Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
\[\huge{a^b=e^{b~Loga}}\]
Here a and b are complex numbers.
This is the definition of complex exponent. Why is it defined so?
 one year ago
 one year ago
\[\huge{a^b=e^{b~Loga}}\] Here a and b are complex numbers. This is the definition of complex exponent. Why is it defined so?
 one year ago
 one year ago

This Question is Closed

bmpBest 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.
 one year ago

2bornot2bBest 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?
 one year ago

estudierBest 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.
 one year ago

bmpBest ResponseYou've already chosen the best response.1
Ah, yeah. That's true. My bad. Hmm.
 one year ago

bmpBest 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.
 one year ago

shivam_bhallaBest 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
 one year ago

2bornot2bBest 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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Even if b were an integer, I couldn't have said that, I think.
 one year ago

2bornot2bBest 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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Please correct me if I am wrong
 one year ago

shivam_bhallaBest 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
 one year ago

2bornot2bBest 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?
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.1
Have a look at this http://math.fullerton.edu/mathews/c2003/ComplexFunComplexPowerMod.html
 one year ago

2bornot2bBest 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"
 one year ago

shivam_bhallaBest 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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Yes, I am looking at that @shivam_bhalla Thanks for that!
 one year ago

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

2bornot2bBest 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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
take log on both sides, it is equivalent.
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
log or Log? a and b are complex by the way
 one year ago

2bornot2bBest 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.
 one year ago

shivam_bhallaBest 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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
And also what is the convention? \[e^{b~Log ~a}=({e^b})^{Log~a}~or~=e^{(b~Log~a)}\]
 one year ago

experimentXBest 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. \( a^b = (e^{\lna + i \arg(a)})^b = e^{b\ln a}\)
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.1
@2bornot2b , Hope you got your answer :D
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.