9.4 The Complex Field 127Proof. Let a = a^1 /", /? = b^1 '^71 => an = a, (3n = b => (a/?)" = an(3n = ab y 0
and n"^1 root is unique => (ab)^1 /™ = a/3. DDefinition 9.3.8 (Extended real numbers) R U {+oo, -oo} 3 preserve
the order in R and Vx £ R, —oo -< x -< oo. R U {+oo, —oo} is an ordered set
and every non-empty subset has supremum/infimum in R U {+oo, —oo}.In R U {+oo, -oo}, we make the following conventions:
i) For x € R, x + oo = -t-oo, x — oo = —oo,
ii) If x -< 0, we have x • (+oo) = —oo, x • (—oo) = +oo,
iii) 0 • (+oo), 0 • (—oo) are undefined.9.4 ' The Complex Field
Let C be the set of all ordered pairs (a, b) of real numbers. We say(a, b) — (c, d) if and only if a — c and b = d.Let x = (a,b), y = (c, d). Definex + y = (a + c,b + d), xy = [ac — bd, ad + be].Under these operations C is a field with (0,0) being the zero element, and
(1,0) being the multiplicative unit.
Define 4>: R H-> C by <j>{a) = (a,0), then (j> is 1-1.(a + b) = (a + 6,0) = (a, 0) + (b, 0) = 0(a) + (&).
0(a&) = (a&,0) = (a,0)(6,0) = 4>{a)
Therefore, R can be identified by means of
way that addition and multiplication are preserved. This identification gives
us the real field as a subfield of the complex field.
Let i = (0,1) => i^2 = (0,1)(0,1) = (-1,0) = <£(-l), i-e. i^2 corresponds to the
real — 1.
Let us introduce some notation.
Hence,
C= {a + ib: a,b£R}.
If z = a + ib G C, we define 1 = a — ib (conjugate of z),
Z ~\~ ~Z Z — ~Z
a = Re{z) = -y-, b = Im(z) = -^r-.If z,w G C ^> z + w = J + w, Jw ='zw.
If z € C => zz = a^2 + b^2 y 0, we define \z\ = \/5i = \/o^2 + 6^2.