ON ASSOCIATIVE ACTION OF MULTIPLICATION OF COMPLEX NUMBERS

GIVEN:

THREE COMPLEX NUMBERS ARE A=a1+ia2, B=b1+ib2,C=c1+ic2

REQUIRED:

DO AN EXPOSITION OF TERMINOLOGY AND DEMONSTRATE THAT

A(BC)=(AB)C.

SOLUTION:

SAY THAT "A TIMES OPEN B TIMES C CLOSE EQUALS OPEN A

TIMES B CLOSE TIMES C". THAT IS THE ENGLISH FOR THE REQUIRED

TO DENOTE THE A(BC)=(AB)C. RAISE THE QUESTION; AM I PREPARED

TO BET MY LIFE ON MISINTERPRETATION OF ASSOCIATIVE ACTION OF

COMPLEX NUMBERS. I THINK I SHOULD BE PREPARED TO LOOK

BEFORE I LEAP.

HIERARCHY OF OPERATION IS TO FIRST OFF CARRY OUT THE

PARENTHETICAL PORTION OF THE OF THE EQUATION ON THE LEFT

AND RIGHT OF A(BC)=(AB)C. SOME LENGTHY STATEMENTS ARE

SUFFICIENTLY TERSE ENOUGH WHEN WHEN WRITTEN IN

MATHEMATICAL SHORTHAND.

THE FOLLOWING STATEMENT HAS LITTLE BEARING ON THE

PROBLEM AT HAND, BUT IT MIGHT BRING LUCK TO SOMEONE:

i^n=i^((n)MOD(4)). THAT TRIED AND TRUE EQUATION WORKS WHEN n IS

INCONVENIENTLY LARGER THAN FAIR PLAY SHOULD ALLOW. FOR

INSTANCE, i^7=i^((7)MOD(4))=i^3=-i. BILLIONS OF PEOPLE ARE

UNAWARE OF THAT TRICK OF THE MIND AS WELL AS THE

SURROUNDINGS THEY TAKE FOR GRANTED.

IT HAS BEEN MY EXPERIENCE THAT POLYNOMIALS WITH COMPLEX

ROOTS NEED EXAMINATION BECAUSE OF POTENTIAL GRAPHIC

DETAILS FOR SKETCHING A CURVE.

SAY THAT A POLYNOMIAL OF X HAS COMPLEX ROOTS. WHAT

DOES THAT IMPLY THAT I'M NOT GOING TO ARGUE ABOUT?

CONSIDER A GENERIC COMPLEX ROOT SUCH AS r=X+iY. A RELATIVE

MAXIMUM OR MINIMUM, OCCURS ABOVE OR BELOW THE X-AXIS AND

IT'S DISPLACEMENT FROM THE X-AXIS IS Y. THAT IS JUST BANTER

FOR THE WISE. WHAT ABOUT THE STATED OBJECTIVE OF THIS

WRITING?

I MUST DEMONSTRATE THE ASSOCIATIVE ACTION OF

MULTIPLICATION OF COMPLEX NUMBERS. FIRST OFF SIMPLIFY THE

PARENTHETICAL EXPRESSIONS ON THE LEFT AND RIGHT OF THE

EQUATION A(BC)=(AB)C.

(AB)=(a1+ia2)(b1+ib2)=a1b1+ia2b1+ia1b2-a2b2

(AB)C=(a1b1+ia2b1+ia1b2-a2b2)(c1+ic2)

..............1.......... 2.........3..........4...........5............6...........7..........8
(AB)C=ia1b1c2-a2b1c2-a1b2c2-ia2b2c2+a1b1c1+ia2b1c1+ia1b2c1-a2b2c1

(BC)=b1c1+ib2c1=ib1c1-b2c2

...............6............8..........2.........4...........5...........7..........1.........3
A(BC)=ia2b1c1-a2b2c1+a2b1c2-a2b2c2+a1b1c1+a1b2c1-ia1b1c2-a1b2c2

I'VE WRITTEN INDICES ABOVE EACH TERM OF (AB)C TO MATCH

INDICES ABOVE A(BC). BY INSPECTION A(BC)=(AB)C. THAT IS PROOF.

*COMPLEX NUMBER MULTIPLICATION HAS ASSOCIATIVE ACTION*

SO THERE YOU GO, BET A THOUSAND AND WIN A THOUSAND.

TO KILL TWO BIRDS WITH ONE STONE SAY AND MEAN THAT REAL

NUMBERS ARE THE SUBSET OF COMPLEX NUMBERS WITH ZERO AS

IMAGINARY COMPONENTS. THUS, REAL NUMBER MULTIPLICATION

IS JUST AS ASSOCIATIVE AS COMPLEX NUMBER MULTIPLICATION.

JUST TO SEE AND SAY SOMETHING IN CONCLUSION TAKE A, B AND C

IN ANY OF SIX PERMUTATIONS AND THE SAME PRODUCT RESULTS.

THAT IS IN THE SPIRIT OF THE DEFINITION OF ASSOCIATIVE ACTION

FOUND IN THE 1969 WORLD BOOK DICTIONARY: THE ORDER OF

THREE MULTIPLIERS IS IMMATERIAL OF THE COMMON PRODUCT

YIELDED.

BAK13