complexities

[Norman Albers]

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

[Shadow]

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

[ajb]

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.

[Norman Albers]

B-B-Bingo! Good answers. [bIG TEETH]

[charmeddvd123]

Thank you for giving us to share this thread