1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

1.3 • THE GEOMETRY OF COMPLEX NUMBERS 19

• EXAMPLE 1.5 To produce an example of which Figure 1.9 is a reason-
  able illustration, we let z1 = 7 + i and z2 = 3 + 5i. Then lz1I = v'49 + 1 =

v'50 and lz2I = v'9 + 25 = v'34. Clearly, z1 + z2 = 10 + 6i; hence lz1 + z2 I =

v'lOO + 36 = v'l36. In this case, we can verify the triangle inequality without

recourse to computation of square roots because lz 1 + Z2I = v'l36 = 2v'34 =

v'34 + v'34 < v'50 + v'34 = lz1 I + lz2 I·

We can also establish other important identities by means of the triangle

inequality. Note that

lzil = l(z 1 + z2) + (-z2)l

~ Jz1 + z2I + l-z2I

= lz1 + z2I + lz2I ·

Subtracting lz2l from the left and right sides of this string of inequalities gives

an important relationship that is used in determining lower bounds of sums of

complex numbers:

(1-24)

From Identity (1-22) and the commutative and associative laws, it follows that

Taking square roots of the terms on the left and right establishes another

important identity:

(1-25)

As an exercise, we ask you to show that

I

zi I - lzd provided z2 =/; 0.

z2 - lz2I'

(1-26)

• EXAMPLE 1.6 If z1 = 1+2i and z2 = 3 + 2i , then Jzil = Vf+4 = v'5 and

lz2I = v'9 + 4 = v'13. Also z1z2 = -1 +Bi; hence lz1z2I = v'l + 64 = v'65 =

v'5v'13 = Jzd lz2I ·

Figure 1.10 illustrates the multiplication shown in Example 1.6. The length

of the z 1 z2 vector apparently equals the product of the lengths of z1 and z2,

confirming Equation (1-25), but why is it located in the second quadrant when

both z 1 and z 2 are in the first quadrant? The answer to this question will become

apparent to you in Section 1.4.