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Example. Let u = e-x(x sin y - y cos y)

(a) Prove u = e-x(x sin y - y cos y) is analytic

(b) Find the conjugate harmonic function of u i.e. find a function v such that f(z) = u + iv is analytic

(c) Find f(z)

(a)     Prove u = e-x(x sin y - y cos y) is analytic

∂u/∂x = (e-x)(sin y) + (-e-x)(x sin y - y cos y) = e-x sin y - xe-x sin y + ye-x cos y

1)        2u/∂x2 = ∂(e-x sin y - xe-x sin y + ye-x cos y)/∂x = -2e-x sin y + xe-x sin y - ye-x cos y

∂u/∂y = e-x (x cos y + y sin y - cos y) = xe-x cos y + ye-x sin y - e-x cos y

2)        2u/∂y2 = ∂(xe-x cos y + ye-x sin y - e-x cos y)/∂y = -xe-x sin y + 2e-x sin y + ye-x cos y

Adding 1) and 2) gives ∂2u/∂x2 + ∂2u/∂y2 = 0 . Thus u is harmonic.

(b)    Find the conjugate harmonic function of u

From the Cauchy-Riemann equations we get

3)        ∂v/∂y = ∂u/∂x = e-x sin y - xe-x sin y + ye-x cos y

4)        ∂v/∂x = - ∂u/∂y = e-x cos y - xe-x cos y - ye-x sin y

We now integrate 3) with respect to y, keeping x constant:

5)        v = - e-x cos y + xe-x cos y + e-x(y sin y + cos y) + F(x)

= ye-x sin y + xe-x cos y + F(x)

where F(x) is an arbitrary real function of x.

We now substitute 5) into 4), taking ∂v/∂x of 5)

-ye-x sin y - xe-x cos y + e-x cos y + F'(x) = e-x cos y - xe-x cos y - ye-x sin y

or

F'(x) = 0 .

Thus F(x) = c, a constant.

Substituting F(x) = c into 5) we get

v = e-x(y sin y + x cos y) + c

(c)      Find f(z)

To find f(z) we employ the following theorem:

Theorem 1. f(z) = u(z, 0) + i v(z, 0)

Derivation.

f(z) = f(x+ iy) = u(x, y) + i v(x, y)

Putting y = 0, one obtains       f(x) = u(x, 0) + i v(x, 0)

Replacing x by z, we get        f(z) = u(z, 0) + i v(z, 0)

Solution. Applying Theorem 1 we get

u(z, 0) = 0

v(z, 0) = ze-z

f(z) = u(z, 0) + i v(z, 0) = ze-z,

apart from an arbitrary additive constant.

If only u (or v) is known another procedure can be used that employs the following theorem:

Theorem 2. If f(z) = u(x, y) + i v(x, y)

f(z) = 2u(z/2, -iz/2) + constant

and

f(z) = 2i v(z/2, -iz/2) + constant

Alternative method for finding f(z) and the conjugate harmonic function. An alternative method for finding the conjugate harmonic function employs the following theorem:

Theorem 3. Let u1 = ∂u/∂x and u2 = ∂u/∂y. Then

f '(z) = u1(z, 0) - i u2(z, 0)

Derivation. We have already shown that

Putting y = 0, we get              f '(x) = u1(x, 0) - i u2(x, 0)

Then replacing x by z we obtain         f '(z) = u1(z, 0) - i u2(z, 0)

Solution. Since u = e-x(x sin y - y cos y) , we have

u1(x, y) = ∂u/∂x = e-x sin y - xe-x sin y + ye-x cos y

u2(x, y) = ∂u/∂y = xe-x cos y + ye-x sin y - e-x cos y

Applying Theorem 3 we get

f '(z) = u1(z, 0) - i u2(z, 0) = 0 - i(ze-z - e-z) = - i(ze-z - e-z)

Integrating with respect to z we obtain, except for a constant,

f(z) = i ze-z

If one then expands the right side in terms of x and y, one obtains

f(z) = e-x(x sin y - y cos y) + i e-x(y sin y + x cos y)

giving

v = e-x(y sin y + x cos y)

Source: Spiegel. Complex Variables (Schaum). p. 73, 74, 85. Prob. 7, 8, 40, 101

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