## 2bornot2b 3 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}}$

1. shivam_bhalla

@2bornot2b , $\huge r = e^{\ln(r)}$ Using this the result is got

2. 2bornot2b

As far as I know $$\large(e^z)^w\ne e^{zw}$$ rather $$\large(e^z)^w=e^{(z+2n\pi i)w}$$

3. 2bornot2b

@shivam_bhalla I am talking about the equality that appears next to the one you have talked about.

4. shivam_bhalla

@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

5. 2bornot2b

Thanks @shivam_bhalla I need this help extensively.

6. shivam_bhalla

I think b is complex number , right ??

7. 2bornot2b

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

8. 2bornot2b

I believe so, but it would be better for you to check it at that link

9. shivam_bhalla

@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)."

10. SmoothMath

Um... isn't it just exponent rules? (a^b)^c = a^(b*c)

11. shivam_bhalla
12. 2bornot2b

Sounds reasonable @shivam_bhalla But the topic starts with the name Computation of $$a^b$$ where both a and b are complex

13. shivam_bhalla

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

14. shivam_bhalla

*specified

15. 2bornot2b

Seems to me you are right, otherwise that line wouldn't have been mentioned at the end. What do others think?

16. 2bornot2b

@Mertsj What do you think? @shivam_bhalla how about editing the page, and refining it for future readers?

17. shivam_bhalla

@2bornot2b , good idea. I think you should go ahead :)

18. 2bornot2b

:)

19. 2bornot2b

Anyway thanks a lot, that really helped me a lot.

20. shivam_bhalla

@2bornot2b , a serious bug at OS. See the attachment See your name is replaced by "Best answer"

21. shivam_bhalla

@cshalvey, check a serious bug above ^^

22. 2bornot2b

Great catch Shivam!

23. 2bornot2b

By the way shivam, can you suggest me a good book on complex functions?

24. 2bornot2b

And I think Officer @shadowfiend is in the department of arresting bugs.

25. shivam_bhalla

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

26. 2bornot2b

Does that book have complex functions?

27. shivam_bhalla

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