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Topic: Complex Numbers
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R. Brown

Posts: 54
From: texas
Registered: 12/29/05
Complex Numbers
Posted: Oct 5, 2006 10:26 AM
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The i-factoring method of determiining Sq. Roots
by Robt W. Brown

Did you know that by just arriving at one unknown ratio (u) we can determine the sq. root of many complex numbers, stated in the form: a + bi....!

I discovered this method by noting that since i^2 = -1 , we could create a
quadratic equivalent of any complex number by simultaneously adding & subtracting the number one. For example, let's first consider the following
pure imaginary:

2i
where:
b = 2 and a= 0
a + bi = y
0 + 2i = y
y= 2i
If we now add i^2 and also (+ 1) to the LHS, we have not changed the value of y:
i^2 + 2i + 1 = y
i^2 + 2i + 1 = y
i^2 + 2i + 1 = y
i^2 + 2i + 1 = 2i ...............eq.101
We now can state eq. 101 as a quadratic:
x^2 + 2x + 1 = 0
Where:
A= 1
B= 2
C= 1
If we arrive at the root(s) of the above equation we will find that in this particular
case: B^2 - 4AC = 0. Consequently the quadratic solution reduces to:

r = -B/2A + /- 0
r = -2/2
r = -1
r' = -1

We arrive at a 'double root', where:
r = r'

Therefore, when we use the roots to recreate the quadratic, we have:
(x- -1) ( x- -1) = x^2 + 2x + 1
(x + 1) ( x + 1) = x^2 + 2x + 1 .............eq. 201
Since the roots (r and r') are equal in this unique instance, we can state eq. 201as:
(x + 1)^2 = x^2 + 2x + 1
Then if we assume:
x= i
We have this equation:
(i + 1)^2 = i^2 + 2i + 1
(i + 1)^2 = 2i
i + 1 = [sqrt]2i
Therefore, by i-factoring with 1, we have arrived at the sq. root of our pure imaginary (2i) stated at the outset.

Now let's look at a problem where we do essentially the same thing, except we
must i-factor with a value that does NOT equal (+/-1)
***********
It works like this. Let's begin with the complex number: 9 + 3i
where:
a= 9
b= 3
Then let's assume that in complex numbers, the following is generally true:
b^{2}/4 = au + u^2
Then by subbing we have:
3^{2}/4= 9u + u^2
Then arranged as a quadratic, we have:
u^2 + 9u - 2.25= 0
We see the constants in the above quad are:
A=1
B= 9
C= -2.25
Then, if we arrive at the two roots, (since u = root) we have two possible values for (u):
namely:
u= .2434167
u'= -9.2434167

We now have enough information to solve the root, in the general form:
([sqrt]u * i ) +/- ([sqrt]{a + u}) = [sqrt] (a + bi)...........eq.301

Now let's see if eq. 301 is in fact a true statement:
([sqrt](.2434167) * i ) +/- ([sqrt]{9 + .2434267}) = [sqrt] (a + bi)...

([sqrt](.2434167) * i ) +/- ([sqrt]{9.2434267}) = [sqrt] (a + bi)...
(.4933727i +/- 3.0402969) = [sqrt]( 9 + 3i)

Since we see the sign: +/- in the LHS of the above equation, we must assume that one of these
signs will balance the equation (but the other won't). So, let's first test:
(.4933727i + 3.0402969) = [sqrt]( 9 + 3i)................eq. 401

(.4933727i + 3.0402969) (.4933727i + 3.0402969) = .2434167i^2 + 1.499999i + 1.499999i + 9.2434167
(.4933727i + 3.0402969) ^2 = . 2434167i ^2 + 2.999999i + 9.2434167
(.4933727i + 3.0402969) ^2 = (. 2434167* -1) + 2.999999i + 9.2434167
(.4933727i + 3.0402969) ^2 = (-. 2434167) + 2.999999i + 9.2434167
(.4933727i + 3.0402969) ^2 = + 2.999999i + 9
(.4933727i + 3.0402969) ^2 = + 3i + 9 ..........eq.401 above proven true!

Since eq. 401 is proven true in this instance, it validates the possibilty that eq.301
is also generally true for one of the 'sign' (and when the power of ( i) is i^1, in the original
statement of a complex number.

finally, we can assume that both possible roots are:
(+.4933727i + 3.0402969) ^2 = + 3i + 9 ..........
and:
(-.4933727i - 3.0402969) ^2 = + 3i + 9 ..........
*********
Postscript statement:
I will now give a brief explanation how I arrived at the above solution.
It all began with an 'idea', which I call (i-factoring). It occurred to me that we could state any complex number, in quadratic form, by adding/subtracting + 1
a+ bi = y
bi + a = y
i^2 + bi + (a+1) = y
therefore:
i^2 + bi + (a+1) = a + bi ...........eq. 501
Then I arrived at the above eq., which is obviously true! We in effect added 1, then
also subtracted 1, from the original equation, therefore eq. 501 is in balance.
*********
I knew at this point that I could state eq. 501 in a more general term by saying:
x^2 + bx + (a+1) = 0
Then our i-factored Quadratic becomes:
A= 1
B= b
C= a + 1
Since, a & b are always 'knowns', it became obvious that if we determine the roots of
any quadratic equation that is built this way, we could factor the quadratic. Knowing:
R= root
R'= 2nd root
The following statement is generally true in quadratics:
A{(x-R)(x - R) }= Ax^2 + Bx + C.............eq.601
Since in my original I-factoring idea, A always equaled (1):
(x - R)(x - R' ) = Ax^2 + Bx + C = a + bi.......eq.701
Since Quadratic roots generally have opposite signs, when applied eq. 601 became:
(x + R ) (x - R)= Ax^2 + Bx + C = a + bi.......eq.701
So at this point it occurred to me that a 'double root' in our quadratic was possible, and when this occurs, we arrive at identical factors:
(x - R)(x - R')= (x - r)^2
(x - R)(x - R')= Ax^2 + Bx + C = a + bi.......eq.801
since: R = R'
(x - R)^2 = Ax^2 + Bx + C = a + bi.......eq.801-A
(x - R)^2 = a + bi.......eq.801-C
Therefore, I knew that when: B^2-4AC = zero, my simple i-factoring method would
produce a "square-root" of the related 'complex number'. The complex number that
had been i-factored into a 'quadratic'!
*********
After this discovery, it became apparent that instead of adding 1 and i^2 to a complex
number, we might be able to use a ' calculated number' instead. A number that would
insure that: B^2- 4A = zero...always! And since the 'calculated number' was going to
cancel from the LHS of the equation, it's numerical magnitude was insignificant. And this is WHY I designed the following equation:
b^{2}/4 = au + u^2..............eq.301-A
The above equation is the equivalent of saying quadratically:
B^2/4 = AC
or:
B^2- 4AC = 0...........eq. 901
********
Finally, it seems possible that we can also find the 4th, 8th, 16th, ect...... sq. roots of a complex number through the i-factoring process. Odds roots seem less likely.

Compliments of Col. Rbtx, the Barnyard Physicist of Texas
*********



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