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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 Specific Problem (fleshed out a bit more) | Comment 7 of 14 |
(In reply to re: The Specific Problem (fleshed out a bit more) by JLo)

Homing in on a simple explanation here is definitely a challenge.  You have done the best job of anyone so far, JLo. However, I still do not feel like I really understand everything as well as I would like. We can't do algebra without knowing its laws, and the laws of exponents almost seem to have deserted us here.  There seems to be no simple replacement in the case of complex values for the law a^(bc)=(a^b)^c that we have learned for positive real a,b,c.  The only thing that seems even remotely safe to say in the complex case is that a^(bc)=(a^b)^c for some interpretation of the multiple values that can be assigned to the right side and for some interpretation of the multiple values that can be assigned to the left side. The practical lesson seems to be that while we can explain things in terms of multiple values, this exponent "law" is pretty much useless in arguments involving complex values.

Edited on August 28, 2006, 12:23 am
  Posted by Richard on 2006-08-26 22:45:33

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