Let's "prove" that every complex number z
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?
The general problem here is that z^(a*b) does not, without proper
restriction, equal (z^a)^b. (You can read "The Powers That Be,"
especially Tristan's example and my reply to it, to see what the story
is here, but that would be cheating!)
Posted by Richard
on 2006-08-24 15:54:55