Lecture 13: The Cauchy-Riemann Equations - Furman University

Lecture 13:
The Cauchy-Riemann Equations
Dan Sloughter
Furman University
Mathematics 39
March 29, 2004
13.1 Necessity of the Cauchy-Riemann Equations
Suppose f is differentiable at a point z 0 ∈ C. If we write z = x + iy,
z 0 = x 0 + iy 0 , ∆z = ∆x + i∆y, and
then
Hence
f(z) = f(x + iy) = u(x, y) + iv(x, y),
f(z 0 + ∆z) − f(z 0 ) = (u(x 0 + ∆x, y 0 + ∆y) + iv(x 0 + ∆x, y 0 + ∆y))
f(z 0 + ∆z) − f(z 0 )
∆z
− (u(x 0 , y 0 ) + iv(x 0 , y 0 ))
= (u(x 0 + ∆x, y 0 + ∆y) − u(x 0 , y ) ))
+ i(v(x 0 + ∆x, y 0 + ∆y) − v(x 0 , y ) ).
= u(x 0 + ∆x, y 0 + ∆y) − u(x 0 , y 0 )
∆z
+ i v(x 0 + ∆x, y 0 + ∆y) − v(x 0 , y 0 )
.
∆z
If we let ∆z → 0 along the real axis, then ∆z = ∆x, and we have
f ′ f(z 0 + ∆z) − f(z 0 )
(z 0 ) = lim
∆z→0 ∆z
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u(x 0 + ∆x, y 0 ) − u(x 0 , y 0 ) v(x 0 + ∆x, y 0 ) − v(x 0 , y 0 )
= lim
+ i lim
∆x→0
∆x
∆x→0
∆x
= u x (x 0 , y 0 ) + iv x (x 0 , y 0 ).
If we let ∆z → 0 along the imaginary axis, then ∆z = i∆y, and we have
f ′ f(z 0 + ∆z) − f(z 0 )
(z 0 ) = lim
∆z→0 ∆z
u(x 0 , y 0 + ∆y) − u(x 0 , y 0 )
= lim
∆y→0 i∆y
= −iu y (x 0 , y 0 ) + v y (x 0 , y 0 ).
It follows that
v(x 0 , y ) + ∆y) − v(x 0 , y 0 )
+ i lim
∆y→0 i∆y
u x (x 0 , y 0 ) = v y (x 0 , y 0 ) and u y (x 0 , y 0 ) = −v x (x 0 , y 0 ).
We call these equations the Cauchy-Riemann equations.
Theorem 13.1. Suppose f is differentiable at z 0 = x 0 + iy 0 and
f(x + iy) = u(x, y) + iv(x, y).
Then
and
u x (x 0 , y 0 ) = v y (x 0 , y 0 ) and u y (x 0 , y 0 ) = −v x (x 0 , y 0 )
f ′ (z 0 ) = u x (x 0 , y 0 ) + iv x (x 0 , y 0 ).
Example 13.1. If f(z) = |z| 2 , then, writing f(x + iy) = u(x, y) + iv(x, y),
Hence
u(x, y) = x 2 + y 2 and v(x, y) = 0.
u x (x, y) = 2x, u y (x, y) = 2y, and v x (x, y) = 0 = v y (x, y).
Hence f satisfies the Cauchy-Riemann equations only at the origin, showing
that, as we have seen before, f is not differentiable at any z ≠ 0.
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13.2 Sufficiency of the Cauchy-Riemann equations
By themselves, the Cauchy-Riemann equations are not sufficient to guarantee
the differentiability of a given function. However, the additional assumption
of continuity of the partial derivatives does suffice to guarantee differentiability.
Theorem 13.2. Suppose f is defined on an ɛ neighborhood U of a point
z 0 = x 0 + y 0 ,
f(x + iy) = u(x, y) + iv(x, y),
and u x , u y , v x , and v y exist on U and are continuous at (x 0 , y 0 ). If
u x (x 0 , y 0 ) = v y (x 0 , y 0 ) and u y (x 0 , y 0 ) = −v x (x 0 , y 0 ),
then f is differentiable at z 0 .
Proof. Let ∆z = ∆x + i∆y,
∆u = u(x 0 + ∆x, y 0 + ∆y) − u(x 0 , y 0 ),
and
∆v = v(x 0 + ∆x, y 0 + ∆y) − v(x 0 , y 0 ).
The key fact we need comes from a theorem of multi-variable calculus: under
the conditions of the theorem, u and v are both differentiable functions. This
means that there exist ɛ 1 and ɛ 2 , depending on ∆x and ∆y, such that
∆u = u x (x 0 , y 0 )∆x + u y (x 0 , y 0 )∆y + ɛ 1 |∆z|
and
∆v = v x (x 0 , y 0 )∆x + v y (x 0 , y 0 )∆y + ɛ 2 |∆z|
and both ɛ 1 → 0 and ɛ 2 → 0 as ∆z → 0. It follows that
f(z 0 + ∆z) − f(z 0 ) = ∆u + i∆v
= u x (x 0 , y 0 )∆x + u y (x 0 , y 0 )∆y + ɛ 1 |∆z|
+ i(v x (x 0 , y 0 )∆x + v y (x 0 , y 0 )∆y + ɛ 2 |∆z|)
= u x (x 0 , y 0 )∆x − v x (x 0 , y ) ∆y + iv x (x 0 , y 0 )∆x
+ iu x (x 0 , y 0 )∆y + (ɛ 1 + iɛ 2 )|∆z|
= u x (x 0 , y 0 )∆z + iv x (x 0 , y 0 )∆z + (ɛ 1 + iɛ 2 )|∆z|.
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Hence
f ′ f(z 0 + ∆z) − f(z 0 )
(z 0 ) = lim
∆z→0
(
∆z
= lim u x (x 0 , y 0 ) + iv x (x 0 , y 0 ) + (ɛ 1 + iɛ 2 ) |∆z| )
∆z→0
z
= u x (x 0 , y 0 ) + iv x (u 0 , y 0 ),
since
and
for all z.
lim (ɛ 1 + ɛ 2 ) = 0
∆z→0
∣ ∣∣∣ |z|
z ∣ = |z|
|z| = 1
Example 13.2. It follows now from the previous example that f(z) = |z| 2
is differentiable at z = 0 but not for any z ≠ 0.
Example 13.3. If f(z) = e z , then
f(x + iy) = e x e iy = e x cos(x) + ie x sin(y).
Hence
Thus
u(x, y) = e x cos(y) and v(x, y) = e x sin(y).
u x (x, y) = e x cos(y) = v y and u y (x, y) = −e x sin(y) = −v x (x, y).
Hence f is differentiable at every z ∈ C. Moreover,
f ′ (z) = u x (x, y) + iv x (x, y) = e x cos(y) + ie x sin(y) = e z = f(z).
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