Lecture 5: the Cauchy–Riemann equations

Last time, we introduced analytic (or holomorphic) functions on a domain as those whose complex derivative is well-defined and continuous at every point in the domain. Today, we will find an explicit way to check, given a complex-valued function, if it is analytic or not.

We work in Cartesian coordinates both on the input and on the output: write $z=x+iy$, and $f(z)=u+iv$. Thinking of $f$ as depending on the two real numbers $x$ and $y$, we can think of each of $u$ and $v$ as real-valued functions of $x$ and $y$:

$$f(x+iy)=u(x,y)+iv(x,y).$$

Let’s study the derivative in this setting, using the formula

$$f^{\prime }(z)=\underset{h\to 0}{lim}\frac{f(z+h)-f(z)}{h}.$$

Here the limit is in the complex sense, so we must be able to take $h$ approach zero from any direction. We will study the cases when $h$ is approaching along the real or imaginary line.

Suppose $h$ is real. Then

${\displaystyle \frac{f(z+h)-f(z)}{h}}$ $={\displaystyle \frac{f(x+h+iy)-f(x+iy)}{h}}$

$={\displaystyle \frac{u(x+h,y)+iv(x+h,y)-u(x,y)-iv(x,y)}{h}}$

$={\displaystyle \frac{u(x+h,y)-u(x,y)}{h}}+{\displaystyle \frac{v(x+h,y)-v(x,y)}{h}}i.$

As $h\to 0$ along the real line, the limit—if it exists—is given by

$$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial x}i.$$

Let’s now take the limit along the imaginary axis; for clarity, we’ll still take $h$ to be a real number, and change $z$ by $ih$. We have

${\displaystyle \frac{f(z+ih)-f(z)}{ih}}$ $={\displaystyle \frac{f(x+i(y+h))-f(x+iy)}{ih}}$

$={\displaystyle \frac{u(x,y+h)+iv(x,y+h)-u(x,y)-iv(x,y)}{ih}}$

$=-{\displaystyle \frac{u(x,y+h)-u(x,y)}{h}}i+{\displaystyle \frac{v(x,y+h)-v(x,y)}{h}}.$

Taking the limit as $h\to 0$, the right-hand side becomes

$$-\frac{\partial u}{\partial y}i+\frac{\partial v}{\partial y}.$$

Now, if $f$ is going to be differentiable at $z$, these limits must both exist and must both agree. So if $f$ is differentiable at $z$, both first-order partial derivatives of $u$ and $v$ must exist; and, equating the expressions above and taking real and imaginary parts, we must have

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$

These are called the Cauchy–Riemann equations.

Theorem.

Let $f=u+iv$ be a complex-valued function on a domain $D$. Then $f$ is analytic on $D$ if and only if $u$ and $v$ have first-order partial derivatives defined and continuous everywhere on $D$ which satisfy the Cauchy–Riemann equations.

We have already essentially shown one direction of this result: if $f$ is differentiable at $z$, then it satisfies the Cauchy–Riemann equations at $z$. (To replace “differentiable” with “analytic,” we only need to add the requirement that the partial derivatives be continuous.) What remains is to prove the converse: if $u$ and $v$ have partial derivatives satisfying the Cauchy–Riemann equations, then $f$ is analytic.

We show this using first-order approximation: if $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ are defined near $(x,y)$, then for small real numbers $j$, $k$ we have

$$u(x+j,y+k)=u(x,y)+\frac{\partial u}{\partial x}\cdot j+\frac{\partial u}{\partial y}\cdot k+R(j,k)$$

where $R$ is some function such that ${lim}_{j\to 0}{lim}_{k\to 0}\frac{R(j,k)}{|(j,k)|}=0$. The same formula holds for $v$ in place of $u$; write $S(j,k)$ for the remainder term in place of $R$. Then we can compute

${\displaystyle \frac{f(z+h)-f(z)}{h}}$ $={\displaystyle \frac{f(x+j+i(y+k))-f(x+iy)}{j+ik}}$

$={\displaystyle \frac{u(x+j,y+k)+iv(x+j,y+k)-u(x,y)-iv(x,y)}{j+ik}}$

$={\displaystyle \frac{\frac{\partial u}{\partial x}\cdot j+\frac{\partial u}{\partial y}\cdot k+R(j,k)+i\left(\frac{\partial v}{\partial x}\cdot j+\frac{\partial v}{\partial y}\cdot k+S(j,k)\right)}{j+ik}}.$

If we assume that the Cauchy–Riemann equations hold, so that $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$, we can rewrite this as

$$\frac{\frac{\partial u}{\partial x}(j+ik)+\frac{\partial u}{\partial y}(k-ij)+R(j,k)+iS(j,k)}{j+ik}=\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}i+\frac{R(j,k)+iS(j,k)}{j+ik}.$$

Taking the limit as $j,k\to 0$, the third term we know tends to zero in absolute value, so we can drop it, and we’re left with the first two terms which by assumption are well-defined and continuous on $D$. Hence $f$ is in fact analytic on $D$ under these assumptions.

We check some examples. For $f(z)=z=x+iy$, we have $u(x,y)=x$ and $v(x,y)=y$, so $\frac{\partial u}{\partial x}=1$, $\frac{\partial u}{\partial y}=0$, $\frac{\partial v}{\partial x}=0$, and $\frac{\partial v}{\partial y}=1$, so the equations hold by inspection.

A slightly more complicated example is $f(z)=e^z$. Writing $z=x+iy$, this is $e^x\cos y+i{e}^x\sin y$ by Euler’s formula, so $u(x,y)=e^x\cos y$ and $v(x,y)=e^x\sin y$. We find

$$\frac{\partial u}{\partial x}=e^x\cos y,\frac{\partial u}{\partial y}=-e^x\sin y,\frac{\partial v}{\partial x}=e^x\sin y,\frac{\partial v}{\partial y}=e^x\cos y,$$

so again the equations hold.

Since the equations are linear, linear combinations of analytic functions are analytic (which we already know by rules of differentiation). A more important property is the following, reflecting a basic principle of integral calculus.

Proposition.

If $f(z)$ is analytic on a domain $D$ and $f^{\prime }(z)=0$ for all $z\in D$, then $f$ is constant on $D$.

Indeed, if $f$ is analytic then it satisfies the Cauchy–Riemann equations, and we computed above that then

$$f^{\prime }(z)=\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}i=0,$$

so taking real and imaginary parts we see that $u$ has both partial derivatives everywhere zero, hence $u$ is constant. By the Cauchy–Riemann equations, both partial derivatives of $v$ are also zero, so $v$ is also constant, hence $f$ is constant.

One can also check other analogous properties. For example, if $f$ is real-valued and analytic on a domain $D$, then it must be constant. This might be surprising, since we’re used to restricting to the real line (or a subset of it) to recover a real-valued function; but note that no (nonempty) subset of the real line is a domain! Indeed, using the Cauchy–Riemann equations, if $v=0$ then its partial derivatives both vanish, so so do those of $u$, so $u$ must be constant, so $f=u+0$ is constant. This immediately tells us that for example $f(z)={|z|}^2=x^2+y^2$ is not an analytic function; this can be verified from the Cauchy–Riemann equation.

Note again that from the apparently weak condition that a function be complex-differentiable, we’ve deduced that it must in fact satisfy a pair of differential equations, a much stronger-looking condition! This relates back to our slogan from last time: the existence (and conditions on the behavior) of complex limits is a much stronger assumption than it looks.

In everything above, we used Cartesian coordinates. However, we could just as well have used polar coordinates: writing $z=r{e}^{i\theta }$ and $f(z)=u+iv$ with $u$ and $v$ viewed as functions of $r$ and $\theta$, by varying either $r$ or $\theta$, one can derive the polar form of the Cauchy–Riemann equations

$$\frac{\partial u}{\partial r}=\frac{1}{r}\cdot \frac{\partial v}{\partial \theta }\frac{\partial u}{\partial \theta }=-r\frac{\partial v}{\partial r}.$$

A slightly different application of the Cauchy–Riemann equations is to showing that inverse functions of analytic functions are analytic, at least after restricting to some neighborhood of a given point (if nothing else to ensure that they are single-valued). Viewing $D\subset ℂ\simeq {ℝ}^2$ and $f:D\to ℂ\simeq {ℝ}^2$ as a function from a region in the plane to the plane, from multivariable calculus the invertibility of $f$ is determined by the Jacobian matrix

In particular, $f$ is invertible when $J_f$ is, equivalently when $det{J}_f$ is nonzero. Using the Cauchy–Riemann equations to substitute for the partial derivatives in $y$, we can write the determinant as

$${\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial v}{\partial x}\right)}^2={\left|\frac{\partial f}{\partial x}\right|}^2.$$

Since $f$ is analytic, the partial derivative with respect to $x$ is just $f^{\prime }$, so this determinant is just ${|f^{\prime }(z)|}^2$.

Therefore if $f^{\prime }(z)\ne 0$, then $f$ is locally invertible at $z$. The standard formula from calculus tells us that the derivative of $f^{-1}$ at $f(z)$ is $\frac{1}{f^{\prime }(z)}$, so since $f^{\prime }$ is nonzero and continuous at $z$ by analyticity, it follows that $f^{-1}$ is also analytic at $z$.

For example, the principal branch $\mathrm{Log}z$ of the logarithm is analytic away from the branch cut at $\theta =\pi$ and has derivative at $z=e^w$ given by $\frac{1}{e^w}=\frac{1}{z}$. Since any other branch of the logarithm differs from $\mathrm{Log}z$ by a constant, their derivatives are the same, so any branch of the logarithm has derivative $\frac{1}{z}$, which is well-defined and continuous away from $z=0$.