Dave's Math Tables: Complexity |
(Math | OddsEnds | Complexity) |
i 2 = -1
1 / i = -i
i 4k = 1; i (4k+1) = i; i (4k+2) = -1; i (4k+3) = -i (k = integer)
( i ) = (1/2)+ (1/2) i
(a + bi) + (c + di) = (a+c) + (b + d) i
(a + bi) (c + di) = ac + adi + bci + bdi 2 = (ac - bd) + (ad +bc) i
1/(a + bi) = a/(a 2 + b 2) - b/(a 2 + b 2) i
(a + bi) / (c + di) = (ac + bd)/(c 2 + d 2) + (bc - ad)/(c 2 +d 2) i
e (i ) = cos + i sin
n (a + bi) = (cos(b ln n) + i sin(b ln n))n a
if z = r(cos + i sin ) then z n = r n ( cos n+ i sin n )(DeMoivre's Theorem)
if w = r(cos + i sin );n=integer. then there are n complex nth roots (z) of w for k=0,1,..n-1:
z(k) = r (1/n) [ cos( (+ 2(PI)k)/n ) + i sin( (+ 2(PI)k)/n ) ]
if z = r (cos + i sin ) thenln(z) = ln r + i
sin(a + bi) = sin(a)cosh(b) + cos(a)sinh(b) i
cos(a + bi) = cos(a)cosh(b) - sin(a)sinh(b) i
tan(a + bi) = ( tan(a) + i tanh(b) ) / ( 1 - i tan(a) tanh(b))
= ( sech 2(b)tan(a) + sec 2(a)tanh(b) i ) / (1 + tan 2(a)tanh 2(b))