MATH 300, ANSWER KEY TO PROBLEM SET 7

5.1.8. Let Cn =∑nj=0 cj and Dn =

∑nj=0 dj . We know that limn→∞ Cn = S and

limn→∞Dn = T .

(a) Note that the partial sum

n∑j=0

cj = Cn .

Since the conjugation function f(z) = z is continuous (check!), we have

limn→∞

Cn = limn→∞

f(Cn) = f(S) = S ,

and part (a) follows.(b) Same argument with f(z) = λz.(c) Here the partial sum

∑nj=0(cj + dj) = Cn +Dn, and

limn→∞

(Cn +Dn) = limn→∞Cn + limn→∞

Dn = S + T,

as desired.

5.1.20. ∣∣∣∣∣∣n∑j=0

zj − 1

1− z

∣∣∣∣∣∣ =

∣∣∣∣ zn+1

1− z

∣∣∣∣Note that limx→1−

xn+1

1−x = ∞. Thus we can choose a real number x in (0, 1) close

enough to 1 so that xn+1

1−x > 1. Consequently, there is no N so that n > N implies∣∣∣∣ zn+1

1− z

∣∣∣∣ < 1

for all z in the open unit disc. We conclude that our series does not convergeuniformly in this disc.

5.2.4. f(z) = eαLog(1+z) is analytic in |z| < 1. Moreover, f(0) = 1,

f (j)(z) = α(α− 1) · · · (α− j + 1)e(α−j)Log(1+z)

f (j)(0) = α(α− 1) · · · (α− j + 1), and the claim follows.

5.2.14. Suppose the Taylor series for f(z) in D is

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + . . .

Then ak =f (k)(z0)

k!= 0 for k = 0, 1, 2, . . .. Thus f(z) is identically zero in D.

5.2.15. Arguing as in the previous exercise (with z0 = 0), we see that

f(z) = a0 + a1z + a2z2 + a3z

3 + . . .

and

ak =f (k)(z0)

j!= 0

whenever k is odd. Thus

f(z) = a0 + a2z2 + a4z

4 + a6z6 + . . .

and consequently, f(−z) = f(z).

5.2.16. Note a typo in the statement of the problem: pn(z) is meant to be p(z).Note also that

p(z) = c0 + c1(z − 1) + · · ·+ cn(z − 1)n

is a Taylor series for p(z) at z = 1. Thus cj =p(j)(1)

j!. In particular,

c0 =p(1)

0!=

n∑j=0

aj

c1 =p′(1)

1!= a1 + 2a2 + . . . nan ,

c2 =p′′(1)

1!= 2 · 1a2 + 3 · 2a3 + · · ·+ n(n− 1)an ,

etc. In general,

p(j)(z) = j · (j − 1) · . . . · 2 · 1 + (j + 1) · j · . . . · 2z+ · · ·+n(n− 1) . . . (n− j + 1)zn−j

and thus

cj =p(j)(1)

j!=

1

j!

(j ·(j−1) · . . . ·2 ·1+(j+1) ·j · . . . ·2+ · · ·+n(n−1) . . . (n−j+1)

).

5.2.18. (a)∣∣∣∣∣ez −n∑k=0

zk

k!

∣∣∣∣∣ =

∣∣∣∣∣∞∑

k=n+1

zk

k!

∣∣∣∣∣ ≤∞∑

k=n+1

∣∣∣∣zkk!

∣∣∣∣ ≤ ∞∑k=n+1

1

k!=

∞∑j=0

1

(n+ j + 1)!=

∞∑j=0

1

(n+ j + 1)(n+ j) · · · (n+ 2)(n+ 1)!≤∞∑j=0

1

(n+ 1)!

(1

n+ 2

)j=

1

(n+ 1)!

∞∑j=0

(1

n+ 2

)j.

Here j = k−n− 1. Using the formula for∑∞j=0 c

j =1

1− cwith c =

1

n+ 2, we see

that∞∑j=0

(1

n+ 2

)j=

(1

1− 1n+2

)=n+ 2

n+ 1= 1 +

1

n+ 1,

and part (a) follows.

(b) ∣∣∣∣∣sin(z)−n∑k=0

(−1)kz2k+1

(2k + 1)!

∣∣∣∣∣ =

∣∣∣∣∣∞∑

k=n+1

(−1)kz2k+1

(2k + 1)!

∣∣∣∣∣ =

∞∑k=n+1

∣∣∣∣(−1)kz2(k+1)

(2k + 1)!

∣∣∣∣ =

∞∑k=n+1

1

(2k + 1)!.

Setting j = k − n− 1, we obtain∞∑

k=n+1

1

(2k + 1)!=

∞∑j=0

1

(2(j + n+ 1) + 1)!=

1

(2n+ 3)!

(1+

1

(2n+ 4)(2n+ 5)+

1

(2n+ 4)(2n+ 5)(2n+ 6)(2n+ 7)+. . .

)≤

1

(2n+ 3)!(1 + c+ c2 + c3 + . . . )

where c =1

(2n+ 4)(2n+ 5). Once again, using the formula

∑∞j=0 c

j =1

1− cfor

the sum of the geometric series, we see that

1 + c+ c2 + c3 + · · · = 1

1− 1(2n+4)(2n+5)

=(2n+ 4)(2n+ 5)

(2n+ 4)(2n+ 5)− 1=

4n2 + 18n+ 20

4n2 + 18n+ 19,

as desired.

5.3.2.

limj→∞

∣∣∣∣aj+1(z − z0)j+1

aj(z − z0)j

∣∣∣∣ = limj→∞

∣∣∣∣aj+1

aj

∣∣∣∣ |z = z0| = L|z − z0|.

By the ratio test (Theorem 2 on page 237), the series is convergent if L|z− z0| < 1(i.e. if|z − z0| < 1/L) and diverges if L|z − z0| > 1 (i.e. if|z − z0| > 1/L). Thisshows that R = 1/L is the radius of convergence.

5.3.4. No because if the series converges at 2 + 3i then the radius of convergenceR is ≥ |2 + 3i− 0| =

√13. Then the distance from the origin to 3− i is

√10, 3− i

would then lie within the disk of convergence, i.e., the series would converge at 3−i.

5.3.6. (a) f(0) = 1 agrees with the series at z = 0. For z 6= 0 divide the Maclaurinexpansion

sin(z) = z − z3

3!+z5

5!− . . .

by z to obtain the desired Taylor expansion for f(z).(b) The series for f(z) in part (a) converges for every complex number z. This

follows from the fact that the Maclaurin series for sin(z) converges for every z (or,alternatively, from the ratio test). By Theorem 10, f(z) is analytic in the entireplane.

(c) f (3)(0) = 3!a3, where a3 is the coefficient of z3 in the power series for f(z)in part (a). Since this series has no z3 term, a3 = 0, and thus f (3)(0) = 0 as well.

Similarly, f (4)(0) = 4!a4 = 4!1

5!=

1

5.

5.3.12. Substituting f(z) =∑∞k=0 akz

k into the equation, we see that a0 = f(0) =1 and kak = ak−1 for all k ≥ 1. Thus a1 = 1, a2 = 1/2, a3 = 1/3a2 = 1/3!,etc. Continuing this way we see that ak = 1/k! for every k ≥ 0 and thus f(z) =∑∞k=0

1k!z

k = ez.(I worked out the details of this calculation in class.)