1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

3c. ~-

3e. l·

3g. ~-

3i. 1.

ANSWERS 595

5.Showthat limsuplc;,I • = l. ( limsuplenl;; ')2.

n-oo n - oo

00

7. The theorem establishes f(k) (z) = L; n (n - 1) · · · (n - k + 1) Cn (z - at-k

n=k

when k = 1. Assume the theorem is true for some k > 1, and set g (z) =

00

L: bn (z -at, where bn = (n + k) (n + k - 1) · · · (n + 1) Cn+k· In other

n=O

words, g (z) = f(k) (z) (confirm this). Applying the case when k = 1 to the

00

function g gives g' (z) = f (k + I} (z) = L: nbn (z - ar-l =

n.=l

00

L: n (b,,) (n + k) (n + k - 1)-· · (n + 1) Cn+k(z -ayn-^1 =

n=l

f n(n-1)··-(n-k+l)(n-k)en(z-at-(k+i) (confirm this also),

n=k+ l

which is what we needed to establish.

9 a. Since s" - t" = (sn- I + s"-^2 t + s"-^3 t^2 + · · · + st"-^2 + t"-^1 ) (s -t) (ver-

ify!), the conclusion follows from division and t he triangle inequality.

11. The series converges for all values of z by the ratio test.

Section 5.1. E lementary Functio ns : page 160

00 00

1. Recall that L: enz" is compact notation for Co+ L: c,.z", and that O! = 1.
   n=O n=l
   00 00
   Then, by definition, exp (0) = L: fhO" = ih + L: ;hon = 1.
   n =O n=I


2. Let n be an integer, and set z = i2mr. Then ei^2 "" = cos (2mr) +i sin (2mr) =
   
   
   1. Conversely, suppose e• = ex+iy = l. Then e"' eiy = e" (cosy + i sin y) =
      1 + Oi. This implies siny = 0. Since e" is always positive and e" cosy= 1,
      this means that y = 2mr for some integer n. This also forces x = 0 , so
      z = x + iy = 0 + i2nn. This establishes Property (5-3). Property (5-
   
   
   
   
   
   
   4. comes from observing that e•^1 = e•• iff e•^1 - z^2 = 1, and app ealing to
      Property (5-3).
   
   
   
   


3. T his follows immediately from Property (5-4).

7a. Following the method of problem 3, e' = -4 iff z = x + iy with

y = (2n+l)n where n is an integer, and e" = 4. T hus, x = ln4, and

z = ln4+i(2n+1) n , where n is an integer.