Theorem3 Let z0 ∈ C and let G be an open subset of C that contains z0. If f(x+iy) = u(x,y)+ iv(x,y) is deﬁned on G and the ﬁrst order partial derivati...

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∂v ∂y (x0 , y0 )

∂u ∂y (x0 , y0 )

∂v = − ∂x (x0 , y0 )

(CR)

and f ′ (x0 + iy0 ) =

∂u ∂x (x0 , y0 )

∂v + i ∂x (x0 , y0 )

The equations (CR) are called the Cauchy–Riemann equations. c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

1

Proof:

By assumption f (z0 + ∆z) − f (z0 ) ∆z→0 ∆z [u(x0 + ∆x, y0 + ∆y) − u(x0 , y0 )] + i[v(x0 + ∆x, y0 + ∆y) − v(x0 , y0 )] = lim ∆z→0 ∆z

f ′ (z0 ) = lim

In particular, by sending ∆z = ∆x + i∆y to zero along the real axis (i.e. setting ∆y = 0 and sending ∆x → 0), we have [u(x0 + ∆x, y0 ) − u(x0 , y0 )] + i[v(x0 + ∆x, y0 ) − v(x0 , y0 )] ∆x→0 ∆x

f ′ (x0 + iy0 ) = lim and hence

u(x0 + ∆x, y0 ) − u(x0 , y0 ) ∆x→0 ∆x v(x + ∆x, y0 ) − v(x0 , y0 ) 0 Im f ′ (z0 ) = lim ∆x→0 ∆x Re f ′ (z0 ) = lim

This tells us that the partial derivatives ∂u ∂x (x0 , y0 )

∂v ∂u (x0 , y0 ), ∂x (x0 , y0 ) ∂x

= Re f ′ (x0 + iy0 )

∂v ∂x (x0 , y0 )

exist and

= Im f ′ (x0 + iy0 )

(1)

This gives the formula for f ′ (x0 + iy0 ) in the statement of the theorem. If, instead, we send ∆z = ∆x + i∆y to zero along the imaginary axis (i.e. set ∆x = 0 and send ∆y → 0), we have [u(x0 , y0 + ∆y) − u(x0 , y0 )] + i[v(x0 , y0 + ∆y) − v(x0 , y0 )] ∆y→0 i∆y [v(x0 , y0 + ∆y) − v(x0 , y0 )] − i[u(x0 , y0 + ∆y) − u(x0 , y0 )] = lim ∆y→0 ∆y

f ′ (x0 + iy0 ) = lim

and hence

v(x0 , y0 + ∆y) − v(x0 , y0 ) ∆x→0 ∆y u(x0 , y0 + ∆y) − u(x0 , y0 ) Im f ′ (z0 ) = − lim ∆x→0 ∆y Re f ′ (z0 ) = lim

This tells us that the partial derivatives ∂v ∂y (x0 , y0 )

∂v (x0 , y0 ), ∂u (x0 , y0 ) ∂y ∂y

= Re f ′ (x0 + iy0 )

∂u ∂y (x0 , y0 )

exist and

= −Im f ′ (x0 + iy0 )

(2)

Comparing (1) and (2) gives (CR).

c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

2

Theorem 2 says that it is necessary for u(x, y) and v(x, y) to obey the Cauchy–Riemann equations in order for f (x + iy) = u(x + iy) + v(x + iy) to be differentiable. The following theorem says that, provided the first order partial derivatives of u and v are continuous, the converse is also true — if u(x, y) and v(x, y) obey the Cauchy–Riemann equations then f (x + iy) = u(x + iy) + v(x + iy) is differentiable. Theorem 3 Let z0 ∈ C and let G be an open subset of C that contains z0 . If f (x + iy) = u(x, y) + iv(x, y) is defined on G and ◦ the first order partial derivatives of u and v exist in G and are continuous at (x0 , y0 ) ◦ u and v obey the Cauchy–Riemann equations at (x0 , y0 ), ∂v (x0 , y0 ) + i ∂x (x0 , y0 ). then f is differentiable at z0 = x0 + iy0 and f ′ (x0 + iy0 ) = ∂u ∂x Proof:

Write

where

f (z0 + ∆z) − f (z0 ) = U (∆z) + iV (∆z) ∆z u(x0 + ∆x, y0 + ∆y) − u(x0 , y0 ) ∆z v(x0 + ∆x, y0 + ∆y) − v(x0 , y0 ) V (∆z) = ∆z U (∆z) =

Our goal is to prove that lim [U (∆z) + iV (∆z)] exists and equals ∆z→0

∂u (x0 , y0 ) + i ∂v (x0 , y0 ). ∂x ∂x

Concentrate on U (∆z). The first step is to rewrite U (∆z) in terms of expressions that 0 ,y0 ) converges will converge to partial derivatives of u and v. For example u(x0 ,y0 +∆y)−u(x ∆y to uy (x0 , y0 ) when ∆y → 0. We can achieve this by adding and subtracting u(x0 , y0 + ∆y): u(x0 + ∆x, y0 + ∆y) − u(x0 , y0 ) ∆z u(x0 + ∆x, y0 + ∆y) − u(x0 , y0 + ∆y) u(x0 , y0 + ∆y) − u(x0 , y0 ) = + ∆z ∆z

U (∆z) =

To express U (∆z) in terms of partial derivatives of u, we use the (ordinary first year Calculus) mean value theorem. Recall that it says that, if F (x) is differentiable everywhere between x0 and x0 + ∆x, then F (x0 + ∆x) − F (x0 ) = F ′ (x∗0 ) ∆x for some x∗0 between x0 and x0 + ∆x. Applying the mean value theorem with F (x) = u(x, y0 + ∆y) to the first half of U (∆z) and with F (y) = u(x0 , y) to the second half gives ux (x∗0 , y0 + ∆y)∆x uy (x0 , y0∗ )∆y U (∆z) = + ∆z ∆z for some x∗0 between x0 and x0 + ∆x and some y0∗ between y0 and y0 + ∆y. Because ux and uy are continuous, ux (x∗0 , y0 + ∆y) is almost ux (x0 , y0 ) and uy (x0 , y0∗ ) is almost uy (x0 , y0 ) c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

3

when ∆z is small. So we write U (∆z) =

ux (x0 , y0 )∆x uy (x0 , y0 )∆y + + E1 (∆z) + E2 (∆z) ∆z ∆z

where the “error terms” are E1 (∆z) = [ux (x∗0 , y0 + ∆y) − ux (x0 , y0 )] E2 (∆z) = [uy (x0 , y0∗ ) − uy (x0 , y0 )]

∆x ∆z

∆y ∆z

Similarly vx (x∗∗ vy (x0 , y0∗∗ )∆y 0 , y0 + ∆y)∆x + ∆z ∆z vx (x0 , y0 )∆x vy (x0 , y0 )∆y = + + E3 (∆z) + E4 (∆z) ∆z ∆z

V (∆z) =

∗∗ for some x∗∗ 0 between x0 and x0 + ∆x, and some y0 between y0 and y0 + ∆y. The error terms are ∆x E3 (∆z) = [vx (x∗∗ 0 , y0 + ∆y) − vx (x0 , y0 )] ∆z ∆y E4 (∆z) = [vy (x0 , y0∗∗ ) − vy (x0 , y0 )] ∆z Now as ∆z → 0 ◦ both x∗0 and x∗∗ 0 (both of which are between x0 and x0 + ∆x) must approach x0 and ∗ ∗∗ ◦ both y0 and y0 (both of which are between y0 and y0 + ∆y) must approach y0 and ≤ 1 and ∆y ≤ 1 ◦ ∆x ∆z ∆z Recalling that ux , uy , vx and vy are all assumed to be continuous at (x0 , y0 ), we conclude that lim E1 (∆z) = lim E2 (∆z) = lim E3 (∆z) = lim E4 (∆z) = 0 ∆z→0

∆z→0

∆z→0

∆z→0

and, using the Cauchy–Riemann equations, f (z0 + ∆z) − f (z0 ) = lim U (∆z) + iV (∆z) ∆z→0 ∆z→0 ∆z h u (x , y )∆x u (x , y )∆y vx (x0 , y0 )∆x vy (x0 , y0 )∆y i x 0 0 y 0 0 = lim + +i +i ∆z→0 ∆z ∆z ∆z ∆z h u (x , y )∆x v (x , y )∆y vx (x0 , y0 )∆x ux (x0 , y0 )∆y i x 0 0 x 0 0 = lim − +i +i ∆z→0 ∆z ∆z ∆z ∆z h ∆x + i∆y ∆x + i∆y i = lim ux (x0 , y0 ) + ivx (x0 , y0 ) ∆z→0 ∆z ∆z = ux (x0 , y0 ) + ivx (x0 , y0 ) lim

as desired. c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

4

Example 4 The function f (z) = z¯ has f (x + iy) = x − iy so that u(x, y) = x and v(x, y) = −y The first order partial derivatives of u and v are ux (x, y) = 1 vx (x, y) = 0 uy (x, y) = 0

vy (x, y) = −1

As the Cauchy–Riemann equation ux (x, y) = vy (x, y) is satisfied nowhere, the function f (z) = z¯ is differentiable nowhere. We have already seen this in Example 1. Example 5 The function f (z) = ez has f (x + iy) = ex+iy = ex cos y + i sin y = u(x, y) + iv(x, y) with u(x, y) = ex cos y and v(x, y) = ex sin y The first order partial derivatives of u and v are ux (x, y) = ex cos y

vx (x, y) = ex sin y

uy (x, y) = −ex sin y

vy (x, y) = ex cos y

As the Cauchy–Riemann equations ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) are satisfied for all (x, y), the function f (z) = ez is entire and its derivative is f ′ (z) = f ′ (x + iy) = ux (x, y) + ivx (x, y) = ex cos y + iex sin y = ez

Example 6 The function f (x + iy) = x2 + y + i(y 2 − x) has u(x, y) = x2 + y and v(x, y) = y 2 − x The first order partial derivatives of u and v are ux (x, y) = 2x vx (x, y) = −1 uy (x, y) = 1

vy (x, y) = 2y

As the Cauchy–Riemann equations ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) are satisfied only on the line y = x, the function f is differentiable on the line y = x and nowhere else. So it is nowhere analytic. c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

5

Example 7 The function f (x + iy) = x2 − y 2 + 2ixy has u(x, y) = x2 − y 2 and v(x, y) = 2xy The first order partial derivatives of u and v are ux (x, y) = 2x

vx (x, y) = 2y

uy (x, y) = −2y

vy (x, y) = 2x

As the Cauchy–Riemann equations ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) are satisfied for all (x, y), this function is entire. There is another way to see this. It suffices to observe that f (z) = z 2 , since (x + iy)2 = x2 − y 2 + 2ixy. So f is a polynomial in z and we already know that all polynomials are differentiable everywhere. Example 8 The function f (x + iy) = x2 + y 2 has u(x, y) = x2 + y 2 and v(x, y) = 0 The first order partial derivatives of u and v are ux (x, y) = 2x

vx (x, y) = 0

uy (x, y) = 2y

vy (x, y) = 0

As the Cauchy–Riemann equations ux (x, y) = vy (x, y), uy (x, y) = −vx (x, y) are satisfied only at x = y = 0, the function f is differentiable only at the point z = 0. So it is nowhere analytic. There is another way to see that f (z) cannot be differentiable at any z 6= 0. Just would observe that f (z) = z¯ z . If f (z) were differentiable at some z0 6= 0, then z¯ = f (z) z also be differentiable at z0 and we already know that this is not case.

c Joel Feldman.

2012. All rights reserved.

January 19, 2012

The Cauchy–Riemann Equations

6