9.4 PROPERTIES OF THE COMPLEX NUMBER FIELD 323An immediate consequence of Definition 3 and the formula (x + yi) - ' =
(X - yi)/(x2 + y2) is the equation z-I = z/(z12, for any nonzero complex
number z. Since this implies zz/lz12 = 1, we have also zz = 1zI2 for any
complex number z, so that, in particular, the product zz (a product of two
complex numbers) is always a nonnegative @ number. This fact is the
basis of a familiar algebraic technique, illustrated in the following example.
EXAMPLE 3 Express 1/(6 + 8i) in the form x + yi.
Solution The technique is multiplication by the complex conjugate of
the denominator divided by itself. Specifically, we have that 1/(6 + 8i)
= [1/(6 + 8i)][(6 - 8i)/(6 - 8i)] = (6 - 8i)/(36 + 64) = (3150) - (2/25)i.
What are we actually doing when we apply this technique? The original
problem could be rephrased, "find z-', where z = 6 + 8i." By the for-
mulas developed following Definition 3, z- ' = z/1zI2 = z/zz, a formula
corresponding precisely to the algebraic technique.
We are now ready to state formally several properties involving Re (z),
Im (z), z, and lzl.
THEOREM 2
Let z be a complex number. Then:
z+ z* = 2 Re (z), so that Re (z) = (&)(z+ z*)
z - z* = 2i Im (z), so that Im (z) = (i/2.)(z* - z)
zz* = lz12
IzI 2 0; IzI =^0 if and only if z =^0
z** = z
14 = IzI
z = z* if and only if z is real
-z = z* if and only if z is imaginary
Im (iz) = Re (z)
Re (iz) = - Im (z)
Partial proof (a) Let z = x + yi. Then z + z* = (x + yi) + (x - yi) =
2x = 2 Re (z). The conclusion Re (z) = (+)(z + z*) follows immediately.
(h) Assume z* = -2, where z = x + yi. This means that x - yi =
-x - yi, so that 2x = 0 and hence x = 0, thus proving that z is purely
imaginary. The converse is left to you.
(i) If z =-x + yi, then iz = xi + yi2 = - y + xi. Hence Im (iz) = x =
Re (2).
The remaining portions of the theorem are left as exercises (see
Exercise 8).
THEOREM 3
Let z, and z, be complex numbers. Then: