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Are all complex numbers real? (Posted on 2006-08-24) Difficulty: 3 of 5
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.)
Hints/Tips The General Problem | Comment 1 of 14
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
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