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2bornot2b

Prove that \[\huge{(e^w)^z=e^{[(w+2ni\pi)(z)]}}\] but \[\huge (e^w)^z\ne e^{(wz)}\] where w and z are complex.

  • one year ago
  • one year ago

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  1. 2bornot2b
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    @shivam_bhalla

    • one year ago
  2. myininaya
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    \[\text{ Do you recall } e^{ i \theta}= \cos(\theta) +i \sin(\theta)\]

    • one year ago
  3. myininaya
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    \[e^{i \theta}=\cos(\theta)+i \sin(\theta)=\cos(\theta+2 n \pi)+ i \sin( \theta + 2 n \pi))=e^{i(\theta+2 n \pi)}\]

    • one year ago
  4. myininaya
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    And of course n is an integer

    • one year ago
  5. myininaya
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    Try to use that in this proof I would write w as a+bi by the way until the end when you write a+bi back as w

    • one year ago
  6. 2bornot2b
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    So you want me to start with \(\large (e^{a+ib})^z\) Sorry for the late reply @myininaya I was away.

    • one year ago
  7. 2bornot2b
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    What do I do next @myininaya ?

    • one year ago
  8. KingGeorge
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    The first thing you need to show is not too bad. Since \(w\) is complex, \(w=w+2ni\pi \). Hence\[\Large (e^w)^z=e^{(w+2ni\pi)~z}\]by using a simple substitution.

    • one year ago
  9. 2bornot2b
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    @KingGeorge, is it just substitution? Perhaps you didn't notice I have replaced the ( )

    • one year ago
  10. KingGeorge
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    Herp derp.

    • one year ago
  11. KingGeorge
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    My ideas aren't panning out :(

    • one year ago
  12. Ishaan94
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    What do you mean by complex? \(w=x+iy\)?

    • one year ago
  13. 2bornot2b
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    @Ishaan94, Yes.

    • one year ago
  14. Ishaan94
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    Does the fact that w and z are complex numbers restrict the use of euler's identity?

    • one year ago
  15. experimentX
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    why \( (e^w)^z \not = e^{wz} \) ???

    • one year ago
  16. 2bornot2b
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    I don't think so @Ishaan94 @experimentX see bullet three after visiting the following link http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities

    • one year ago
  17. 2bornot2b
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    Certainly you can use euler's identity

    • one year ago
  18. Ishaan94
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    I know that \(e^{x} =e^{y}\) doesn't imply that x equals y, but I still don't understand it. Hmm okay maybe I know why. \(e^{i\theta} = \cos \theta +i\sin \theta\). \[\sin 30 = \sin150,\,\text{but}\, 150\neq30.\]\[\sin \theta=\sin (2n\pi + \theta)\]

    • one year ago
  19. Ishaan94
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    You can only have one unique value of theta in a period by adding 2npi you change the period. Or, you complete the circle and reach the point where you started from.

    • one year ago
  20. 2bornot2b
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    But what does it have to do with this complex identity ?

    • one year ago
  21. estudier
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    hmm, maybe you have to go back to the definition in terms of complex log

    • one year ago
  22. 2bornot2b
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    @estudier Do you mean the definition of \(a^b\) where both a and b are complex?

    • one year ago
  23. estudier
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    Problem seems to be you cannot equate a multi value f with a single value f....

    • one year ago
  24. Ishaan94
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    A lot. This is how you represent a complex number in complex plane. |dw:1336309768854:dw|You can only get the same theta value if you complete circle.

    • one year ago
  25. estudier
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    z^az^b = e^(alnz)e^(blnz) = e^a(lnz +2pi i n) etc

    • one year ago
  26. 2bornot2b
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    @estudier Let me tell you the problem I am facing. I was going through the definition of \(a^b\) which is defined as \(e^{b Loga}\). Now I already know that \(a^b=(e^{Loga})^b\) Does that mean \(e^{Loga~b}=(e^{Loga})^b\)

    • one year ago
  27. estudier
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    Then z^(a+b) = e^(a+b) ln z = e^(a+b)( ln z + 2 pi i k) = e^(a+b)ln ze^2pi i k(a+b) -> z^(a+b) is in z^az^b (might be same might not)

    • one year ago
  28. estudier
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    I really don't like thi stuff.........:-(

    • one year ago
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