17.5 Cauchy–Riemann Equations

INTRODUCTION

In the preceding section we saw that a function f of a complex variable z is analytic at a point z when f is differentiable at z and differentiable at every point in some neighborhood of z. This requirement is more stringent than simply differentiability at a point because a complex function can be differentiable at a point z but yet be differentiable nowhere else. A function f is analytic in a domain D if f is differentiable at all points in D. We shall now develop a test for analyticity of a complex function f(z) = u(x, y) + iv(x, y).

A Necessary Condition for Analyticity

In the next theorem we see that if a function f(z) = u(x, y) + iv(x, y) is differentiable at a point z, then the functions u and v must satisfy a pair of equations that relate their first-order partial derivatives. This result is a necessary condition for analyticity.

THEOREM 17.5.1 Cauchy–Riemann Equations

Suppose f(z) = u(x, y) + iv(x, y) is differentiable at a point z = x + iy. Then at z the first-order partial derivatives of u and v exist and satisfy the Cauchy–Riemann equations

(1)

PROOF:

Since f′(z) exists, we know that

(2)

By writing f(z) = u(x, y) + iv(x, y) and Δz = Δx + iΔy, we get from (2)

(3)

Since this limit exists, Δz can approach zero from any convenient direction. In particular, if Δz → 0 horizontally, then Δz = Δx, and so (3) becomes

(4)

Since f′(z) exists, the two limits in (4) exist. But by definition the limits in (4) are the first partial derivatives of u and v with respect to x. Thus, we have shown that

(5)

Now if we let Δz → 0 vertically, then Δz = iΔy and (3) becomes

(6)

which is the same as

(7)

Equating the real and imaginary parts of (5) and (7) yields the pair of equations in (1). ≡

If a complex function f(z) = u(x, y) + iv(x, y) is analytic throughout a domain D, then the real functions u and v must satisfy the Cauchy–Riemann equations (1) at every point in D.

EXAMPLE 1 Using the Cauchy–Riemann Equations

The polynomial f(z) = z2 + z is analytic for all z and f(z) = x2 − y2 + x + i(2xy + y). Thus, u(x, y) = x2 − y2 + x and v(x, y) = 2xy + y. For any point (x, y), we see that the Cauchy–Riemann equations are satisfied:

≡

EXAMPLE 2 Using the Cauchy–Riemann Equations

Show that the function f(z) = (2x2 + y) + i(y2 − x) is not analytic at any point.

SOLUTION

We identify u(x, y) = 2x2 + y and v(x, y) = y2 − x. Now from

we see that ∂u/∂y = −∂v/∂x but that the equality ∂u/∂x = ∂v/∂y is satisfied only on the line y = 2x. However, for any point z on the line, there is no neighborhood or open disk about z in which f is differentiable. We conclude that f is nowhere analytic. ≡

Important.

By themselves, the Cauchy–Riemann equations are not sufficient to ensure analyticity. However, when we add the condition of continuity to u and v and the four partial derivatives, the Cauchy–Riemann equations can be shown to imply analyticity. The proof is long and complicated and so we state only the result.

THEOREM 17.5.2 Criterion for Analyticity

Suppose the real-valued functions u(x, y) and v(x, y) are continuous and have continuous first-order partial derivatives in a domain D. If u and v satisfy the Cauchy–Riemann equations at all points of D, then the complex function f(z) = u(x, y) + iv(x, y) is analytic in D.

EXAMPLE 3 Using Theorem 17.5.2

For the function f(z) = we have

In other words, the Cauchy–Riemann equations are satisfied except at the point where x2 + y2 = 0; that is, at z = 0. We conclude from Theorem 17.5.2 that f is analytic in any domain not containing the point z = 0. ≡

The results in (5) and (7) were obtained under the basic assumption that f was differentiable at the point z. In other words, (5) and (7) give us a formula for computing f′(z):

(8)

For example, we know that f(z) = z2 is differentiable for all z. With u(x, y) = x2 − y2, ∂u/∂x = 2x, v(x, y) = 2xy, and ∂v/∂x = 2y, we see that

f′(z) = 2x + i2y = 2(x + iy) = 2z.

Recall that analyticity implies differentiability but not vice versa. Theorem 17.5.2 has an analogue that gives sufficient conditions for differentiability:

If the real-valued functions u(x, y) and v(x, y) are continuous and have continuous first-order partial derivatives in a neighborhood of z, and if u and v satisfy the Cauchy–Riemann equations at the point z, then the complex function f(z) = u(x, y) + iv(x, y) is differentiable at z and f′ (z) is given by (8).

The function f(z) = x2 − y2i is nowhere analytic. With the identifications u(x, y) = x2 and v(x, y) = −y2, we see from

that the Cauchy–Riemann equations are satisfied only when y = −x. But since the functions u, ∂u/∂x, ∂u/∂y, v, ∂v/∂x, and ∂v/∂y are continuous at every point, it follows that f is differentiable on the line y = −x and on that line (8) gives the derivative f′(z) = 2x = −2y.

Harmonic Functions

We saw in Chapter 13 that Laplace’s equation ∂2u/∂x2 + ∂2u/∂y2 = 0 occurs in certain problems involving steady-state temperatures. This partial differential equation also plays an important role in many areas of applied mathematics. Indeed, as we now see, the real and imaginary parts of an analytic function cannot be chosen arbitrarily, since both u and v must satisfy Laplace’s equation. It is this link between analytic functions and Laplace’s equation that makes complex variables so essential in the serious study of applied mathematics.

DEFINITION 17.5.1 Harmonic Functions

A real-valued function ϕ(x, y) that has continuous second-order partial derivatives in a domain D and satisfies Laplace’s equation is said to be harmonic in D.

THEOREM 17.5.3 A Source of Harmonic Functions

Suppose f(z) = u(x, y) + iv(x, y) is analytic in a domain D. Then the functions u(x, y) and v(x, y) are harmonic functions.

We will see in Chapter 18 that an analytic function possesses derivatives of all orders.

PROOF:

In this proof we shall assume that u and v have continuous second-order partial derivatives. Since f is analytic, the Cauchy–Riemann equations are satisfied. Differentiating both sides of ∂u/∂x = ∂v/∂y with respect to x and differentiating both sides of ∂u/∂y = −∂v/∂x with respect to y then give

With the assumption of continuity, the mixed partials are equal. Hence, adding these two equations gives

This shows that u(x, y) is harmonic.

Now differentiating both sides of ∂u/∂x = ∂v/∂y with respect to y and differentiating both sides of ∂u/∂y = −∂v/∂x with respect to x and subtracting yield

≡

Harmonic Conjugate Functions

If f(z) = u(x, y) + iv(x, y) is analytic in a domain D, then u and v are harmonic in D. Now suppose u(x, y) is a given function that is harmonic in D. It is then sometimes possible to find another function v(x, y) that is harmonic in D so that u(x, y) + iv(x, y) is an analytic function in D. The function v is called a harmonic conjugate function of u.

EXAMPLE 4 Harmonic Function/Harmonic Conjugate Function

(a) Verify that the function u(x, y) = x3 − 3xy2 − 5y is harmonic in the entire complex plane.

(b) Find the harmonic conjugate function of u.

SOLUTION

(a) From the partial derivatives

we see that u satisfies Laplace’s equation:

(b) Since the harmonic conjugate function v must satisfy the Cauchy–Riemann equations, we must have

(9)

Partial integration of the first equation in (9) with respect to y gives v(x, y) = 3x2y − y3 + h(x). From this we get

Substituting this result into the second equation in (9) gives h′(x) = 5, and so h(x) = 5x + C. Therefore, the harmonic conjugate function of u is v(x, y) = 3x2y − y3 + 5x + C. The analytic function is f(z) = x3 − 3xy2 − 5y + i(3x2y − y3 + 5x + C). ≡

REMARKS

Suppose u and v are the harmonic functions that comprise the real and imaginary parts of an analytic function f(z). The level curves u(x, y) = c1 and v(x, y) = c2 defined by these functions form two orthogonal families of curves. See Problem 32 in Exercises 17.5. For example, the level curves generated by the simple analytic function f(z) = z = x + iy are x = c1 and y = c2. The family of vertical lines defined by x = c1 is clearly orthogonal to the family of horizontal lines defined by y = c2. In electrostatics, if u(x, y) = c1 defines the equipotential curves, then the other, and orthogonal, family v(x, y) = c2 defines the lines of force.

17.5 Exercises Answers to selected odd-numbered problems begin on page ANS-45.

In Problems 1 and 2, the given function is analytic for all z. Show that the Cauchy–Riemann equations are satisfied at every point.

  1. f(z) = z3
  2. f(z) = 3z2 + 5z − 6i

In Problems 3–8, show that the given function is not analytic at any point.

  1. f(z) = Re(z)
  2. f(z) = y + ix
  3. f(z) = 4z − + 3
  4. f(z) =
  5. f(z) = x2 + y2

In Problems 9–14, use Theorem 17.5.2 to show that the given function is analytic in an appropriate domain.

  1. f(z) = ex cos y + iex sin y
  2. f(z) = x + sin x cosh y + i(y + cos x sinh y)
  3. f(z) = cos 2xy + i sin 2xy
  4. f(z) = 4x2 + 5x − 4y2 + 9 + i(8xy + 5y − 1)

In Problems 15 and 16, find real constants a, b, c, and d so that the given function is analytic.

  1. f(z) = 3x − y + 5 + i(ax + by − 3)
  2. f(z) = x2 + axy + by2 + i(cx2 + dxy + y2)

In Problems 17–20, show that the given function is not analytic at any point but is differentiable along the indicated curve(s).

  1. f(z) = x2 + y2 + 2xyi; x-axis
  2. f(z) = 3x2y2 − 6x2y2i; coordinate axes
  3. f(z) = x3 + 3xy2 − x + i(y3 + 3x2y − y); coordinate axes
  4. f(z) = x2 − x + y + i(y2 − 5y − x); y = x + 2
  5. Use (8) to find the derivative of the function in Problem 9.
  6. Use (8) to find the derivative of the function in Problem 11.

In Problems 23–28, verify that the given function u is harmonic. Find v, the harmonic conjugate function of u. Form the corresponding analytic function f(z) = u + iv.

  1. u(x, y) = x
  2. u(x, y) = 2x − 2xy
  3. u(x, y) = x2 − y2
  4. u(x, y) = 4xy3 − 4x3y + x
  5. u(x, y) = loge(x2 + y2)
  6. u(x, y) = ex(x cos y − y sin y)
  7. Sketch the level curves u(x, y) = c1 and v(x, y) = c2 of the analytic function f(z) = z2.
  8. Consider the function f(z) = 1/z. Describe the level curves.
  9. Consider the function f(z) = z + 1/z. Describe the level curve v(x, y) = 0.
  10. Suppose u and v are the harmonic functions forming the real and imaginary parts of an analytic function. Show that the level curves u(x, y) = c1 and v(x, y) = c2 are orthogonal. [Hint: Consider the gradient of u and the gradient of v. Ignore the case where a gradient vector is the zero vector.]