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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.)
I think this is wrong because... | Comment 4 of 14 |
...exponentiation of complex number x and y is not unique. It is defined as x^y:=e^ln(x)y. If the natural logarithm ln was well-defined, this expression would be well-defined because exponentiation with real basis (in this case e) can be defined via the well-known taylor expansion of the exp function f(x)=e^x which works for every complex x.

The natural logarithm unfortunately is not uniquely defined because ln(y) is basically defined as the solution of e^x=y, which has in fact an infinite number of solutions. This BTW is also the reason why there are two square roots of every complex number except 0.

  Posted by JLo on 2006-08-24 19:24:54
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