## Monday, October 28, 2013

### On Reduction, 6 of 11

Reciprocal Complex Numbers

Reciprocal complex numbers are of the form a<+>ib, where i equals the square root of negative one. This equals:

a <+> ib   =      aib     =   iab(a-ib)
a + ib        a2 + b2

=          ab        ( b + ia )
a2 + b2

=    (  b  <+>  a  )  ( b + ia )
a       b
Conversely,

a + ib     =          ab        ( b <+> ia )
a2 <+> b2

=    (  b  +  a  )  ( b <+> ia )
a     b

They follow the usual rules of complex arithmetic, in reductions:

(a <+> ib) <+> (c <+> id)   =   (a<+>c) <+> i(b<+>d)

(a <+> ib) <-> (c <+> id)   =   (a<->c) <+> i(b<->d)

(a <+> ib) * (c <+> id)   =   (ac<->bd) <+> i(ad<+>bc)

1    / (c <+> id)   =   (c/(c2<+>d2)) <-> i(d/(c2<+>d2))

The reciprocal form of the cis function is  sec(a)<+> i csc(a). Call this “sic(a)”; these equations apply:

sic(a)     =    sec(a)<+> i csc(a)

=    (1/cos(a))<+> (1/(-isin(a))

=    1/ ( cos(a) - isin(a) )

=    (cos(a) + isin(a))/(cos2(a)+ sin2(a))

=    ( cos(a) + i sin(a) )   =  cis(a)

So sic equals cis, and has the same identities:

sic(a+b)  =  sic(a) * sic(b)

sic(a-b)  =  sic(a) / sic(b)

sic(Na)  =   sic(a)N

In general  a <+> ib  equals

ab   (b + ia)
a2 + b2

=        ab    sqrt(a2 + b2) (cos(q) + i sin(q))
a2 + b2

=          ab     ( cos(q)  +  i sin(q) )
sqrt(a2+b2)

=      sqrt(a2<+>b2)  *  ( sec(q)  <+>  i csc(q) )

-   where q is the angle that a<+>ib makes with the positive real axis.

From sic and reciprocal-complex algebra we can derive these and other reciprocal-trigonometric identities:

sec2(a)<+> csc2(a)    =    1

sec(a+b)         =    sec(a)*sec(b)  <->  csc(a)*csc(b)

csc(a+b)         =    sec(a)*csc(b)  <+>  csc(a)*sec(b)

sec(2a)         =    sec2(a)  <->  csc2(a)

csc(2a)         =    ½ sec(a)*csc(a)

sec(3a)         =    sec3(a)   <->  (1/3) sec(a)csc2(a)

csc(3a)         =    (1/3) sec2(a)*csc(a) <-> csc3(a)