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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?

  • one year ago
  • one year ago

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  1. bmp
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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.

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

    • one year ago
  3. estudier
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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.

    • one year ago
  4. bmp
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    Ah, yeah. That's true. My bad. Hmm.

    • one year ago
  5. bmp
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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.

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

    • one year ago
  7. shivam_bhalla
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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

    • one year ago
  8. 2bornot2b
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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

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

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

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

    • one year ago
  12. shivam_bhalla
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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

    • one year ago
  13. 2bornot2b
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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?

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

    • one year ago
  15. 2bornot2b
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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"

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

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

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

    • one year ago
  19. 2bornot2b
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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?

    • one year ago
  20. experimentX
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    take log on both sides, it is equivalent.

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

    • one year ago
  22. 2bornot2b
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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.

    • one year ago
  23. shivam_bhalla
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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

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

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

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

    • one year ago
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