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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.)
re(2): The General Problem | Comment 3 of 14 |
(In reply to re: The General Problem by Federico Kereki)

Depends on what your definition of ^(1/2) is. It is possible to define the 1/2 power to make (-1)^1=((-1)^2)^(1/2). In fact, there are ways to define things so that z^(a*b)=(z^a)^b for complex values of z, a, and b as long as the definition is being followed. You cannot, however, just define the powers however you feel like, or absurd results like the alleged one of this problem will be obtained.  The question is, then, how does one for complex values define z^w in such a way that the exponent law z^(a*b)=(z^a)^b holds?
  Posted by Richard on 2006-08-24 17:32:19

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