Key to Demonstration 6

Key to Demonstration 6

I.Solution First let's show that e^iz = cosz+isinz. We know that

e^iz = X∞

n=0

(iz)ⁿ

n! = X

n: even

(iz)ⁿ

n! +X

n: odd

(iz)ⁿ n! =

X∞

n=0

(iz)^2k (2k)!+

X∞

n=0

(iz)^2k+1 (2k+ 1)! =

X∞

n=0

i^2kz^2k (2k)!+

X∞

n=0

i^2k+1z^2k+1 (2k+ 1)! . Since i^2k = (i²)^k = (−1)^k and i^2k+1 =i·i^2k=i(−1)^k, then

e^iz = X∞

n=0

(−1)^k z^2k (2k)!+i

X∞

n=0

(−1)^k z^2k+1

(2k+ 1)! = cosz+isinz.

Then, we show that e^−iz = cosz−isinz. We know that e^−iz =

X∞

n=0

(−iz)ⁿ

n! = X

n: even

(−iz)ⁿ

n! + X

n: odd

(−iz)ⁿ n!

= X∞

n=0

(−iz)^2k (2k)! +

X∞

n=0

(−iz)^2k+1 (2k+ 1)!

= X∞

n=0

(−1)^2ki^2kz^2k (2k)! +

X∞

n=0

(−1)^2k+1i^2k+1z^2k+1 (2k+ 1)! . Since i^2k = (i²)^k = (−1)^k and i^2k+1 =i·i^2k=i(−1)^k, then

e^iz = X∞

n=0

(−1)^k z^2k (2k)! −i

X∞

n=0

(−1)^k z^2k+1

(2k+ 1)! = cosz−isinz.

From the above two results, we can immediately get cosz = 1

2(e^iz+e^−iz),and sinz = 1

2i(e^iz −e^−iz).

II.Solution

d

dzcosz = d dz

h1

2(e^iz+e^−iz) i

= 1

2i(e^iz−e^−iz) =−sinz.

III.Solution We have known that sin(z+ς) = sinzcosς+ coszsinς. On one hand, d

dς sin(z+ς) = cos(z+ς);

one the other hand, d

dς(sinzcosς+ coszsinς) = sinz(−sinς) + coszcosς = coszcosς−sinzsinς.

IV.Solution Let

L= lim sup pⁿ

|a_n|= inf

k sup

n≥k

pn

|a_n|= lim

k→∞sup{p^k

|a_k|, ^k+1p

|a_k+1|, . . .}.

1

2

Given ε >0, there is an N such that pⁿ

|a_n|< L+ε for all n ≥N. Suppose that

|z−z₀|<(1−ε)/(L+ε), then |a_n||z −z₀|ⁿ < (1−ε)ⁿ for all n ≥ N, so

X∞

0

a_n(z−z₀)ⁿ is absolutely convergent.

Hence,R ≥(1−ε)/(L+ε)for eachε >0. Therefore, R≥1/L. Conversely, suppose that

|z−z₀|>1/(L−ε)for eachε >0. There are innitely many indicesnwith pⁿ

|a_n|> L−ε. Hence, |a_n||z−z₀|ⁿ >1 for innitely many indicesn, so

X∞

0

a_n(z−z₀)ⁿ diverges. Hence, R≤1/(L−ε) for each ε >0, so R≤1/L. Hence,R = 1/L.