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(1526-1573)
was the one who finally managed to settle the problem with imaginary numbers.
In Algebra 1569, Bombelli solves equations, using
the method of del Ferro/Tartaglia, introduces +i and -i
and describes
what happens when these are multiplied in various combinations.
That
the solution of imaginary numbers created difficulties and needed
knowledge about imaginary numbers is here shown with two examples:
The
equation x3 = 15x +4
has the real root x=4 which easily can
be seen by testing: 43 = 15*4 +4, 64 = 60 +4.
Solving
the equation using the method of del Ferro/Tartaglia leads to:
(u+v) 3 =15(u+v) +4 gives u3 +v3
= 4 and u3*v3 =125
which
gives vp +q = 4 and p*q =125
having the solution:
p = 2+  (-121)
and q = 2- (-121)
Hence x = 3p - 3q = 3(2 + (-121)) + 3(2-(-121)) =
= 3(2 + 11i) +3(2 - 11i)
and we know that this expression has
to be = 4. To show this requires quite a lot of
work, and some
cleverness.
It is
reasonable to start with assuming that 2+11i
can be expressed as a cube. And it can easily be shown that this
is true as
2 +11i = (2 + i) 3 and also that
2 - 11i = (2 - 1) 3 and hence:
(2 - 1) 3 = 8 -3*4i-3*2*1+i = 2-11i
And (2
+ 1) 3 = 8+3*4i-3*2*1- i = 2+11i
And
we get = ((2+i) 3 ) 1/3 + ((2-i) 3 )
1/3 = 2+2=4
Another
example, in order to demonstrate a more general method:
Examining
the equation x 3 + 6x = 20 it can easily be seen
that there
is an answer x = 2, but using the formula of Tartaglia leads to
the
answer x = 3(108 + 10) - 3( 108 - 10) which then
obviously
must be equal to 2.
It is
reasonable to assume that 10 + 108 and 10 + 108 can be
expressed as cubes as in the previous example. In order to find that cube we
can assume that there is a number a+b such that
(a + b) 3 = 10 + 108
and
(a - b) 3 = 10 - 108 so a 3
+ 3ab2= 10 , 3a2b + b 3 = 108
and since (a+b) + (a-b) is to be 2, we should have a=1,
so the
equations become 1 + 3b2 = 10
and 3b + b3 = 108 = 63
which do have the common solution b = 3
so all that is required is to check that
(1 + 3) 3 = 10 + 108, whence of course
(1 - 3) 3 = 10 - 108
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