Norman Albers Posted April 9, 2009 Share Posted April 9, 2009 (edited) Can you find a complex number which, when multiplied by its complex conjugate, is -1? Today, I tried to find one. Edited April 10, 2009 by Norman Albers Link to comment Share on other sites More sharing options...
Shadow Posted April 10, 2009 Share Posted April 10, 2009 Does this have a solution? I'm a complete amateur, but if you plug this into an equation, you get [math](a+bi)\cdot(a-bi) = -1[/math] where [math] a, b \in R[/math] [math]a^2+b^2=-1[/math] And as far as I know, there are no real numbers which satisfy this equation...then again, there is the pretty big chance that I'm completely off, so don't take this too seriously... Cheers, Gabe Link to comment Share on other sites More sharing options...
ajb Posted April 10, 2009 Share Posted April 10, 2009 Let [math]z \in \mathbb{C}[/math] be a complex number. Then (by thinking geometrically) you can show that [math]z \overline{z} = |z|^{2}[/math], as shadow has done (even if he did not realise it). So it has to be positive. Or in Shadows words [math]a^{2}+ b^{2} = -1[/math] has no real solutions. Link to comment Share on other sites More sharing options...
Norman Albers Posted April 10, 2009 Author Share Posted April 10, 2009 B-B-Bingo! Good answers. [bIG TEETH] Link to comment Share on other sites More sharing options...
charmeddvd123 Posted July 28, 2009 Share Posted July 28, 2009 Thank you for giving us to share this thread Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now