1 2πi Z ∂D f (w)dw w − z + 1 2πi Z D ∂f ∂w dw ∧ dw w − z

1. Cauchy-integral Formula

Theorem 1.1. Let D be a disk in C and f be a smooth function on D. For each z ∈ D, f (z) = 1

2πi Z

∂D

f (w)dw w − z + 1

2πi Z

D

∂f

∂w

dw ∧ dw w − z .

Proof. The proof is based on the Stokes theorem. 

2. ∂-Poincare Lemma

Let D be unit disk {z ∈ C : |z| < 1} in C and g be a smooth function on D, i.e. g can be extended to an open set U containing D such that g is smooth on U. Define

f (z) = Z

D

g(w)

w − zdw ∧ dw.

Theorem 2.1. The function f is smooth on D and

∂f

∂z = g.

Let z0 ∈ D. Choose  > 0 so that D(z0, ) ⊂ D. Decompose g = g1 + g2 where g1 is smooth vanishing outside D(z₀, ) and g₂ is smooth vanishing inside D(z₀, /2). Define

f₂(z) = 1 2πi

Z

D

g₂(w)

w − zdw ∧ dw

= 1 2πi

Z

D\D(z0,₂^)

g2(w)

w − zdw ∧ dw + 1 2πi

Z

D(z0,^₂)

g2(w)

w − zdw ∧ dw

= 1 2πi

Z

D\D(z0,₂^)

g₂(w)

w − zdw ∧ dw.

Now if z ∈ D(z0, ), w − z 6= 0 for all w ∈ D \ D(z0,₂^). Hence

∂

∂z 1 w − z = 0 for all w ∈ D \ D(z0,₂^). Hence

∂f2

∂z (z) = 1 2πi

Z

D\D(z0,^₂)

∂

∂z g2(w)

w − zdw ∧ dw = 0 for all z ∈ D(z₀,₂^). On the other hand, define

f1(z) = 1 2πi

Z

D

g1(w)

w − zdw ∧ dw.

Then f = f1+ f2 on D. We see that f1(z) = 1

2πi Z

D

g1(w)dw ∧ dw w − z .

Since g1 has support in D(z0, ), we can rewrite the above integral in the form f1(z) = 1

2πi Z

C

g1(w)dw ∧ dw w − z . Let w = z + re^iθ for 0 ≤ r < ∞ and θ ∈ [0, 2π]. Then

f1(z) = −1 π

Z 2π 0

Z ∞ 0

g1(z + re^iθ)e^−iθdr ∧ dθ.

1

2

Since g1 is smooth with compact support, using the Lebesgue dominated convergence the- orem, we know that f₁ is smooth on C. Then

∂f₁

∂z = −1 π

Z

C

∂g₁

∂z (z + re^iθ)e^−iθdr ∧ dθ.

We see that on D,

∂f1

∂z = 1 2πi

Z

C

∂g1(w)

∂w

dw ∧ dw w − z . Using the Cauchy integral formula, we know that on D, one has

g1(z) = 1 2πi

Z

∂D

g1(w)

w − zdw + 1 2πi

Z

D

∂g1(w)

∂w

dw ∧ dw w − z . Since g₁ vanishes on ∂D, the first integral vanishes. Hence

g1(z) = 1 2πi

Z

D

∂g1(w)

∂w

dw ∧ dw w − z = ∂f1

∂z (z) on D. Since g2(z₀) = 0, using the previous results, we find

∂f

∂z(z₀) = ∂f₁

∂z (z₀) +∂f₂

∂z (z₀)

= g₁(z₀)

= g(z₀).

Since z₀ is arbitrary in D, the differential equation holds on D.