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2bornot2b
Group Title
\[\huge{a^b=e^{b~Loga}}\]
Here a and b are complex numbers.
This is the definition of complex exponent. Why is it defined so?
 2 years ago
 2 years ago
2bornot2b Group Title
\[\huge{a^b=e^{b~Loga}}\] Here a and b are complex numbers. This is the definition of complex exponent. Why is it defined so?
 2 years ago
 2 years ago

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bmp Group TitleBest ResponseYou've already chosen the best response.1
Can't we write \( \LARGE e^{b \log{a}}\) as \(\LARGE e^{\log{a} ^ b} = a^b \). Log here is the natural log.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@bmp are you sure that you made the above statement keeping in mind that a and b are complex numbers?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Isn't it because you are at the same time defining the complex log? In analogy with the usual exp and log functions.
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.1
Ah, yeah. That's true. My bad. Hmm.
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.1
I think @estudier is correct. The complex log will be of the form ln(z) = lnz + 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.
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.1
Typo: \( 2 \pi i n \)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Is this true ?\[e^{Log~a^b}=e^{b~Log~a}\] Please note we are talking about complex a and b
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Even if b were an integer, I couldn't have said that, I think.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
However I could have said that without any hesitation if b was some 1/m where m is an integer
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Please correct me if I am wrong
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
If a were a complex number and b were an integer, even then we couldn't have said that. Isn't it?
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
Have a look at this http://math.fullerton.edu/mathews/c2003/ComplexFunComplexPowerMod.html
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@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"
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@2bornot2b , the b statement is better :D and have a look at the link which I gave above
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Yes, I am looking at that @shivam_bhalla Thanks for that!
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
An example to prove that the statement you gave above in your question is valid is \[\large i^i = e^{i\ln i} = e^{i\ln\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)} = e^{i\ln e^{\frac{i\pi}{2}}} = e^{\frac{\pi}{2}}\]
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
@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?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
take log on both sides, it is equivalent.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
log or Log? a and b are complex by the way
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Can you tell me one thing, is the product \(w~log a\) commutative. If so, then I believe the problem is solved.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@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
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
And also what is the convention? \[e^{b~Log ~a}=({e^b})^{Log~a}~or~=e^{(b~Log~a)}\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
let a be a complex number, then a is equivalent to, \( a = e^{\lna + i \arg(a)}\) \( \ln a = lna + i \arg(a) \) They are equivalent. \( a^b = (e^{\lna + i \arg(a)})^b = e^{b\ln a}\)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@2bornot2b , Hope you got your answer :D
 2 years ago
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