1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

16 CHAPTER 1 • COMPLEX NUMBERS

12. Verify that if z = (x, y), with x and y not both 0, then z-^1 = <\^0 > (i.e., z -^1 = ~}.
    Hint: Let z = (x, y) and use the (ordered pair} definition for division to compute
    z-^1 = [!:~~·Then, with tbe result you obtained, use the (ordered pair} definition
    for multiplication to confirm that zz-• = (1, O} = 1.

13. Ftom Exercise 12 and basic cancellation laws, it follows that z-^1 = ~ = !i. The

numerator here, z, is trivial to calculate and, as the denominator zz is a real

number (Exercise 3}, computing the quot ient -!x should be rather straightforward.

Use this fact to compute z-^1 if z = 2 + 3i and again if z = 7 -5i.

14. Show, by equating the real numbers x 1 and x2 witb (x 1 , 0) and (x 2 , O}, respec-
    tively, that the complex definition for division is consistent with the real definition
    for division. Hint: Mimic the argument given in the text for multiplication.

1.3 The Geometry of Complex Numbers

Complex numbers are ordered pairs of real numbers, so they can be represented
by points in the plane. In this section we show the effect that algebraic operations
on complex numbers have on their geometric representations.

We can represent the nwnber z = x + iy = (x, y) by a position vector in t he

xy plane whose tail is at the origin and whose head is at the point (x, y). When

the xy plane is used for displaying complex numbers, it is called the complex

plane, or more simply, the z plane. R ecall that Re(z) = x and Im(z) = y.

Geometr ically, Re(z) is the projection of z = (x, y) onto the x-axis, and Im(z)

is the projection of z onto the y-axis. It makes sense, then, to call t he x-axis the

real axis and the y-axis t he imaginary axis, as Figure 1.3 illustrates.

Addition of complex numbers is analogous to addition of vectors in the

plane. As we saw in Section 1.2, the sum of z1 = X1 + iy 1 = (x1> y 1 ) and

Z2 = x2 + iy2 = (x2, Y2) is (x1 + x2, YI + Y2). Hence z1 + z2 can be obtained

vectorially by using the "parallelogram law," where the vector sum is the vector

represented by the diagonal of the parallelogram formed by the two original

vectors. Figure 1.4 illustrates this notion.

The difference z 1 - z 2 can be represented by the displacement vector from

the point z2 = (x2, v2) to the point z 1 = (x1, Y1), as Figure 1.5 shows.

Imaginary axis

y

x + iy = , z ....... y

x

Figure 1.3 The complex plane.

C

Copy of vector z 1

Y (positioned at the tail of vector zv

Zz. • ,,.., Z1 + Z2

• :-.__J Copy of vector z 2
  ~z 1 l (positioned at the tail of vector Ztl

-!""=---"'"--x

Figure 1.4 T he sum z1 + z2.