In other words, if x and y are complex (each have the form a+bi), show that x^y can have a real value, or prove that it is impossible.

Note: i is the imaginary value defined as the number that yields -1 when squared. a and b are any real numbers, but b is not 0.

Actually, the link did help somewhat. I already had the mild idea that it was proven that way. But then you'd need to prove that e^n=n^0/0!+n^1/1!+n^2/2!... You don't actually need to though, I think that what is done is enough, let's just assume that it's given. Though, Charlie brings up a good point; this solution might be considered invalid anyway.

I like complex numbers, they make everything more open-minded, and they're so cool! I happen to have submitted a complex puzzle too. There's a simple calculator program that calculates quadratic roots, but only the real ones. That program annoys me so much because someone wasted their time making so it wouldn't show complex roots! It's like they don't like complex numbers or something!

Posted by Tristan on 2003-11-26 18:13:58