Re: [MLUG - DISCUSSION] Wolfram Alpha

[Stephen Montgomery-Smith <EMAIL:PROTECTED>]

So i was created to be one of the square roots of -1.  But what about 
the square root of i?  At first glance, you would think that you have to 
invent yet another number, and then you realize that (1+i)/sqrt(2) works 
just fine!

Indeed, using the Euler identity, you see that the nth roots of 1 are 
exp(2pi i k/n) for 0<=k<n.  For example, the cube roots of 1 are 1, -1/2 
+ i sqrt(3)/2, and -1/2 - i sqrt(3)/2.

In fact, it turns out to be that ANY polynomial equation can be solved 
with a complex number.  So, for example,

x^6 + x + 24 = 0

has no real roots, but it has six complex roots.  Mathematica gives 
their approximate values as:

-1.46083 - 0.831482 I
-1.46083 + 0.831482 I
-0.0200238 - 1.69874 I
-0.0200238 + 1.69874 I
1.48085 - 0.86619 I
1.48085 + 0.86619 I

That ANY polynomial equation has a complex root is considered so 
important that it is called the "fundamental theorem of algebra:"

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

(If you are interested in the proofs, by far and away the most intuitive 
proof is the "topological proof" listed in this article.  But it is also 
one of the hardest to make rigorous.)


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