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Topic: [HM] Bombelli: Cube root of Complex Nos.
Replies: 8   Last Post: Feb 19, 2001 5:31 AM

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Milan Bozic

Posts: 10
Registered: 12/3/04
Re: [HM] Bombelli: Cube root of Complex Nos.
Posted: Feb 14, 2001 11:27 PM
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Bonnie Shulman wrote:

>
> However, my question is, how could we, using current methods,
> see that the cube root of 2 + 11i is 2 + i ? I can certainly
> find the three cube roots in terms of cube roots of unity,
> and sum the conjugates to get the three real
> roots of the original cubic. BUT, in actually computing
> the real and imaginary parts of 2 + 11i, I am left evaluating
> cos(arctan 11/2) (or sin of same). Is there a way to compute
> this analytically (series?) so as to get exactly 2 or 1, rather
> than a numerical approximation? [I suspect so, as my calculator
> gives the exact answers.]
>


Of course that we can. First of all it is absolutely unnecessary to
bother oneself with Cardano formula in the case of the equation x^3=15x+4 as
it is obvious that if it has an integer (rational is the same) solution then
this solution must be the divisor of 4. So, the only candidates are 1,2,4
(-1,-2,-4 were not allowed in Cardano epoch). By inspection we obtain x=4.
On the other hand we have Cardano formula which gives x = (2 +
11i)^(1/3) + (2 - 11i)^(1/3). What to do with it? The same which we are
always doing with the equations with integer coefficients. More or less the
same which I proposed to apply to the starting equation in previous
paragraph.
As (2+11i)^(1/3) equals a+bi for some a and b, we obtain a^3-3ab^2=2 and
3a^2b-b^3=11 (2 and 11 are irellevant, they can be any integers). So, if a
and b are integers, a must be divisor of 2 (or real part of complex number
obtained from Cardano formula) and b must be a divisor of 11 (or imaginary
part of same complex number). The list is finite and by inspection we obtain
a=2 and b-1.
Here must be stated that this problem (of determining "Gaus integer"
roots of Gauss integer complex number) is always recusive, so problem is
decidable.
If a and b are not integers the we must aply some numerical method for
obtaining, digit by digit, digits of a and b. Any standard numerical method
can be applied and they all work and thay are also decidable!
So, your amusement, I apologize if I am implying too much, comes from
the difference in numerical methods which we are applying. For example
"everybody" knows how to obtain, digit by digit any amount of digits of
a^(1/2) as we learned in school standard method of "extracting the square
root". Provided that we have had a stubborn teacher which insisted in
learning a lot of numerical methods many other decidable "integer" like
equation problem would be common to us.
Imagine the question wether 234698374973082625647841^(1/2) is an
integer? You know how to solve it in principle but a lot of calculations
must be applied before you get the right answer,
Finally it looks that Bombelli somehow realy invented the general
method. His only mistake was to start with assumption a=2. It is enough to
start with question wether both a and b are integers and inspect all cases.

Milan Bozic
milan@sezampro.yu






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