2.6: Cauchy-Riemann Equations

The Cauchy-Riemann equations are our first consequence of the fact that the limit defining \(f(z)\) must be the same no matter which direction you approach \(z\) from. The Cauchy-Riemann equations will be one of the most important tools in our toolbox.

Partial Derivatives as Limits

Before getting to the Cauchy-Riemann equations we remind you about partial derivatives. If \(u(x, y)\) is a function of two variables then the partial derivatives of \(u\) are defined as

\[\dfrac{\partial u}{\partial x} (x, y) = \lim_{\Delta x \to 0} \dfrac{u(x + \Delta x, y) - u(x, y)}{\Delta x}, \nonumber \]

i.e., the derivative of \(u\) holding \(y\) constant.

\[\dfrac{\partial u}{\partial y} (x, y) = \lim_{\Delta y \to 0} \dfrac{u(x, y + \Delta y) - u(x, y)}{\Delta y}, \nonumber \]

i.e., the derivative of \(u\) holding \(x\) constant.

Cauchy-Riemann Equations

The Cauchy-Riemann equations use the partial derivatives of \(u\) and \(v\) to allow us to do two things: first, to check if \(f\) has a complex derivative and second, to compute that derivative. We start by stating the equations as a theorem.

Theorem \(\PageIndex{1}\): Cauchy-Riemann Equations

If \(f(z) = u(x, y) + iv(x, y)\) is analytic (complex differentiable) then

\[\begin{align*} f'(z) &= \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} \\[4pt] &= \dfrac{\partial v}{\partial y} - i \dfrac{\partial u}{\partial y} \end{align*} \]

In particular,

\[\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y} \label{CR1}\]

and

\[\dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}. \label{CR2} \]

These partial differential equations (Equations \ref{CR1} and \ref{CR2}) are what is usually meant by the Cauchy-Riemann equations. Here is the short form of the Cauchy-Riemann equations:

\[u_x = v_y \nonumber \]

\[u_y = -v_x \nonumber \]

Proof

Let's suppose that \(f(z)\) is differentiable in some region \(A\) and

\[f(z) = f(x + iy) = u(x, y) + iv(x, y). \nonumber \]

We'll compute \(f'(z)\) by approaching \(z\) first from the horizontal direction and then from the vertical direction. We’ll use the formula

\[f'(z) = \lim_{\Delta z \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z}, \nonumber \]

where \(\Delta z = \Delta x + i \Delta y\).

Horizontal direction: \(\Delta y = 0, \Delta z = \Delta x\)

\[\begin{align*} f'(z) & =  \lim_{\Delta z \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z} \\[4pt] &= \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x + iy) - f(x + iy)}{\Delta x} \\[4pt] & = \lim_{\Delta x \to 0} \dfrac{(u (x + \Delta x, y) + iv(x + \Delta x, y)) - (u(x, y) + iv(x, y))}{\Delta x} \\[4pt] & = \lim_{\Delta x \to 0} \dfrac{u(x + \Delta x, y) - u(x, y)}{\Delta x} + i \dfrac{v(x + \Delta x, y) - v(x, y)}{\Delta x} \\[4pt] & = \dfrac{\partial u}{\partial x} (x, y) + i \dfrac{\partial v}{\partial x} (x, y) \end{align*} \]

Vertical direction: \(\Delta x = 0\), \(\Delta z = i \Delta y\) (We'll do this one a little faster.)

\[\begin{align*} f'(z) & = \lim_{\Delta z \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z} \\[4pt] & = \lim_{\Delta y \to 0} \dfrac{(u(x, y + \Delta y) + iv (x, y + \Delta y)) - (u(x, y) + iv (x, y))}{i \Delta y} \\[4pt] & = \lim_{\Delta y \to 0} \dfrac{u(x, y+ \Delta y) - u(x, y)}{i \Delta y} + i \dfrac{v(x, y + \Delta y) - v(x, y)}{i \Delta y} \\[4pt]  & = \dfrac{1}{i} \dfrac{\partial u}{\partial y} (x, y) + \dfrac{\partial v}{\partial y} (x, y) \\[4pt] & = \dfrac{\partial v}{\partial y} (x, y) - i \dfrac{\partial u}{\partial y} (x, y) \end{align*} \]

We have found two different representations of \(f'(z)\) in terms of the partials of \(u\) and \(v\). If put them together we have the Cauchy-Riemann equations:

\[f'(z) = \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} = \dfrac{\partial v}{\partial y} - i \dfrac{\partial u}{\partial y} \ \ \Rightarrow \ \ \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}, \text{ and } -\dfrac{\partial u}{\partial y} = \dfrac{\partial v}{\partial x}. \nonumber \]

It turns out that the converse is true and will be very useful to us.

Theorem \(\PageIndex{2}\)

Consider the function \(f(z) = u(x, y) + iv (x, y)\) defined on a region \(A\). If \(u\) and \(v\) satisfy the Cauchy-Riemann equations and have continuous partials then \(f(z)\) is differentiable on \(A\).

Proof

The proof of this is a tricky exercise in analysis. It is somewhat beyond the scope of this class, so we will skip it. If you’re interested, with a little effort you should be able to grasp it.

Using the Cauchy-Riemann equations

The Cauchy-Riemann equations provide us with a direct way of checking that a function is differen- tiable and computing its derivative.

Example \(\PageIndex{1}\)

Use the Cauchy-Riemann equations to show that \(e^z\) is differentiable and its derivative is \(e^z\).

Solution

We write

\[e^z = e^{x + iy} = e^x \cos (y) + ie^x \sin (y). \nonumber \]

So

\[u(x, y) = e^x \cos (y) \nonumber \]

and

\[v(x, y) = e^x \sin(y). \nonumber \]

Computing partial derivatives we have

• \(u_x = e^x \cos (y)\),
• \(u_y = -e^x \sin (y)\),
• \(v_x = e^x \sin (y)\),
• \(v_y = e^x \cos (y) \).

We see that \(u_x = v_y\) and \(u_y = -v_x\), so the Cauchy-Riemann equations (Equations \ref{CR1} and \ref{CR2}) are satisfied. Thus, \(e^z\) is differentiable and

\[\dfrac{d}{dz} e^z = u_x + iv_x = e^x \cos (y) + ie^x \sin (y) = e^z. \nonumber \]

Example \(\PageIndex{2}\)

Use the Cauchy-Riemann equations to show that \(f(z) = \overline{z}\) is not differentiable.

Solution

\(f(x + iy) = x - iy\), so \(u(x, y) = x, v(x, y) = -y\). Taking partial derivatives

\(u_x = 1\), \(u_y = 0\), \(v_x = 0\), \(v_y = -1\)

Since \(u_x \ne v_y\) the Cauchy-Riemann equations (Equations \ref{CR1} and \ref{CR2}) are not satisfied and therefore \(f\) is not differentiable.

Theorem \(\PageIndex{3}\)

If \(f(z)\) is differentiable on a disk and \(f'(z) =0\) on the disk then \(f(z)\) is constant.

Proof

Since \(f\) is differentiable and \(f'(z) \equiv 0\), the Cauchy-Riemann equations show that

\[u_x (x, y) = u_y (x, y) = v_x (x, y) = v_y (x, y) = 0 \nonumber \]

We know from multivariable calculus that a function of \((x, y)\) with both partials identically zero is constant. Thus \(u\) and \(v\) are constant, and therefore so is \(f\).

\(f'(z)\) as a \(2 \times 2\) matrix

Recall that we could represent a complex number \(a + ib\) as a \(2 \times 2\) matrix

\[a + ib \ \leftrightarrow \ \begin{bmatrix} a & -b \\ b & a \end{bmatrix}. \nonumber \]

Now if we write \(f(z)\) in terms of \((x, y)\) we have

\[f(z) = f(x + iy) = u(x, y) + iv(x, y) \ \leftrightarrow \ f(x, y) = (u(x, y), v(x, y)). \nonumber \]

We have

\[f'(z) = u_x + iv_x, \nonumber \]

so we can represent \(f'(z)\) as

\[\begin{bmatrix} u_x & -v_x \\ v_x & u_x \end{bmatrix}. \label{1}\]

Using the Cauchy-Riemann equations we can replace \(-v_x\) by \(u_y\) and \(u_x\) by \(v_y\) which gives us the representation

\[f'(z) \ \leftrightarrow \ \begin{bmatrix} u_x & u_y \\ v_x & v_y \end{bmatrix}, \nonumber \]

i.e, \(f'(z)\) is just the Jacobian of \(f(x, y)\).

For me, it is easier to remember the Jacobian than the Cauchy-Riemann equations. Since \(f'(z)\) is a complex number I can use the matrix representation in Equation \(\ref{1}\) to remember the Cauchy-Riemann equations!