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2bornot2b Group Title

Can someone explain why the last equality is possible? \[\large{a^b=(re^{i\theta })^b}\]\[\large{=(e^{ln~r+i\theta})^b=e^{(ln~r+i\theta)b}} \]

  • 2 years ago
  • 2 years ago

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  1. shivam_bhalla Group Title
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    @2bornot2b , \[\huge r = e^{\ln(r)}\] Using this the result is got

    • 2 years ago
  2. 2bornot2b Group Title
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    As far as I know \(\large(e^z)^w\ne e^{zw}\) rather \(\large(e^z)^w=e^{(z+2n\pi i)w} \)

    • 2 years ago
  3. 2bornot2b Group Title
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    @shivam_bhalla I am talking about the equality that appears next to the one you have talked about.

    • 2 years ago
  4. shivam_bhalla Group Title
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    @2bornot2b .OK. I got what you are asking for. Let me research on this and get back to you. Looks like you are doing some extensive research on complex numbers @2bornot2b

    • 2 years ago
  5. 2bornot2b Group Title
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    Thanks @shivam_bhalla I need this help extensively.

    • 2 years ago
  6. shivam_bhalla Group Title
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    I think b is complex number , right ??

    • 2 years ago
  7. 2bornot2b Group Title
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    By the way, the result that I typed in my question is from the link http://en.wikipedia.org/wiki/Exponential_function#Computation_of_ab_where_both_a_and_b_are_complex

    • 2 years ago
  8. 2bornot2b Group Title
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    I believe so, but it would be better for you to check it at that link

    • 2 years ago
  9. shivam_bhalla Group Title
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    @2bornot2b , I suggest you read it again properly. It says in the case you have taken that b is integer. Read this last statement "However, when b is not an integer, this function is multivalued, because θ is not unique (see failure of power and logarithm identities)."

    • 2 years ago
  10. SmoothMath Group Title
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    Um... isn't it just exponent rules? (a^b)^c = a^(b*c)

    • 2 years ago
  11. shivam_bhalla Group Title
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    Here:http://en.wikipedia.org/wiki/Exponential_function#Computation_of_ab_where_both_a_and_b_are_complex

    • 2 years ago
  12. 2bornot2b Group Title
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    Sounds reasonable @shivam_bhalla But the topic starts with the name Computation of \(a^b\) where both a and b are complex

    • 2 years ago
  13. shivam_bhalla Group Title
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    The way the article is written is very confusing. They should have clearly spacified that b is integer in this case and "a" is a complex number in consideration

    • 2 years ago
  14. shivam_bhalla Group Title
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    *specified

    • 2 years ago
  15. 2bornot2b Group Title
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    Seems to me you are right, otherwise that line wouldn't have been mentioned at the end. What do others think?

    • 2 years ago
  16. 2bornot2b Group Title
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    @Mertsj What do you think? @shivam_bhalla how about editing the page, and refining it for future readers?

    • 2 years ago
  17. shivam_bhalla Group Title
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    @2bornot2b , good idea. I think you should go ahead :)

    • 2 years ago
  18. 2bornot2b Group Title
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    :)

    • 2 years ago
  19. 2bornot2b Group Title
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    Anyway thanks a lot, that really helped me a lot.

    • 2 years ago
  20. shivam_bhalla Group Title
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    @2bornot2b , a serious bug at OS. See the attachment See your name is replaced by "Best answer"

    • 2 years ago
  21. shivam_bhalla Group Title
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    @cshalvey, check a serious bug above ^^

    • 2 years ago
  22. 2bornot2b Group Title
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    Great catch Shivam!

    • 2 years ago
  23. 2bornot2b Group Title
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    By the way shivam, can you suggest me a good book on complex functions?

    • 2 years ago
  24. 2bornot2b Group Title
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    And I think Officer @shadowfiend is in the department of arresting bugs.

    • 2 years ago
  25. shivam_bhalla Group Title
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    Seriously, I didn't refer any book for complex numbers. I just listened to what my teacher taught in my class. That's it :D. If you are looking for some books, here is a book recommended by my friend HIGHER ALGEBRA- By Hall & Knight

    • 2 years ago
  26. 2bornot2b Group Title
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    Does that book have complex functions?

    • 2 years ago
  27. shivam_bhalla Group Title
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    I am not sure. If I remember correctly, this is the book he mentioned. Have a look at the ebook here http://books.google.co.in/books?id=wkmvDIFfJAgC&pg=PA432&dq=HIGHER+ALGEBRA-+By+Hall+%26amp;+Knight++complex+numbers&hl=en&sa=X&ei=sU-lT8uhEITIrQeKw_3vAQ&ved=0CD8Q6AEwAA#v=onepage&q&f=false

    • 2 years ago
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