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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

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

    • 2 years ago
  3. estudier Group Title
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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.

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

    • 2 years ago
  5. bmp Group Title
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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.

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

    • 2 years ago
  7. shivam_bhalla Group Title
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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

    • 2 years ago
  8. 2bornot2b Group Title
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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

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

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

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

    • 2 years ago
  12. shivam_bhalla Group Title
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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

    • 2 years ago
  13. 2bornot2b Group Title
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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?

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

    • 2 years ago
  15. 2bornot2b Group Title
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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"

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

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

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

    • 2 years ago
  19. 2bornot2b Group Title
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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?

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

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

    • 2 years ago
  22. 2bornot2b Group Title
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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.

    • 2 years ago
  23. shivam_bhalla Group Title
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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

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

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

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

    • 2 years ago
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