1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

S82 ANSWERS

Sa. Since z 1 is a root of the polynomial P, P(zi) = O. Use properties (1-12)

through (1-14) of Theorem 1.1 to show that P (z1) = P(z1). This implies

P (z 1 ) = 0. Next show that if P (z1) = 0, then P (z1) = 0, confirming that

z 1 i s also a root of P.

Sc. Find a polynomial for part a, another for part b, and multiply them together.

1. Use the (ordered pair) definition for multiplication to verify that if z = (x, y)
   is any complex number, then (x, y)(l, 0) = (x, y).

9a. We would want to find a number (=(a, b) such that for any z = (x, y ) we

have z * ( = z. Obviously, if ( = (1, 1}, then according to the definition of*

we would have z * ( = (x, y) * (1, 1) = (x, y) = z. Thus, the multiplicative

identity in this case would have to be ( = (1, I).

9b. For any complex number w = (x, y) we would have (0, a)'* (x, y) = (O, ay),

which can't possibly equal (1, 1).

11. Let z1 = (x1, Y1), z2 = (x2, y2), and z3 = (xa, Ya) be arbitrary complex
    numbers. Then
    z1(z2 +za) =(xi, Y1) [(x2, y2) + (xs, Ya)}= (x1, Y1) [(x2 + x3, Y2 +ya)}=
    (x1(x2 + xa) - y1(Y2 +Ya), X1(Y2 + y3) + (x2 + xa)Y1) = · · · =
    (x 1 X2 - Y1Y2, X1Y2 + X2Y1) + (x1X3 - Y1Y3, XtY3 + X2Ya) = z1z2 + Z1Z3.
    Complete the missing steps in · · · above using the distributive and ot her
    laws for real numbers.

(^13). + (2 3 t .)-1 -- 13 2 - L3. ai, (7 - 5')-i 1 -- 74 7 + 74 5 i..
Section 1.3. The Geometry of Complex Numbers: page 20
la. JIO.
le. 225.
le. J(x - 1)^2 + y^2.
3a. Inside, since I(~ +i) -ii=~, which is less than 2.
3c. Outside, since 1 (2 + 3i) - i i = 12 + 2il = JS, which is greater than 2.

s. Let z1 = (x1, Y1) and z2 = (x2, y2). Since neither z1 nor z2 equals zero, they

are perpendicular iff their dot product is zero. But their dot product is

(xi, Y1) • (x2, Y2) = X1X2 + Y1Y2, which is precisely Re(z1z2).

1. Let z = x + iy. Then v'21z l 2 IRe(z)I + IIm(z)I iff v'2izl 2 lxl + lyliff 2lz l^2 2
   lx^2 1+ 2lxl IYl+IY^2 1iff2x^2 +2y^22 lx^21 +2lx l lv l+ IY^21 iffx^2 -2lxl lvl+y^2 2 0
   iff (Ix! - lyl)^2? 0, which is clearly true. A proper argument will start with
   this last inequality and work backwards to the appropriate conclusion.