46
From (l..12-2) and (1.12-5) we obtain
x^2 = %(r +a),;; = ± ./%(r +a),
Chapter 1y^2 = %(r - a)
y = ± .;---1/2-( r---a-)These solutions must be combined in pairs so as to satisfy equation (1.12-3).
If b > 0, then xy > 0, which implies either x > 0, y > 0 or x < 0, y < 0.
Hence we obtain the two roots
./%(r +a)+ i./%(r - a) and
If b < O, then xy < O, which implies either x > O, y < 0 or x < O, y > 0,
and we obtain the roots
./%(r +a)-i./%(r - a) and
If b = 0, then xy = O, which implies either x = O, or y = 0, or x = y = 0.
Since r = lal when b = O, we have y = 0 if a > O, x = 0 if a < O, and
x = y == 0 if a = 0.
By using the notation *Va + bi to denote the set of both square roots
of a+ bi, and the notation sgnb (signum of b) to denote +1 if b > 0 and
-1 if b < 0, we may summarize the foregoing results as follows:
!
*Va+ bi= {± [./^1 Mr +a) +i(sgnb)./%(r - a)]},
*'/a+ bi= { y'a, -Va}, b = 0, a> 0
*/a+ bi= {iy'=a, -iv'=ci}, b = O, a< 0
*/a+bi= {O,O}, b=O, a=O
Example If a+ bi = 15 - 8i, then r = 17 and
*V15 - si = {±(4 - i)}(1.12-6)By expressing the complex number w =fa 0 in trigonometric or exponen-
tial form, it is possible to give a simple general expression for the nth roots
of w. Suppose that w = rei^8 and let z = Reioi. Substituting in (1.12-1),
we have, taking (1.11-2) into account,
Rn eina = reiewhich implies that
and na = () + 2k7r
so thatand a=
() + 2br
n