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!