Cauchy-Riemann Equations

From Complex Derivatives to Real Partial Derivatives

Complex Differentiability established that if $f$ is complex differentiable at a point, then its real and imaginary components satisfy certain relations among their partial derivatives there. These are the Cauchy–Riemann equations. They express a necessary compatibility between changes in the two real coordinates: a complex derivative must give the same first-order behavior whether an increment is taken horizontally, vertically, or in any other direction.

The equations are useful as a test, but the hypotheses matter. Holding at a single point does not, by itself, guarantee a complex derivative there. In contrast, if the first partial derivatives are continuous in a neighborhood and the equations hold throughout that neighborhood, they do guarantee complex differentiability. We will establish this sufficient condition and use it to calculate derivatives.

Write $z=x+iy$ and express $f$ as $f(z)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real-valued functions. The symbols $u_x,u_y,v_x,v_y$ denote their real partial derivatives with respect to the coordinates $x$ and $y$.

Definition (Cauchy–Riemann Equations): The Cauchy–Riemann equations for $f=u+iv$ at $(x,y)$ are $$ u_x(x,y)=v_y(x,y),\qquad u_y(x,y)=-v_x(x,y). $$ When these equations hold at every point of a set, we say that $f$ satisfies the Cauchy–Riemann equations on that set.

Why the Equations Are Necessary

The equations can be seen by taking the complex difference quotient along the two coordinate directions. If $f$ is complex differentiable at $z_0=x_0+iy_0$, then increments $h$ along the real axis and increments $ik$ along the imaginary axis must produce the same derivative. Complex Differentiability established the resulting theorem, so we recall it here rather than prove it again.

Theorem (Cauchy–Riemann Equations as Necessary Conditions): Suppose $f=u+iv$ is complex differentiable at $z_0=x_0+iy_0$. Then the real partial derivatives of $u$ and $v$ exist at $(x_0,y_0)$, and $$ u_x(x_0,y_0)=v_y(x_0,y_0),\qquad u_y(x_0,y_0)=-v_x(x_0,y_0). $$

When the equations hold and the complex derivative exists, they also identify its value. Along real increments, the difference quotient tends to $u_x+iv_x$. Along imaginary increments, it tends to $v_y-iu_y$. The equations make these expressions equal, since $v_y=u_x$ and $-u_y=v_x$. Thus the derivative, when it exists, is $u_x+iv_x$. The next result shows when this candidate is indeed the derivative.

Worked Example: Checking the Equations for a Polynomial

Let $f(z)=z^2$. Expanding $(x+iy)^2$ gives

$$ f(z)=x^2-y^2+i(2xy), $$

so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$. Their partial derivatives are

u_x=2x,\qquad u_y=-2y,\qquad v_x=2y,\qquad v_y=2x.

Therefore $u_x=v_y=2x$, and $u_y=-2y=-v_x$. Both equations hold everywhere. The partial derivatives are continuous, so the sufficient condition proved below applies on all of $\mathbb{C}$. It follows that $f$ is holomorphic, with derivative $f'(z)=u_x+iv_x=2x+2iy=2z$.

A Sufficient Condition for Complex Differentiability

The key additional assumption is continuity of the first partial derivatives in a neighborhood—not just their existence at the point in question. Continuity lets us approximate the change in each component by its linear terms, with an error that becomes negligible compared with the size of the increment.

Theorem (Cauchy–Riemann Sufficiency Criterion): Let $U\subseteq\mathbb{C}$ be open, and let $f=u+iv:U\to\mathbb{C}$. Suppose the first partial derivatives $u_x,u_y,v_x,v_y$ exist and are continuous on $U$. If $f$ satisfies the Cauchy–Riemann equations at every point of $U$, then $f$ is holomorphic on $U$, and $$ f'(x+iy)=u_x(x,y)+iv_x(x,y). $$

Proof. Fix an arbitrary point $z_0=x_0+iy_0\in U$. Since $U$ is open, a rectangle around $(x_0,y_0)$ lies in $U$. Let $h,k\in\mathbb{R}$ be small enough that the points used below remain in this rectangle, and set $r=\sqrt{h^2+k^2}$.

We first obtain a real first-order approximation for $u$. Split its change into a horizontal change followed by a vertical one:

\begin{aligned} u(x_0+h,y_0+k)-u(x_0,y_0) ={}&u(x_0+h,y_0+k)-u(x_0,y_0+k)\\ &+u(x_0,y_0+k)-u(x_0,y_0). \end{aligned}

By the one-variable mean value theorem applied to each difference, the right-hand side equals $h\,u_x(\xi,y_0+k)+k\,u_y(x_0,\eta)$ for some $\xi$ between $x_0$ and $x_0+h$, and some $\eta$ between $y_0$ and $y_0+k$. This statement also holds when $h=0$ or $k=0$, with the corresponding difference equal to zero. As $(h,k)\to(0,0)$, both derivative values tend to their values at $(x_0,y_0)$, by continuity. Consequently,

$$ u(x_0+h,y_0+k)-u(x_0,y_0) =u_x(x_0,y_0)h+u_y(x_0,y_0)k+o(r). $$

The same argument applied to $v$ gives

$$ v(x_0+h,y_0+k)-v(x_0,y_0) =v_x(x_0,y_0)h+v_y(x_0,y_0)k+o(r). $$

Combining these expansions, and abbreviating the partial derivatives at $(x_0,y_0)$, yields

\begin{aligned} f(z_0+h+ik)-f(z_0) &=(u_x+iv_x)h+(u_y+iv_y)k+o(r). \end{aligned}

The Cauchy–Riemann equations give $u_y=-v_x$ and $v_y=u_x$. Hence $u_y+iv_y=-v_x+iu_x=i(u_x+iv_x)$, and therefore

$$ f(z_0+h+ik)-f(z_0) =(u_x+iv_x)(h+ik)+o(r). $$

The complex increment is $\Delta z=h+ik$, and $|\Delta z|=r$. For nonzero increments, divide by $\Delta z$. The error term divided by $\Delta z$ tends to zero because its magnitude is $o(r)/r$. Thus the complex difference quotient tends to $u_x(x_0,y_0)+iv_x(x_0,y_0)$. The point $z_0$ was arbitrary, so $f$ is complex differentiable at every point of $U$, hence holomorphic there, with the stated derivative. $\square$

Using the Criterion

The theorem provides a practical route from real-variable calculations to holomorphicity: express the real and imaginary parts, calculate their first partial derivatives, check the equations, and verify continuity of those partial derivatives on the open set. The derivative formula then follows from the proof.

Worked Example: The Reciprocal Function

On $U=\mathbb{C}\setminus\{0\}$, write $f(z)=1/z$ in real coordinates. Multiplying numerator and denominator by $x-iy$ gives

f(x+iy)=\frac{x-iy}{x^2+y^2} =\frac{x}{x^2+y^2}+i\left(-\frac{y}{x^2+y^2}\right).

Thus $u=x/(x^2+y^2)$ and $v=-y/(x^2+y^2)$. The denominator is nonzero on $U$, so these functions have continuous first partial derivatives there. Direct differentiation gives

u_x=\frac{y^2-x^2}{(x^2+y^2)^2},\qquad v_y=\frac{y^2-x^2}{(x^2+y^2)^2},

u_y=\frac{-2xy}{(x^2+y^2)^2},\qquad v_x=\frac{2xy}{(x^2+y^2)^2}.

Therefore $u_x=v_y$, and $u_y=-v_x$, throughout $U$. The sufficiency criterion shows that $1/z$ is holomorphic on the punctured plane and that

f'(z)=u_x+iv_x =\frac{y^2-x^2+2ixy}{(x^2+y^2)^2} =-\frac{1}{z^2}.

For the last identity, $z^2=(x^2-y^2)+2ixy$, so $-\overline{z}^{\,2}/(x^2+y^2)^2$ has numerator $y^2-x^2+2ixy$. To check the displayed derivative carefully, note that $iv_x=2ixy/(x^2+y^2)^2$, while $-1/z^2=-(x-iy)^2/(x^2+y^2)^2=(-x^2+y^2+2ixy)/(x^2+y^2)^2$. These expressions agree.

Why a Pointwise Check Is Not Enough

The sufficiency criterion assumes continuous first partial derivatives on an open set, along with the equations throughout that set. Removing these hypotheses can invalidate the conclusion. In particular, even if all four partial derivatives exist at one point and the Cauchy–Riemann equations hold there, the complex derivative need not exist.

Worked Example: The Equations Hold at the Origin but No Derivative Exists

Define $f(0)=0$, and for $z=x+iy\neq0$, define

f(x+iy)=\frac{x^3+iy^3}{x^2+y^2}.

Write $f=u+iv$. At the origin, the coordinate-axis values are

u(t,0)=t,\qquad u(0,t)=0,\qquad v(t,0)=0,\qquad v(0,t)=t.

Using the definitions of the partial derivatives, these identities give $u_x(0,0)=1$, $u_y(0,0)=0$, $v_x(0,0)=0$, and $v_y(0,0)=1$. In particular, $u_x(0,0)=v_y(0,0)$ and $u_y(0,0)=-v_x(0,0)$: the Cauchy–Riemann equations hold at the origin.

Nevertheless, the complex difference quotient has different limits along two paths. For nonzero real $t$, substituting $x=t,y=0$ into the definition gives $f(t)=t^3/t^2=t$, so $f(t)/t=1$. On the line $z=t(1+i)$, substituting $x=y=t$ gives

f(t(1+i))=\frac{t^3+it^3}{2t^2} =\frac{t(1+i)}{2}, \qquad \frac{f(t(1+i))}{t(1+i)}=\frac12.

The two quotient values are $1$ and $1/2$, so the difference quotient has no limit as $z\to0$. Thus $f$ is not complex differentiable at the origin, despite satisfying the equations there. This is why the pointwise equations are necessary conditions, not a standalone sufficiency test.

Worked Example: Complex Conjugation Fails the Equations

For $f(z)=\overline{z}=x-iy$, the components are $u(x,y)=x$ and $v(x,y)=-y$. Their partial derivatives are $u_x=1$, $u_y=0$, $v_x=0$, and $v_y=-1$. The first Cauchy–Riemann equation would require $1=-1$, so it fails at every point.

This failure agrees with the difference quotient: for a nonzero real increment $h$, $\overline{h}/h=1$; for a nonzero purely imaginary increment $h=ik$, $\overline{ik}/(ik)=-1$. Since these values differ, $\overline{z}$ is not complex differentiable at any point.

What the Equations Do—and Do Not—Tell Us

The Cauchy–Riemann equations translate a complex differentiability question into relations among real partial derivatives. If a function is holomorphic, the equations must hold wherever its component partial derivatives are considered. Conversely, the equations together with continuous first partial derivatives on an open set guarantee holomorphicity there. In that setting, the derivative can be read off as $u_x+iv_x$.

A common pitfall is to check only the equations at one point and conclude that a complex derivative exists. The example at the origin shows why that inference fails: different directions can still give different difference-quotient limits. The neighborhood continuity assumption in the sufficiency criterion rules out this failure by ensuring a valid first-order approximation in every direction.

Check Your Understanding

Use the definitions, examples, and theorem in this tutorial to answer the following questions.

State the two Cauchy–Riemann equations for $f=u+iv$.
What additional hypotheses make the Cauchy–Riemann equations a sufficient condition for holomorphicity on an open set?
When the sufficiency criterion applies, how can $f'$ be expressed using partial derivatives of $u$ and $v$?
For $f(z)=\overline z$, which Cauchy–Riemann equation fails?
In the example defined by $f(x+iy)=(x^3+iy^3)/(x^2+y^2)$ away from zero, what two difference-quotient values are obtained along the real axis and the line $z=t(1+i)$?

What You'll Learn