Appendix B.
Solution of Polynomials with Complex Coefficients -
Examples.
B.1 Same Sign Roots, (Imaginary Component).
Consider the equation,
|
y = ( x + 2 ){ x + ( 1 + j2 ) }{ x + ( 3 + j4 ) } |
| (B.1) |
This multiplies out to,
|
y = x3 + ( 6 + j6 )x2 + ( 3 + j2 )x + ( - 10 + j20 ) |
| (B.2) |
To determine the roots of (B.2) using Bairstow, first multiply (B.2) by its
complex conjugate equation
|
yc = x3 + ( 6 - j6 )x2 + ( 3 - j2 )x + ( - 10 - j20 ) |
| (B.3) |
To give
|
yyc = x6 + 12x5 + 78x4 + 280x3 + 621x2 + 820x + 500 |
| (B.4) |
Equating (B.4) to zero and inserting its coefficients into the Bairstow
spreadsheet yields the roots as,
|
yyc = ( x + 2 )2{ x + ( 1 + j2 ) }{ x + ( 1 - j2 ) }{ x + ( 3 + j4 ) }{ x + ( 3 - j4 ) } |
| (B.5) |
Consequently, within (B.5), the roots of (B.2), from an application
Descartes' Rule of Signs to the imaginary components of (B.2), are
determined to be those of (B.1).
B.2 Opposite Sign Roots, (Imaginary Components).
If (B.1) were
|
y = ( x + 2 ){ x + ( 1 + j2 ) }{ x + ( 3 - j4 ) } |
| (B.6) |
To give
|
y = x3 + ( 6 - j2 )x2 + ( 19 - j2 )x + ( 22 + j4 ) |
| (B.7) |
Multiplying (B.7) by its complex conjugate again gives (B.4) and
subsequently (B.5). Descartes' Rule of Signs applied to (B.7) then shows
complex roots with both positive and negative imaginary components. The
correct ones are then identified by inserting the two complex root
combination possibilities, i.e. (1 + j2) with (3 - j4) or (1 - j2) with (3 + j4),
together with the single real root, into the Polynomial Construction
spreadsheet to determine which gives the correct coefficients of (B.7).
M3 Version 1.0.0
Ó
P.G.Bass, January 2010
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