1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

ANSWERS 583

9. By the triangle inequality, lz1 - z2I = lz1 + (-z2)I :S lz1l+[-.z:i] = lz1l+lz2I ·


10. Let z =(a, b). Then z =(a, -b), - z = (-a, -b), and - z =(-a, b). The
    line segment from z to z is perpendicular to the line segment from z to -z
    since the vector from z to z is z - z = (0, -2b). The vector from z to - z is
    (-2a, 0), and the dot product of these is clearly zero. A similar argument
    works for the other line segments. It is also easy to show that the diagonals
    intersect at the origin, establishing symmetry there.

13. This is simply an equivalent form of the vector equation between the points

z1 = (x1, Y1) and z2 = (x2, Y2). Explain!

15. By repeated application of equation (1-25), we have lz1z2z3I = l(z1z2) z3I =

lz1z2l lz3I = lz1l lz2l lz3I.

17. lz - wl^2 = (z -w) (z -w) = (z - w) (z -w) = lz l^2 - zw - zw + lwl^2.
    ll -zwl^2 = (1-zw) (1- zw) = (1- zw) (1-zw) = 1-zw-zw+lzl^2 lwl^2 •

If lz l = 1, lz - wl

2

reduces to 1 - zw - zw + lwl

2

, and ll - zwl

2

becomes

1 -zw - zw + lwl^2. T hus, lz - w l^2 ;= I 1 -zwl^2 , and the conclusion follows.

Similarly, if lw I = 1, we get the same result.

19. By inequality (1-24), we see that lz1l - lz2I $ lz1 - z2I -Also, lz2l - lz1l $

lz:i - zil = lz1 - z2I, so that lz d - lz2I 2 - lz1 - z2I · Putting these two

inequalit ies together gives - lz1 - z2I $ lz1 I - lz2I $ lz1 - z2I, from whence

the conclusion follows.

21. Let z1 = (xi, Y1) and z2 = (x2, Y2). Re(z1.z:i) = X1X2 + Y1Y2· lz1z2I =

V(x1x2 + Y1Y2)^2 + (- X1Y2 + X2Y1)^2. If either z1 or z2 equals 0, then clearly

Re(z 1 z 2 ) = lz 1 z2I · If neither equals 0, the two quantities are equal precisely

when -X 1 Y2 + X2Y1 = 0 and X1X2 + Y1Y2 2 0. This occurs when the points

z 1 and z 2 lie on a straight line through the origin. Show the details for this

last statement.

23. The inequality If: Zkl $ E lzkl is clearly true when n = 1. Suppose that,
    k=l k=l
    for some j > 1, It zkl $ t lzkl. Then, using the triangle inequality and
    k=l k=l

ourinductionassumption, Ii£ zkl = l(t zi.) +zi+II :s I t Zkl+lzj+d $

k=I k=l k=l

(t

1

lzkl) + lz;+d = :~ lzi.I ·

25a. By definition, an ellipse is the locus of points t he sum of whose distances
from two fixed points is constant. Since lz - z1I gives the distance from

the point z to the point z1, the set {z: lz - z1I + lz - z2I = K} is precisely

those points that satisfy the d efinition of an ellipse.