1616 CHAPTER 51. FUNDAMENTALS OF COMPLEX ANALYSIS

Then f is analytic if and only if u,v are C1 (Ω) and

∂u∂x

=∂v∂y

,∂u∂y

=−∂v∂x

.

Furthermore,

f ′ (z) =∂u∂x

(x,y)+ i∂v∂x

(x,y) .

Proof: Suppose f is analytic first. Then letting t ∈ R,

f ′ (z) = limt→0

f (z+ t)− f (z)t

=

limt→0

(u(x+ t,y)+ iv(x+ t,y)

t− u(x,y)+ iv(x,y)

t

)=

∂u(x,y)∂x

+ i∂v(x,y)

∂x.

But also

f ′ (z) = limt→0

f (z+ it)− f (z)it

=

limt→0

(u(x,y+ t)+ iv(x,y+ t)

it− u(x,y)+ iv(x,y)

it

)1i

(∂u(x,y)

∂y+ i

∂v(x,y)∂y

)=

∂v(x,y)∂y

− i∂u(x,y)

∂y.

This verifies the Cauchy Riemann equations. We are assuming that z→ f ′ (z) is continuous.Therefore, the partial derivatives of u and v are also continuous. To see this, note that fromthe formulas for f ′ (z) given above, and letting z1 = x1 + iy1∣∣∣∣∂v(x,y)

∂y− ∂v(x1,y1)

∂y

∣∣∣∣≤ ∣∣ f ′ (z)− f ′ (z1)∣∣ ,

showing that (x,y)→ ∂v(x,y)∂y is continuous since (x1,y1)→ (x,y) if and only if z1→ z. The

other cases are similar.Now suppose the Cauchy Riemann equations hold and the functions, u and v are

C1 (Ω) . Then letting h = h1 + ih2,

f (z+h)− f (z) = u(x+h1,y+h2)

+iv(x+h1,y+h2)− (u(x,y)+ iv(x,y))

We know u and v are both differentiable and so

f (z+h)− f (z) =∂u∂x

(x,y)h1 +∂u∂y

(x,y)h2+