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2bornot2b

  • 2 years 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?

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

  2. 2bornot2b
    • 2 years ago
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    @bmp are you sure that you made the above statement keeping in mind that a and b are complex numbers?

  3. estudier
    • 2 years ago
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    Isn't it because you are at the same time defining the complex log? In analogy with the usual exp and log functions.

  4. bmp
    • 2 years ago
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    Ah, yeah. That's true. My bad. Hmm.

  5. bmp
    • 2 years ago
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    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.

  6. bmp
    • 2 years ago
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    Typo: \( 2 \pi i n \)

  7. shivam_bhalla
    • 2 years ago
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    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

  8. 2bornot2b
    • 2 years ago
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    Is this true ?\[e^{Log~a^b}=e^{b~Log~a}\] Please note we are talking about complex a and b

  9. 2bornot2b
    • 2 years ago
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    Even if b were an integer, I couldn't have said that, I think.

  10. 2bornot2b
    • 2 years ago
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    However I could have said that without any hesitation if b was some 1/m where m is an integer

  11. 2bornot2b
    • 2 years ago
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    Please correct me if I am wrong

  12. shivam_bhalla
    • 2 years ago
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    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

  13. 2bornot2b
    • 2 years ago
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    If a were a complex number and b were an integer, even then we couldn't have said that. Isn't it?

  14. shivam_bhalla
    • 2 years ago
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    Have a look at this http://math.fullerton.edu/mathews/c2003/ComplexFunComplexPowerMod.html

  15. 2bornot2b
    • 2 years ago
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    @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"

  16. shivam_bhalla
    • 2 years ago
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    @2bornot2b , the b statement is better :D and have a look at the link which I gave above

  17. 2bornot2b
    • 2 years ago
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    Yes, I am looking at that @shivam_bhalla Thanks for that!

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

  19. 2bornot2b
    • 2 years ago
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    @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?

  20. experimentX
    • 2 years ago
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    take log on both sides, it is equivalent.

  21. 2bornot2b
    • 2 years ago
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    log or Log? a and b are complex by the way

  22. 2bornot2b
    • 2 years ago
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    Can you tell me one thing, is the product \(w~log a\) commutative. If so, then I believe the problem is solved.

  23. shivam_bhalla
    • 2 years ago
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    @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

  24. 2bornot2b
    • 2 years ago
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    And also what is the convention? \[e^{b~Log ~a}=({e^b})^{Log~a}~or~=e^{(b~Log~a)}\]

  25. experimentX
    • 2 years ago
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    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. \( a^b = (e^{\ln|a| + i \arg(a)})^b = e^{b\ln a}\)

  26. shivam_bhalla
    • 2 years ago
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    @2bornot2b , Hope you got your answer :D

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