Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

Typo: \( 2 \pi i n \)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Please correct me if I am wrong

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

@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"

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

@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?

Answer 20

take log on both sides, it is equivalent.

Answer 21

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

Answer 22

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

Answer 23

@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

Answer 24

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

Answer 25

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.
\( a^b = (e^{\ln|a| + i \arg(a)})^b = e^{b\ln a}\)

Answer 26

@2bornot2b , Hope you got your answer :D