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

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

2bornot2bBest ResponseYou've already chosen the best response.1
As far as I know \(\large(e^z)^w\ne e^{zw}\) rather \(\large(e^z)^w=e^{(z+2n\pi i)w} \)
 one year ago

2bornot2bBest 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.
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
@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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
Thanks @shivam_bhalla I need this help extensively.
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
I think b is complex number , right ??
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
I believe so, but it would be better for you to check it at that link
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
@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)."
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
Um... isn't it just exponent rules? (a^b)^c = a^(b*c)
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
Here:http://en.wikipedia.org/wiki/Exponential_function#Computation_of_ab_where_both_a_and_b_are_complex
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
Sounds reasonable @shivam_bhalla But the topic starts with the name Computation of \(a^b\) where both a and b are complex
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
Seems to me you are right, otherwise that line wouldn't have been mentioned at the end. What do others think?
 one year ago

2bornot2bBest 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?
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
@2bornot2b , good idea. I think you should go ahead :)
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
Anyway thanks a lot, that really helped me a lot.
 one year ago

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

shivam_bhallaBest ResponseYou've already chosen the best response.2
@cshalvey, check a serious bug above ^^
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
By the way shivam, can you suggest me a good book on complex functions?
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
And I think Officer @shadowfiend is in the department of arresting bugs.
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.1
Does that book have complex functions?
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
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=sUlT8uhEITIrQeKw_3vAQ&ved=0CD8Q6AEwAA#v=onepage&q&f=false
 one year ago
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