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?
...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.
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Posted by JLo
on 2006-08-24 19:24:54 |