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2bornot2b
 4 years ago
Can someone explain why the last equality is possible?
\[\large{a^b=(re^{i\theta })^b}\]\[\large{=(e^{ln~r+i\theta})^b=e^{(ln~r+i\theta)b}} \]
2bornot2b
 4 years ago
Can someone explain why the last equality is possible? \[\large{a^b=(re^{i\theta })^b}\]\[\large{=(e^{ln~r+i\theta})^b=e^{(ln~r+i\theta)b}} \]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@2bornot2b , \[\huge r = e^{\ln(r)}\] Using this the result is got

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1As far as I know \(\large(e^z)^w\ne e^{zw}\) rather \(\large(e^z)^w=e^{(z+2n\pi i)w} \)

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1@shivam_bhalla I am talking about the equality that appears next to the one you have talked about.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Thanks @shivam_bhalla I need this help extensively.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think b is complex number , right ??

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1By 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

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1I believe so, but it would be better for you to check it at that link

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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)."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Um... isn't it just exponent rules? (a^b)^c = a^(b*c)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here: http://en.wikipedia.org/wiki/Exponential_function#Computation_of_ab_where_both_a_and_b_are_complex

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Sounds reasonable @shivam_bhalla But the topic starts with the name Computation of \(a^b\) where both a and b are complex

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Seems to me you are right, otherwise that line wouldn't have been mentioned at the end. What do others think?

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1@Mertsj What do you think? @shivam_bhalla how about editing the page, and refining it for future readers?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@2bornot2b , good idea. I think you should go ahead :)

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Anyway thanks a lot, that really helped me a lot.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@2bornot2b , a serious bug at OS. See the attachment See your name is replaced by "Best answer"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@cshalvey, check a serious bug above ^^

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1By the way shivam, can you suggest me a good book on complex functions?

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1And I think Officer @shadowfiend is in the department of arresting bugs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Seriously, 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

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Does that book have complex functions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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=sUlT8uhEITIrQeKw_3vAQ&ved=0CD8Q6AEwAA#v=onepage&q&f=false
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