1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

S92 ANSWERS

Section 3.3. Harmonic Functions: page 119

l a. u is harmonic for all values of (x, y).

3. c = - a.

Sa. v(x,y)=x^3 - 3xy^2 +c.

Sc. u(x,y) = -eYcosx +c.

7. By the chain rule, U,, (x, y) = u., (x, -y), Uy (x, y) = -uy (x, -y),

U.,,,(x,y) = u.,,,(x,-y), Uyy(x,y) = uyy(x,- y). Hence, U.,.,(x,y) +

U 11 y (x, y) = u""" (x, -y) + u 1111 (x, -y) = 0.

9. The funct ion f = u +iv must be analytic, hence so is j^2 = u^2 - v^2 + i (2uv).
   By T heorem 3.8, the result follows.

1 1. U8 = -rVr implies U99 = -rVr8· and U8r = -rVrr - Vr. Also, V9 = TUr

implies V9 9 = ru, 9 and V8r = rUrr +Ur· From this we get r^2 Urr + rur +

U88 = (rV8r -ru,) + (rur) + (-rvre) = O.

13a. J (z) = ,..;iy = .,>~y" +i,.27y':I·

1s. T he equipotentials are concent ric circles with radii l, 2, 3, and 4. T he

streamlines are lines from the origin making an angle of ¥ radians for

k = 0, 1,... ' 7.

S ection 4.1. Sequences and Se r ies : page 131

l a. O.

le. i.

3. Let e > 0 be given. Since Jim Zn = z 0 , there exists N, such that if n > N,
   n-->oo
   then Zn E De (zo), i.e., lzn -zol < e-But since lzn -zol = lzn - zol, this
   implies that if n > N,, then Zn ED, (zo).

s. This is a "telescoping sum" and we have for the nth part ial sum Sn =

• t + n~i (show t he details for this). Then JSn = J (- t + n~i) =

n~~ (i + n';-:il) = i + 0 = i.

7. No. In polar form we have Jim (eift = lim ei"f. These points oscillate

n-oo n-oo

a.round the eight roots of unity as example ( 4-2) indicated.

00

9. Since L:: Zn converges, lim Sn = S , where S is a complex number. But then

n=l n - oo

lim Sn-I = S, SO lim Zn= lim (Sn - Sn- d = Um Sn - Jim Sn- I = 0.

n-oo n-oo n - oo n~oo n-oo