**If the real part of a complex function is harmonic, then there exists a harmonic imaginary part, so that the function is analytic.** The imaginary part is known as the harmonic conjugate of the real part. In other words, the harmonic conjugate to a real-valued function is an imaginary function, such that the whole function is differentiable.

**A harmonic conjugate satisfies both the Cauchy–Riemann equations and Laplace’s equation.**

## Theorem

Let’s say the complex function is f(z) and the real part is u(x,y) and the imaginary part is v(x,y). Then we can write the function as,

f(z) = u(x,y) + vi(x,y)

**If u(x,y) of the function is harmonic, then there exists a harmonic v(x,y) part, so f(z) is differentiable. To prove a function is harmonic, you need to prove that the complex function is differentiable.**

To continue, we need to understand how can a complex function be differentiable.

## Conditions to be a differentiable complex function

A complex function is only differentiable when it satisfies two following conditions.

#1 The complex function should be continues.

#2 It should satisfy Cauchy-Reimann (CR) Equations. If you need to learn about those equations, read MechCollege’s article on Cauchy-Reimann Equations.

∂u/∂x = ∂v/∂y ——–(CR Eq. 1)

∂u/∂y = – ∂v/∂x ——–(CR Eq. 2)

## Solved Problems

### Example #1 | Basic example for finding harmonic conjugate

**Find the harmonic conjugate of u(x,y) = x**^{2}** – y**^{2}** – x + y . And u(x,y) is a harmonic function and f(z) = u(x,y) + vi(x,y).**

So, we’re given the real part of the equation. So, now we need to find the imaginary part v(x,y), in way that the complex function f(z) is differentiable.

f(z) = u(x,y) + vi(x,y)

To be the f(z) differentiable, according to Cauchy-Reimann equations, the real part should satisfy the following equations.

∂u/∂x = ∂v/∂y

∂u/∂y = – ∂v/∂x

Lets get the partial differentiation of u(x,y) with respect to x and y.

∂u/∂x = 2x – 1

∂u/∂y = -2y + 1

So,

∂v/∂y = ∂u/∂x = 2x – 1

∂v/∂x = – ∂u/∂y = 2y – 1

There are two methods to solve the rest of the problem.

**Method #1**

Partially integrating ∂v/∂x with respect to x gives,

v = 2xy – x + h(x)

Partially integrating ∂v/∂y with respect to y gives,

v = 2xy – y + g(x)

So, the h(x) = – y and g(x) = – x

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) = 2xy – x – y

**Method #2**

Partially integrating ∂v/∂x with respect to y gives,

v = 2xy – y + g(x)

By finding the partial differentiation of v with respect to x,

∂v/∂x = 2y + g'(x)

Since ∂v/∂x = 2y – 1,

2y – 1 = 2y + g'(x)

g'(x) = -1

Partially integrating g'(x) with respect to x gives,

g(x) = – x

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) = 2xy – x – y

### Example #2 | The harmonic conjugate of an exponential function

**Show that u(x,y) = e**^{x}**sin(y) is a harmonic function and find the harmonic conjugate v(x,y) such that f(z) = u(x,y) + vi(x,y) is analytic.**

In this question, we can see something we didn’t talk about earlier. How to prove a function is harmonic? To that, you need to prove that the function has continuous second order partial derivatives.

Lets get the partial differentiation of u(x,y) with respect to x and y.

∂u/∂x = e^{x}sin(y)

∂u/∂y = e^{x}cos(y)

Since we need to prove u(x,y) = e^{x}sin(y) is a harmonic function, lets take the second degree partial derivatives.

∂^{2}u/∂x^{2} = e^{x}cos(y)

∂^{2}u/∂y^{2} = – e^{x}sin(y)

Since it has continues second order partial derivatives, u(x,y) = e^{x}sin(y) is a harmonic function.

Now, let’s find the harmonic conjugate. According Cauchy-Reimann equations,

∂v/∂y = ∂u/∂x = e^{x}sin(y)

∂v/∂x = – ∂u/∂y = – e^{x}cos(y)

Partially integrating ∂v/∂y with respect to y gives,

v = – e^{x}cos(y) + g(x)

By finding the partial differentiation of v with respect to x,

∂v/∂x = – e^{x}cos(y) + g'(x)

Since ∂v/∂x =- e^{x}cos(y),

– e^{x}cos(y) = – e^{x}cos(y) + g'(x)

g'(x) = 0

Partially integrating g'(x) with respect to x gives,

g(x) = C

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) = – e^{x}cos(y) + C