51.4. EXERCISES 1627

2. Abel’s theorem says that if ∑∞n=0 an (z−a)n has radius of convergence equal to 1 and

if A = ∑∞n=0 an, then limr→1−∑

∞n=0 anrn = A. Hint: Show

∞

∑k=0

akrk =∞

∑k=0

Ak

(rk− rk+1

)where Ak denotes the kth partial sum of ∑a j. Thus

∞

∑k=0

akrk =∞

∑k=m+1

Ak

(rk− rk+1

)+

m

∑k=0

Ak

(rk− rk+1

),

where |Ak−A|< ε for all k≥m. In the first sum, write Ak = A+ek and use Problem1. Use this theorem to verify that arctan(1) = ∑

∞k=0 (−1)k 1

2k+1 .

3. Find the integrals using the Cauchy integral formula.

(a)∫

γsinzz−i dz where γ (t) = 2eit : t ∈ [0,2π] .

(b)∫

γ1

z−a dz where γ (t) = a+ reit : t ∈ [0,2π]

(c)∫

γcoszz2 dz where γ (t) = eit : t ∈ [0,2π]

(d)∫

γ

log(z)zn dz where γ (t) = 1 + 1

2 eit : t ∈ [0,2π] and n = 0,1,2. In this prob-lem, log(z) ≡ ln |z|+ iarg(z) where arg(z) ∈ (−π,π) and z = |z|eiarg(z). Thuselog(z) = z and log(z)′ = 1

z .

4. Let γ (t) = 4eit : t ∈ [0,2π] and find∫

γz2+4

z(z2+1)dz.

5. Suppose f (z) = ∑∞n=0 anzn for all |z|< R. Show that then

12π

∫ 2π

0

∣∣∣ f (reiθ)∣∣∣2 dθ =

∞

∑n=0|an|2 r2n

for all r ∈ [0,R). Hint: Let

fn (z)≡n

∑k=0

akzk,

show1

2π

∫ 2π

0

∣∣∣ fn

(reiθ)∣∣∣2 dθ =

n

∑k=0|ak|2 r2k

and then take limits as n→ ∞ using uniform convergence.

6. The Cauchy integral formula, marvelous as it is, can actually be improved upon. TheCauchy integral formula involves representing f by the values of f on the boundaryof the disk, B(a,r) . It is possible to represent f by using only the values of Re f onthe boundary. This leads to the Schwarz formula . Supply the details in the followingoutline.