2bornot2b
  • 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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • 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
  • 2bornot2b
@bmp are you sure that you made the above statement keeping in mind that a and b are complex numbers?
anonymous
  • 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
  • anonymous
Ah, yeah. That's true. My bad. Hmm.
anonymous
  • 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
  • anonymous
Typo: \( 2 \pi i n \)
anonymous
  • 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
  • 2bornot2b
Is this true ?\[e^{Log~a^b}=e^{b~Log~a}\] Please note we are talking about complex a and b
2bornot2b
  • 2bornot2b
Even if b were an integer, I couldn't have said that, I think.
2bornot2b
  • 2bornot2b
However I could have said that without any hesitation if b was some 1/m where m is an integer
2bornot2b
  • 2bornot2b
Please correct me if I am wrong
anonymous
  • 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
  • 2bornot2b
If a were a complex number and b were an integer, even then we couldn't have said that. Isn't it?
anonymous
  • anonymous
Have a look at this http://math.fullerton.edu/mathews/c2003/ComplexFunComplexPowerMod.html
2bornot2b
  • 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
  • anonymous
@2bornot2b , the b statement is better :D and have a look at the link which I gave above
2bornot2b
  • 2bornot2b
Yes, I am looking at that @shivam_bhalla Thanks for that!
anonymous
  • 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
  • 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
  • experimentX
take log on both sides, it is equivalent.
2bornot2b
  • 2bornot2b
log or Log? a and b are complex by the way
2bornot2b
  • 2bornot2b
Can you tell me one thing, is the product \(w~log a\) commutative. If so, then I believe the problem is solved.
anonymous
  • 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
  • 2bornot2b
And also what is the convention? \[e^{b~Log ~a}=({e^b})^{Log~a}~or~=e^{(b~Log~a)}\]
experimentX
  • 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
  • anonymous
@2bornot2b , Hope you got your answer :D

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