(1)Differentiability and Power Series Definition Let S ⊆ C be a subset of C, and let f : S → C be a complex-valued function defined on S

Differentiability and Power Series

Definition Let S ⊆ C be a subset of C, and let f : S → C be a complex-valued function defined on S. We say that f is complex-differentiable (or differentiable) at an interior point z of S if

h→0lim

f (z + h) − f (z) h

exists in C. In that case, the limit is denoted f⁰(z).

Remark If f is differentiable at z = x + yi then, by identifying z = x + iy with (x, y), we have f⁰(z) = lim

h∈C, h→0

f (z + h) − f (z) z + h − z

= lim

h∈R, h→0

f ((x + h) + yi) − f (x + yi) (x + h) + yi − (x + yi)

= lim

h∈R, h→0

f (x + h, y) − f (x, y) h

= f_x(z)

= lim

h∈R, h→0

f (x + (y + h)i) − f (x + yi) x + (y + h)i − (x + yi)

= lim

h∈R, h→0

f (x, y + h) − f (x, y) ih

= −ify(z).

Proposition Let U ⊂ C and let f, g : U → C be differentiable at z ∈ U. Then (a) f + g is differentiable at z with (f + g)⁰(z) = f⁰(z) + g⁰(z).

(b) f g is differentiable at z with (f g)⁰(z) = f⁰(z)g(z) + f (z)g⁰(z).

(c) f

g is differentiable at z with  f g

0

(z) = f⁰(z)g(z) − f (z)g⁰(z)

g²(z) provided that g(z) 6= 0.

Definition Let S ⊆ C be a subset of C, and let f : S → C be a complex-valued function defined on S. We say thatf is analytic (or holomorphic) at zif f is differentiable in an open neighborhood of z. Hence, we say that f is analytic on a set S ⊆ U if f is differentiable at all points of some open set containing S. If additionally S = C, the analytic function f : C → C is called an entire function.

Examples

(a) Let P (z) = a₀+a₁z+· · ·+a_nzⁿbe a polynomial of degree n ≥ 0. Show that P is differentiable at all points z and P⁰(z) = a₁+ 2a₂z + · · · + na_nzⁿ⁻¹.

(b) Show that the function f (z) = ¯z is not complex-differentiable.

Proposition Let u and v be real-valued functions and let f (x, y) = u(x, y) + iv(x, y) :=

(u(x, y), v(x, y)).

(a) If f = u + vi is differentiable at z = x + yi, then the Cauchy-Riemann equations hold at z f_y(z) = if_x(z)

⇐⇒ u_y + iv_y = i(u_x+ iv_x) = −v_x+ iu_x

⇐⇒ u_x = v_y; u_y = −v_x. (b) If f is differentiable at z = x + iy = (x, y), then

|f⁰(z)| = |fx(z)| = |fy(z)|,

=⇒ |f⁰(z)| =q

u²_x(x, y) + u²_y(x, y) =q

v_x²(x, y) + v_y²(x, y), and det Df (x, y)=

u_x u_y v_x v_y    

=    

u_x u_y

−u_y u_x    

= u²_x(x, y) + u²_y(x, y)= |f⁰(z)|², by the Cauchy-Riemann equations.

(c) Consider the Jacobian of f : R² → R²

J_f :=u_x u_y v_x v_y



and the rotation matrices

cos θ − sin θ sin θ cos θ

 .

Since multiplication by i is equivalent to multiplication by the rotation matrix (by θ = π/2) I =0 −1

1 0

 , and since

J_fI =u_y −u_x v_y −v_x



and IJ_f =−v_x −v_y u_x u_y

 ,

theJacobian matrix commutes with the rotation by π/2 is equivalent to the Cauchy Riemann equations, i.e.

J_fI = IJ_f ⇐⇒ u_x = v_y; u_y = −v_x.

Proposition A polynomial P (x, y) is analytic on C if and only if Py = iP_x on C.

Proof The necessity of the condition follows from the preceding Proposition part (a).

To show that it is also sufficient, note that if

P_y = iP_x,

the condition must be met separately by the terms of any fixed degree. Suppose that P has n-th degree terms of the form

Q(x, y) = C₀xⁿ+ C₁xⁿ⁻¹y + C₂xⁿ⁻²y²+ · · · + C_nyⁿ. Since Q_y = iQ_x,

C₁xⁿ⁻¹+ 2C₂xⁿ⁻²y + · · · + nC_nyⁿ⁻¹

= i[nC₀xⁿ⁻¹+ (n − 1)C₁xⁿ⁻²y + · · · + C_n−1yⁿ⁻¹].

Comparing coefficients,

C₁ = i n C₀ = in 1

 C₀ C₂ = i(n − 1)

2 C₁ = i² n(n − 1)

2 C₀ = i²n 2

 C₀, and in general

C_k = i^kn k

 C₀ so that

Q(x, y) =

n

X

k=0

C_kx^n−ky^k= C₀

n

X

k=0

n k



x^n−k(iy)^k= C₀(x + yi)ⁿ= C₀zⁿ which is analytic on C. Thus P is analytic on C.

Examples

(a) A non-constant analytic polynomial cannot be real-valued, for then both P_x and P_y would be real and the Cauchy-Riemann equations would not be satisfied.

(b) Using the Cauchy-Riemann equations, one can verify that x²− y²+ 2ixy is analytic every- where while x²+ y²− 2ixy is not.

(c) Let

f (x, y) =







xy(x + iy)

x²+ y² if z 6= 0

0 if z = 0

Show that ∂f

∂x(0, 0) = 0 = ∂f

∂y(0, 0) and thus the Cauchy-Riemann equations are satisfied at z = 0, but f is not complex-differentiable. Note that ∂f

∂x and ∂f

∂y are not continuous at (0, 0).

Theorem Let f_n : [a, b] → C be a sequence of continuously differentiable functions defined on [a, b]. If fn(x0) converges for some x0 ∈ [a, b] and if f_n⁰ : [a, b] → C converges uniformly on [a, b], then f_n converges uniformly on [a, b] to a function f.

Proof For any x ∈ [a, b] and for any m, n ∈ N, there exists a y depending on m, n such that [f_m(x) − f_n(x)] − [f_m(x₀) − f_n(x₀)] = [f_m⁰ (y) − f_n⁰(y)](x − x₀)

=⇒ sup

x∈[a,b]

|f_m(x) − f_n(x)| ≤ |f_m(x₀) − f_n(x₀)| + (b − a) sup

y∈[a,b]

|f_m⁰ (y) − f_n⁰(y)|.

Hence f_n converges uniformly on [a, b] to a function f and f is continuous on [a, b].

Remark In the proof, we have applied the following Mean Value Theorem to the real and imaginary parts of f_m(x) − f_n(x).

Mean Value Theorem If f, g are continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that:

[f (b) − f (a)]g⁰(c) = [g(b) − g(a)]f⁰(c).

Note if g(x) = x, this gives the Mean Value Theorem.

Proof For x ∈ [a, b], let

h(x) = [f (b) − f (a)]g(x) − [g(b) − g(a)]f (x).

Since h(a) = h(b), there is a c ∈ (a, b) such that h⁰(c) = 0 and the Mean Value Theorem follows from Rolle’s Theorem.

Theorem (Term-by-Term Differentiation) For each k ∈ N, let fk be a real-valued function on I = [a, b] which has a derivative f_k⁰ on I. Suppose that

(i)

∞

X

k=0

f_k converges at x₀ ∈ I,

(ii)

∞

X

k=0

f_k⁰ converges uniformly on I,

there exists a real-valued function f on I such that (a)

∞

X

k=0

f_k converges uniformly on I to f,

(b) f is differentiable on I and f⁰ =

∞

X

k=0

f_k⁰ on I.

Proof

(a) Note that for each x ∈ I and for any m, n ∈ N, by the Mean Value Theorem, there exists y lying between x and x0 such that

sm(x) − sn(x) = sm(x0) − sn(x0) + (x − x0)(s⁰_m(y) − s⁰_n(y)).

Since

∞

X

k=0

f_k converges at x₀ ∈ I and

∞

X

k=0

f_k⁰ converges uniformly on I, i.e.

s_n(x₀) =

n

X

k=0

f_k(x₀) converges and s⁰_n=

n

X

k=0

f_k⁰ converges uniformly on I,

s_n =

n

X

k=0

f_k converges uniformly on I to f =

∞

X

k=0

f_k.

(b) Suppose that

∞

X

k=0

f_k⁰ converges uniformly on I to g.For each x 6= c ∈ I and for any m, n ∈ N, by the Mean Value Theorem, there exists y lying between x and c such that

s_m(x) − s_n(x) = s_m(c) − s_n(c) + (x − c)(s⁰_m(y) − s⁰_n(y))

=⇒

sm(x) − sm(c)

x − c −sn(x) − sn(c) x − c

≤ ks⁰_m− s⁰_nk_I = sup

y∈I

|s⁰_m(y) − s⁰_n(y)|.

For each ε > 0, since s⁰_n converges uniformly on I, there exists a M (ε) such that if m, n ≥ M (ε) then ks⁰_m− s⁰_nk_I < ε. Taking the limit with respect to m, we get

f (x) − f (c)

x − c − s_n(x) − s_n(c) x − c

≤ ε when n ≥ M (ε).

Next, since g(c) = lim

n→∞s⁰_n(c), there exists an N (ε) such that if n ≥ N (ε), then

|s⁰_n(c) − g(c)| < ε when n ≥ N (ε).

Let L = max{M (ε), N (ε)}. Since s_L is differentiable at c ∈ I, there exists a δ_L(ε) > 0 such that if 0 < |x − c| < δ_L(ε), then

s_L(x) − s_L(c)

x − c − s⁰_L(c)    

< ε.

Therefore, it follows that if 0 < |x − c| < δ_L(ε), then 

f (x) − f (c)

x − c − g(c)    

< 3ε.

This shows that f⁰(c) exists and equals g(c).

Definition A power series in z is an infinite series of the form

∞

X

n=0

c_kz^k.

To study the convergence of a power series, we recall the notion of thelim sup = limof a positive real-valued sequence. That is,

n→∞lima_n= lim

n→∞

 sup

k≥n

a_k

 . Since sup

k≥n

a_k= sup{a_k | k ≥ n} is a non-increasing function of n, the limit always exists or equals

∞.

Also note that if lim

n→∞an = L, then

(i) for each N and for each ε > 0, there exists some k > N such that a_k > L − ε.

(ii) for each ε > 0, there is some N such that a_k≤ L + ε for all k > N.

(iii) lim

n→∞c a_n= c L for any nonnegative constant c.

Theorem Suppose lim

k→∞|c_k|^1/k = L.

1. If L = 0, then

∞

X

k=0

c_kz^k converges (absolutely) for all z ∈ C.

2. If L = ∞, then

∞

X

k=0

c_kz^k converges for z = 0 only.

3. If 0 < L < ∞, set R = 1/L. Then

∞

X

k=0

c_kz^k converges for |z| < R and diverges for |z| > R.

(R is calledthe radius of convergence of the power series.) Proof

1. L = 0 Since lim

k→∞|ck|^1/k = 0, lim

k→∞|ck|^1/k|z| = 0 for all z ∈ C. Thus for each z ∈ C, there exists N ∈ N such that if k > N, then

|c_k|^1/k|z| ≤ 1

2 =⇒ |c_kz^k| ≤ 1 2^k =⇒

∞

X

k=0

c_kz^k converges (absolutely) for all z ∈ C.

2. L = ∞

For any z 6= 0, there exist infinitely many values of k such that

|c_k|^1/k ≥ 1

|z| =⇒ |c_kz^k| ≥ 1 =⇒ lim sup

k→∞

|c_kz^k| 6= 0 =⇒

∞

X

k=0

c_kz^k diverges.

3. 0 < L < ∞, R = 1/L.

Assume first that |z| < R and set |z| = R(1 − 2δ). Then since lim

k→∞|c_k|^1/k|z| = |z|

R = 1 − 2δ, there exists N ∈ N such that if k > N, then

|c_k|^1/k|z| < 1 − δ < 1 =⇒

∞

X

k=0

c_kz^k is absolutely convergent.

On the order hand, if |z| > R, since lim

k→∞|c_k|^1/k|z| = |z|

R > 1, so that for infinitely many values of k, c_kz^k has absolute value greater than 1 and

∞

X

k=0

c_kz^k diverges.

Examples 1. Since lim

k→∞k^1/k = 1,

∞

X

k=1

k z^k−1 the power series converges for |z| < 1 and diverges for |z| > 1.

Also the series diverges for |z| = 1 since lim

k→∞|k z^k−1| = ∞.

2. The power series

∞

X

k=1

(z^k/k²) has radius of convergence equal to 1. In this case, however, the series converges for all points z on the unit circle since

z^k k²    

= 1

k² for |z| = 1.

3.

∞

X

k=1

(z^k/k) has radius of convergence equal to 1. In this case, the series converges for all points on the unit circle except z = 1. [Hint: Let z = cis θ and analyze the real and imaginary parts of the series separately.]

4.

∞

X

k=0

(z^k/k!) converges for all z ∈ C since

k→∞lim 1

(k!)^1/k = 0.

5.

∞

X

k=0

[1 + (−1)^k]^kz^k has radius of convergence ¹₂ since lim sup

k→∞

[1 + (−1)^k] = 2.

Example Show that

∞

X

k=0

z^k= 1

1 − z for |z| < 1.

Theorem Suppose f (z) =

∞

X

k=0

c_kz^k converges for |z| < R. Then f⁰(z) exists and equals to

∞

X

k=0

kc_kz^k−1 throughout |z| < R.

Proof For each 0 < δ < R, since

∞

X

k=0

|c_k||R − δ|^k converges, f (z) =

∞

X

k=0

c_kz^k converges uniformly on |z| ≤ R − δ.

Since

∞

X

k=1

kc_kz^k−1 has the same radius of convergence as that of

∞

X

k=0

c_kz^k,

∞

X

k=1

kc_kz^k−1 converges uniformly on |z| ≤ R − δ.

By the Term-by-Term Differentiation Theorem, f⁰(z) exists and equals to

∞

X

k=0

kc_kz^k−1throughout

|z| ≤ R − δ. Since 0 < δ < R is arbitrary, f⁰(z) exists and equals to

∞

X

k=0

kckz^k−1 throughout

|z| < R.

Example Show that

∞

X

k=0

(k + 1)z^k = 1

(1 − z)² for |z| < 1.

Corollary Inside its radius of convergence, a power series is infinitely differentiable, with the expected derivatives as convergent power series.

Corollary If R > 0, then c_k = f^(k)(0) k! .

Lemma If f (z) =

∞

X

k=0

c_kz^k is a convergent power series and f (z_n) = 0 for a sequence {z_n}^∞_n=1 with lim

n→∞z_n = 0 and z_n 6= 0 for all n ∈ N, then ck= 0 for all k.

Proposition If f (z) =

∞

X

k=0

a_kz^k, g(z) =

∞

X

k=0

b_kz^k and these agree on some set accumulating to 0, then a_k= b_k for all k and f (z) = g(z).

Proof Consider

∞

X

k=0

(a_k − b_k)z^k and apply the Lemma. Note that if lim

k→∞|a_k|^1/k ≤ L and

k→∞lim |bk|^1/k ≤ L, then lim

k→∞|ak− bk|^1/k ≤ lim

k→∞2^1/kmax{|ak|, |bk|}^1/k ≤ L.