The Cauchy-Riemann Equations
Theorem 5: Let f (z) = u(x, y) + iv(x, y) be defined
on an open subset U of the complex plane C.
(a) If f (z) is analytic in U , then u and v are differentiable and satisfy the Cauchy-Riemann equations:
∂u
∂v
∂u
∂v
(1)
=
and
= −
in U.
∂x
∂y
∂y
∂x
(b) “Conversely,” if the partial derivatives of u and
v exist and are continuous in U , and if the C-R
equations (1) hold in U , then f (z) is analytic in U.
Remarks: 1) If f (z) is analytic in U, then
∂u
∂v
0
f (z) =
(x, y) + i (x, y).
(2)
∂x
∂x
2) Note that part (b) is not quite the converse of
part (a). To be a true converse, we would need to
add to part (a) the assertion:
f (z) analytic ⇒ the partials of u, v are continuous.
This is true, but will only be established later (in
§4.5). It follows from the following remarkable fact:
f (z) analytic ⇒ f 0(z) analytic.
2
Proof Sketch of Theorem 5b: Fix z0 = x0 + iy0 and write
∆z = ∆x + i∆y.
Step 1: Use the mean value theorem and the continuity of
the partials to obtain the approximations
∂u
+ ε1 ,
u(x0 + ∆x, y0 + ∆y) − u(x0, y0 + ∆y) = ∆x
∂x
∂u
u(x0, y0 + ∆y) − u(x0, y0) = ∆y
+ ε2 ,
∂y
∂v
v(x0 + ∆x, y0 + ∆y) − v(x0, y0 + ∆y) = ∆x
+ ε3 ,
∂x
∂v
+ ε4
v(x0, y0 + ∆y) − v(x0, y0) = ∆y
∂y
∂u
where ∂u
=
∂x
∂x (x0 , y0 ), etc. and εk = εk (∆x, ∆y) → 0 as
∆x → 0 and ∆y → 0, for k = 1, . . . , 4.
Step 2: Substitute these in the differential quotient
q(∆z) :=
=
f (z0 + ∆z) − f (z0)
∆z
h
i
∂u
∂v
∂v
∆x ∂x + ε1 + i ∂x
+ iε3 + ∆y ∂u
+
ε
+
i
2
∂y
∂y + iε4
∆x + i∆y
∂u
∂u
∂v
∂v
λ
(CR) ∆x ∂x + i ∂x + i∆y ∂x + i ∂x
=
+
∆x + i∆y
∆x + i∆y
∂u
∂v
λ
∂u
∂v
check!
=
+i +
→
+i
as ∆z → 0,
∂x
∂x ∆z
∂x
∂x
where λ := ∆x(ε1 + iε3) + ∆y(ε2 + iε4).
3
Remark: The hypothesis that u and v have continuous derivatives is not necessary for the validity of
Theorem 5(b). This was discovered by Goursat (1900).
However, it is much more difficult to prove this refinement.
Theorem 6: If f (z) is analytic on a domain D and if
f 0(z) = 0, for all z ∈ D, then f is constant in D.
Corollary: If f (z)and g(z) are two analytic functions
on a domain D, then
f 0(z) = g 0(z) on D ⇔ f (z) = g(z) + c, with c ∈ C.