is the convention? \[e^{b~Log ~a}=({e^b})^{Log~a}~or~=e^{(b~Log~a)}\]

- experimentX

let a be a complex number, then a is equivalent to,
\( a = e^{\ln|a| + i \arg(a)}\)
\( \ln a = ln|a| + i \arg(a) \)
They are equivalent.
\( a^b = (e^{\ln|a| + i \arg(a)})^b = e^{b\ln a}\)

- anonymous

@2bornot2b , Hope you got your answer :D

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