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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.)
Solution | Comment 12 of 14 |
The solution to this problem is that 1^(fractional power) is not in general a real number, whereas the last step in your argument assumes that we select the real root.

For example, take z=i so that r=1 and ?=?/2.  We can write z=i=r*(1)^(?/2?)=1^(1/4).  However, it is incorrect to infer from this that i=1 because i is also a fourth root of 1: i^4 = 1.

Your argument is correct up until the last step (? z=r ) and you have shown that z/r=1^(?/2?)  i.e. z/r is always a root of unity.

  Posted by Danny on 2007-01-06 22:29:02
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