All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Are all complex numbers real? (Posted on 2006-08-24)
Let's "prove" that every complex number z is real.

If z=0 it's obvious. For all other complex numbers z=r*e^(θi), where r is a real number, and i=√-1.

Now, z= r*e^(θi)= r*(e^(2πi))^(θ/2π). Now as we know that e^(2πi)=1 we can write z =r*(1)^(θ/2π) → z=r.

What's wrong with this?

 No Solution Yet Submitted by atheron Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 a possible solution Comment 14 of 14 |
exp^(ab)=(exp^a)^b does not apply to complex no.it is only for real numbers.if it apply to complex no then say b is complex

(exp^(a))^b=(cos a +sin a)^b=cos ab + sin ab

quantity ab is complex, so  cosine or sine of it is meaningless

exp^(θ*i)=cos θ + i sin θ   is not equal to

cos (θ*i)  + sin (θ*i)  as  your  reasoning  apply

 Posted by zohaib ahmed on 2007-03-05 11:57:52
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information