1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

586 ANSWERS

13. Let C: z(t) = (x (t), y (t)), as ts b be any curve joining -2 and 2. Then

x (a)= - 2, and x (b) = 2. By the intermediate value theorem, there is some

t* E (a, b) such t hat x (t* ) = 0. But this means z (t*) = (0, y (t*)) is not in

the set in question. Explain why!

15 a. We prove the contrapositive. Suppose zo is accumulation point of S, but
that zo does not belong to S. By definition of an accumulat ion point, every
deleted neighborhood, D; (zo), contains at least one point of S. Therefore,
every (nondeleted) neighborhood De (zo) also contains at least one point of
Sand at least one point not in S (namely, zo). This condition implies that
zo, which does not belong to S, is a boundary point of S. (Show the details
for this last assertion.) Thus, the set S is not closed.

Section 2 .1. Func tions and Linear Mappings: page 61

la. 6 + !i.

le. 2i.

3 a. u(x, y) = x^3 - 3xy^2 ; v (x, y) = 3x^2 y -y3.

( ) _ :i:

(^2) -11 (^2) ( ) _ -20:11
3 c. U x, Y - x•+2xiy^2 +11•' V x, Y - x^4 +2:i:^2 11~+11^4 •
5a. 1.

5 c. - 2 I + i ·fl 2 •

5e. -1.

7a. 0.

7c. lnv'2+ii,or !In2+ii.

7e. Yes, because if f (zi) = f (z2) (where z1 = r 1 e'^61 and z2 = r2ei^6 2, and 91 and

92 are the a rguments of z1 and z2, respectively), then In r 1 +i91 =In r2 +i92.

Equating real and imaginary parts gives In r1 = In r2, so r1 = r 2 (because t he

function In is one-to-one). Also, i8 1 = i82, so r 1 ei^91 = r2e'^9 >, i.e., z 1 = z2.

9a. H E (z) = z~zo, then with rods at the points zo = 0, 1 - i, and 1 + i , each

carrying a charge of ~ coulombs per unit length, the total charge at z will be

-.~o + •-(~-i) + •-(~+i). Combining terms and solving (us ing the q uadratic

formula) for when the numerator equals zero (tedious, but good for you!)

reveals the total charge to be zero when z = j ± i ~. Be sure to show the

details of your calculations.

lla. Clearly, f is onto, because if w E B, then by definition of B there exists
a point z E A such that f (z) = w. Suppose that f (z1) = f (z2) for some
values z 1 and z2 in A. T hen, because A is a subset of D , z 1 and z2 both

belong to D. But f is one-to-one on D. Therefore, z1 = z2.