The Central Idea
Let
\[ z=x+iy \]and write
\[ f(z)=u(x,y)+iv(x,y). \]If \(f\) is complex differentiable, then its real and imaginary components satisfy
\[ \boxed{u_x=v_y} \]and
\[ \boxed{u_y=-v_x}. \]These are the Cauchy–Riemann relationships.
Why These Relationships Appear
The complex derivative is
\[ f'(z) = \lim_{h\to0} \frac{f(z+h)-f(z)}{h}. \]Because \(h\) may approach \(0\) from any direction in the complex plane, the same derivative must result whether we move horizontally, vertically, or along any other path.
Comparing horizontal and vertical changes produces the Cauchy–Riemann equations.
Approach Along the Real Direction
Let the increment be real:
\[ h=\Delta x. \]Then
\[ f(z+h) = u(x+\Delta x,y) + iv(x+\Delta x,y). \]The difference quotient becomes
\[ \frac{ u(x+\Delta x,y)-u(x,y) }{\Delta x} + i \frac{ v(x+\Delta x,y)-v(x,y) }{\Delta x}. \]Taking the limit gives
\[ \boxed{ f'(z)=u_x+iv_x. } \]Approach Along the Imaginary Direction
Now let
\[ h=i\Delta y. \]Then
\[ f(z+h) = u(x,y+\Delta y) + iv(x,y+\Delta y). \]The difference quotient is
\[ \frac{ u(x,y+\Delta y)-u(x,y) + i[v(x,y+\Delta y)-v(x,y)] }{i\Delta y}. \]Since
\[ \frac1i=-i, \]the limiting expression becomes
\[ \boxed{ f'(z)=v_y-iu_y. } \]Equating the Two Directional Forms
If the complex derivative exists, both directional calculations must agree:
\[ u_x+iv_x = v_y-iu_y. \]Equal complex numbers have equal real and imaginary parts.
Therefore,
\[ \boxed{u_x=v_y} \]and
\[ \boxed{v_x=-u_y}. \]Equivalently,
\[ \boxed{u_y=-v_x}. \]The Standard Cartesian Form
The Cauchy–Riemann relationships are most commonly written
\[ \boxed{ u_x=v_y, \qquad u_y=-v_x. } \]This compact pair should become immediately recognizable when studying complex differentiability.
A Necessary Condition for Differentiability
If \(f\) is complex differentiable at a point and the relevant partial derivatives exist there, then the Cauchy–Riemann relationships must hold at that point.
Therefore, if either equation fails,
\[ u_x\neq v_y \]or
\[ u_y\neq -v_x, \]then
\[ \boxed{ f\text{ is not complex differentiable there}. } \]A Practical Sufficient Condition
Suppose \(u_x,u_y,v_x,v_y\) exist and are continuous in a neighborhood of \(z_0\).
If the Cauchy–Riemann relationships hold throughout that neighborhood, then \(f\) is complex differentiable there.
Thus, under these regularity assumptions,
\[ \boxed{ u_x=v_y, \qquad u_y=-v_x } \]provides an efficient test for analyticity.
An Important Nuance
Merely checking the Cauchy–Riemann equations at one isolated point is not always sufficient to prove differentiability there.
Additional regularity of the component functions is needed for the standard sufficient theorem.
Therefore, distinguish carefully between:
- necessary conditions;
- sufficient conditions;
- conditions holding at one point;
- conditions holding throughout a neighborhood.
Computing the Derivative from \(u\) and \(v\)
When \(f\) is complex differentiable,
\[ \boxed{ f'(z)=u_x+iv_x. } \]Using the Cauchy–Riemann equations, the same derivative can also be written
\[ \boxed{ f'(z)=v_y-iu_y. } \]These equivalent formulas provide a useful internal check.
Example: Verify the Relationships for \(f(z)=z^2\)
Write
\[ z=x+iy. \]Then
\[ z^2=x^2-y^2+i(2xy). \]Thus,
\[ u=x^2-y^2, \qquad v=2xy. \]Compute the partial derivatives:
\[ u_x=2x, \qquad u_y=-2y, \] \[ v_x=2y, \qquad v_y=2x. \]Therefore,
\[ u_x=v_y \]and
\[ u_y=-v_x. \]The relationships hold everywhere.
Recovering the Derivative of \(z^2\)
Using
\[ f'(z)=u_x+iv_x, \]we obtain
\[ f'(z) = 2x+i(2y). \]Since
\[ z=x+iy, \]this becomes
\[ \boxed{f'(z)=2z}. \]Example: Verify the Relationships for \(e^z\)
Recall
\[ e^{x+iy} = e^x\cos y + i e^x\sin y. \]Thus,
\[ u=e^x\cos y \]and
\[ v=e^x\sin y. \]Compute:
\[ u_x=e^x\cos y, \qquad u_y=-e^x\sin y, \] \[ v_x=e^x\sin y, \qquad v_y=e^x\cos y. \]Therefore,
\[ u_x=v_y \]and
\[ u_y=-v_x. \]Hence the relationships hold everywhere, consistent with \(e^z\) being entire.
Nonexample: \(f(z)=\overline{z}\)
If
\[ z=x+iy, \]then
\[ \overline{z}=x-iy. \]Therefore,
\[ u=x, \qquad v=-y. \]The partial derivatives are
\[ u_x=1, \qquad v_y=-1. \]Since
\[ u_x\neq v_y, \]the Cauchy–Riemann relationships fail everywhere.
Therefore,
\[ \boxed{ \overline{z} \text{ is nowhere complex differentiable}. } \]Nonexample: \(f(z)=\operatorname{Re}(z)\)
Here,
\[ u=x, \qquad v=0. \]Thus,
\[ u_x=1, \qquad v_y=0. \]The first Cauchy–Riemann equation fails:
\[ 1\neq0. \]Hence,
\[ \boxed{ \operatorname{Re}(z) \text{ is nowhere complex differentiable}. } \]Example: \(f(z)=z\overline{z}\)
Since
\[ z\overline{z}=x^2+y^2, \]we have
\[ u=x^2+y^2, \qquad v=0. \]Therefore,
\[ u_x=2x, \qquad u_y=2y, \] \[ v_x=0, \qquad v_y=0. \]The Cauchy–Riemann equations require
\[ 2x=0 \]and
\[ 2y=0. \]Thus they hold only at
\[ \boxed{(x,y)=(0,0)}. \]Indeed, the function is differentiable at \(z=0\), but nowhere else.
A Systematic Cartesian Test
For
\[ f(z)=u(x,y)+iv(x,y), \]use the following structure:
\[ \begin{array}{c|cc} & x & y\\ \hline u & u_x & u_y\\ v & v_x & v_y \end{array} \]Then compare:
\[ \boxed{u_x\overset{?}=v_y} \]and
\[ \boxed{u_y\overset{?}=-v_x}. \]The Jacobian Matrix
The real two-dimensional derivative of
\[ (x,y)\mapsto(u(x,y),v(x,y)) \]is represented by the Jacobian matrix
\[ J_f = \begin{pmatrix} u_x & u_y\\ v_x & v_y \end{pmatrix}. \]When the Cauchy–Riemann relationships hold,
\[ J_f = \begin{pmatrix} u_x & -v_x\\ v_x & u_x \end{pmatrix}. \]Why the Jacobian Has This Form
Multiplication by a complex number
\[ a+ib \]corresponds in real coordinates to the matrix
\[ \boxed{ \begin{pmatrix} a & -b\\ b & a \end{pmatrix}. } \]The Cauchy–Riemann equations force the Jacobian of a differentiable complex function into exactly this structure.
This explains why the derivative acts locally like complex multiplication.
Geometric Meaning
If
\[ f'(z_0)=a+ib\neq0, \]then the local linear map has matrix
\[ \begin{pmatrix} a & -b\\ b & a \end{pmatrix}. \]This matrix performs a uniform scaling and rotation.
Thus the Cauchy–Riemann structure is the real-coordinate signature of local complex multiplication.
Local Scale Factor
If
\[ f'(z_0)=a+ib, \]then the local scale factor is
\[ \boxed{ |f'(z_0)| = \sqrt{a^2+b^2}. } \]Small lengths are approximately multiplied by this amount.
Local Rotation
The local rotation angle is
\[ \boxed{ \arg f'(z_0). } \]Every sufficiently small direction near \(z_0\) is rotated by approximately the same angle.
Connection to Conformal Mapping
When \(f\) is analytic and
\[ f'(z_0)\neq0, \]the local map behaves like nonzero complex multiplication.
Therefore, angles between sufficiently smooth intersecting curves are preserved locally.
This is the foundation of conformal mapping.
Connection to Harmonic Functions
Suppose
\[ f=u+iv \]is analytic and the necessary second partial derivatives are continuous.
From
\[ u_x=v_y \]and
\[ u_y=-v_x, \]we can derive
\[ \boxed{ u_{xx}+u_{yy}=0 } \]and
\[ \boxed{ v_{xx}+v_{yy}=0. } \]Laplace's Equation
The equation
\[ \boxed{ \phi_{xx}+\phi_{yy}=0 } \]is called Laplace's equation.
A twice-differentiable real-valued function satisfying it is called harmonic.
Thus the real and imaginary parts of analytic functions are harmonic under the standard smoothness assumptions.
Harmonic Conjugates
If
\[ f(z)=u(x,y)+iv(x,y) \]is analytic, then \(u\) and \(v\) are called harmonic conjugates.
The Cauchy–Riemann equations allow one component to be reconstructed from the other, subject to domain and consistency conditions.
Finding a Harmonic Conjugate
Suppose \(u(x,y)\) is known and we seek \(v(x,y)\).
From
\[ u_x=v_y, \]integrate \(u_x\) with respect to \(y\):
\[ v(x,y) = \int u_x\,dy + \phi(x). \]Then use
\[ u_y=-v_x \]to determine the remaining function \(\phi(x)\).
Example: Find a Harmonic Conjugate
Let
\[ u(x,y)=x^2-y^2. \]Then
\[ u_x=2x. \]Since
\[ v_y=u_x=2x, \]integrate with respect to \(y\):
\[ v(x,y)=2xy+\phi(x). \]Now
\[ u_y=-2y. \]The second Cauchy–Riemann equation requires
\[ -v_x=-2y. \]Since
\[ v_x=2y+\phi'(x), \]we get
\[ -2y-\phi'(x)=-2y. \]Thus
\[ \phi'(x)=0. \]Therefore,
\[ \boxed{ v(x,y)=2xy+C. } \]Recovering the Analytic Function
With
\[ u=x^2-y^2 \]and
\[ v=2xy+C, \]we obtain
\[ f(z) = x^2-y^2+i(2xy+C). \]Since
\[ z^2=x^2-y^2+i2xy, \]the function is
\[ \boxed{ f(z)=z^2+iC. } \]Finding \(u\) from a Known \(v\)
If \(v(x,y)\) is known, use
\[ u_x=v_y \]to integrate with respect to \(x\):
\[ u(x,y) = \int v_y\,dx + \psi(y). \]Then apply
\[ u_y=-v_x \]to determine \(\psi(y)\).
The Gradients of \(u\) and \(v\)
The gradients are
\[ \nabla u = (u_x,u_y) \]and
\[ \nabla v = (v_x,v_y). \]Using the Cauchy–Riemann equations,
\[ \nabla v = (-u_y,u_x). \]Thus \(\nabla v\) is obtained by rotating \(\nabla u\) by \(90^\circ\).
Orthogonal Level Curves
Because
\[ \nabla u\cdot\nabla v = u_xv_x+u_yv_y, \]the Cauchy–Riemann equations give
\[ \nabla u\cdot\nabla v=0. \]Therefore, where the gradients are nonzero, level curves of \(u\) and \(v\) intersect at right angles.
This provides another geometric manifestation of analyticity.
Equal Gradient Magnitudes
Using
\[ v_x=-u_y, \qquad v_y=u_x, \]we obtain
\[ |\nabla v|^2 = v_x^2+v_y^2 = u_y^2+u_x^2. \]Therefore,
\[ \boxed{ |\nabla u|=|\nabla v|. } \]The component fields have equal local rate magnitudes but directions rotated by \(90^\circ\).
Cauchy–Riemann Relationships in Polar Coordinates
Sometimes a complex function is more naturally written using
\[ z=re^{i\theta}. \]If
\[ f(z)=u(r,\theta)+iv(r,\theta), \]the Cartesian Cauchy–Riemann equations can be transformed into polar form.
The Polar Cauchy–Riemann Equations
For \(r>0\), the polar form is
\[ \boxed{ u_r=\frac1r v_\theta } \]and
\[ \boxed{ v_r=-\frac1r u_\theta. } \]Equivalently,
\[ \boxed{ r u_r=v_\theta, \qquad r v_r=-u_\theta. } \]The Derivative in Polar Coordinates
When the polar Cauchy–Riemann equations hold appropriately,
\[ f'(z) = e^{-i\theta} \left( u_r+iv_r \right). \]Another equivalent form can be obtained from the angular derivatives.
This formulation is useful for functions naturally expressed in terms of modulus and argument.
Example: \(f(z)=z^n\) in Polar Form
Write
\[ z=re^{i\theta}. \]Then
\[ z^n = r^n \left( \cos n\theta+i\sin n\theta \right). \]Thus,
\[ u=r^n\cos n\theta \]and
\[ v=r^n\sin n\theta. \]Their radial and angular derivatives satisfy the polar Cauchy–Riemann equations.
A Preview of Wirtinger Derivatives
Complex functions can also be studied using the formal differential operators
\[ \frac{\partial}{\partial z} = \frac12 \left( \frac{\partial}{\partial x} - i\frac{\partial}{\partial y} \right) \]and
\[ \frac{\partial}{\partial\overline{z}} = \frac12 \left( \frac{\partial}{\partial x} + i\frac{\partial}{\partial y} \right). \]These are called Wirtinger derivatives.
The Cauchy–Riemann Equations in Wirtinger Form
Under suitable smoothness assumptions, a function is holomorphic precisely when
\[ \boxed{ \frac{\partial f}{\partial\overline{z}}=0. } \]This compact statement packages the two Cauchy–Riemann equations into one complex equation.
It is especially useful in advanced complex analysis and partial differential equations.
Common Error: Losing the Negative Sign
The equations are
\[ u_x=v_y \]and
\[ \boxed{u_y=-v_x}. \]The negative sign in the second relationship is essential.
Omitting it destroys the rotational structure of complex multiplication.
Common Error: Checking Only One Equation
Both relationships must hold:
\[ u_x=v_y \]and
\[ u_y=-v_x. \]Satisfying only one is not enough.
Common Error: Proving Analyticity from One Point
If the Cauchy–Riemann equations hold at one point, that may help establish differentiability there under additional conditions.
But analyticity requires complex differentiability throughout an open neighborhood.
Always state whether the claim concerns:
- one point;
- a neighborhood;
- an entire region.
Common Error: Assuming Every Harmonic Function Automatically Has a Global Conjugate
A harmonic function may possess a harmonic conjugate locally, but global existence can depend on the topology of the domain.
Domain shape matters, especially when holes or excluded points are present.
Common Error: Ignoring Continuity of Partial Derivatives
In the standard practical theorem, continuity of the first partial derivatives in a neighborhood is part of the sufficient condition.
The equations alone should not be applied mechanically without checking the hypotheses of the theorem being used.
A Reliable Procedure for Testing the Cauchy–Riemann Relationships
- Write \(z=x+iy\).
- Rewrite the function as \(u(x,y)+iv(x,y)\).
- Compute \(u_x\) and \(u_y\).
- Compute \(v_x\) and \(v_y\).
- Check \(u_x=v_y\).
- Check \(u_y=-v_x\).
- Determine where both equations hold.
- Check the regularity assumptions needed for the differentiability conclusion.
- State precisely where the function is differentiable or analytic.
A Reliable Procedure for Finding a Harmonic Conjugate
- Start with the known component \(u\) or \(v\).
- Compute the appropriate partial derivatives.
- Use one Cauchy–Riemann equation to integrate for the missing component.
- Include an unknown function of the variable held constant during integration.
- Use the second Cauchy–Riemann equation to determine that function.
- Verify both equations.
- Combine \(u+iv\) and simplify in terms of \(z\) when possible.
Worked Example: Test \(f(z)=z^3\)
Expand
\[ (x+iy)^3. \]This gives
\[ x^3-3xy^2 + i(3x^2y-y^3). \]Therefore,
\[ u=x^3-3xy^2 \]and
\[ v=3x^2y-y^3. \]Compute:
\[ u_x=3x^2-3y^2, \qquad u_y=-6xy, \] \[ v_x=6xy, \qquad v_y=3x^2-3y^2. \]Therefore,
\[ u_x=v_y \]and
\[ u_y=-v_x \]everywhere.
Hence \(z^3\) is entire.
Worked Example: Test \(f(z)=x^2+y^2+i0\)
Let
\[ u=x^2+y^2, \qquad v=0. \]Then
\[ u_x=2x, \qquad u_y=2y, \] \[ v_x=0, \qquad v_y=0. \]The equations require
\[ x=0 \]and
\[ y=0. \]So they hold only at the origin.
This function is not analytic on any open neighborhood.
Worked Example: Verify Harmonicity
Consider
\[ u(x,y)=x^2-y^2. \]Then
\[ u_{xx}=2 \]and
\[ u_{yy}=-2. \]Therefore,
\[ u_{xx}+u_{yy}=0. \]Hence,
\[ \boxed{ u(x,y)=x^2-y^2 \text{ is harmonic}. } \]Worked Example: Build an Analytic Function from Its Real Part
Suppose
\[ u(x,y)=e^x\cos y. \]Then
\[ u_x=e^x\cos y. \]Since
\[ v_y=u_x, \]integrate:
\[ v=e^x\sin y+\phi(x). \]Now
\[ u_y=-e^x\sin y. \]The second Cauchy–Riemann equation requires
\[ -v_x=-e^x\sin y. \]Since
\[ v_x=e^x\sin y+\phi'(x), \]we obtain
\[ \phi'(x)=0. \]Thus
\[ v=e^x\sin y+C. \]Therefore,
\[ f(z) = e^x(\cos y+i\sin y)+iC. \]So
\[ \boxed{ f(z)=e^z+iC. } \]Practice 1: Verify the Relationships
For
\[ f(z)=z^2, \]verify the Cauchy–Riemann relationships directly.
Practice 2: Test a Nonexample
Determine whether
\[ f(z)=\overline{z} \]satisfies the Cauchy–Riemann relationships anywhere.
Practice 3: Find the Derivative
Suppose
\[ u=x^2-y^2, \qquad v=2xy. \]Use
\[ f'(z)=u_x+iv_x \]to find the derivative.
Practice 4: Find Where the Equations Hold
Let
\[ f(z)=x^2+y^2+i(x-y). \]Determine where the Cauchy–Riemann relationships hold.
Practice 5: Harmonic Test
Determine whether
\[ u(x,y)=x^3-3xy^2 \]is harmonic.
Practice 6: Harmonic Conjugate
Find a harmonic conjugate \(v(x,y)\) for
\[ u(x,y)=x^2-y^2. \]Practice 7: Recover an Analytic Function
Using the harmonic conjugate from Practice 6, write the corresponding analytic function in terms of \(z\).
Practice 8: Exponential Components
Verify the Cauchy–Riemann relationships for
\[ u=e^x\cos y, \qquad v=e^x\sin y. \]Practice 9: Geometric Meaning
If
\[ f'(z_0)=2+2i, \]find the local scale factor and rotation angle.
Practice 10: Polar Form
State the Cauchy–Riemann relationships in polar coordinates for \(r>0\).
Practice Answers
Practice 1
For
\[ z^2=x^2-y^2+i2xy, \]we have
\[ u_x=2x, \qquad v_y=2x, \]and
\[ u_y=-2y, \qquad -v_x=-2y. \]Therefore,
\[ \boxed{ u_x=v_y, \qquad u_y=-v_x. } \]Practice 2
For
\[ \overline{z}=x-iy, \]we have
\[ u_x=1 \]and
\[ v_y=-1. \]Since these are never equal,
\[ \boxed{ \overline{z} \text{ satisfies the equations nowhere}. } \]Practice 3
\[ u_x=2x, \qquad v_x=2y. \]Therefore,
\[ f'(z) = 2x+i2y = \boxed{2z}. \]Practice 4
Here,
\[ u=x^2+y^2, \qquad v=x-y. \]Then
\[ u_x=2x, \qquad u_y=2y, \] \[ v_x=1, \qquad v_y=-1. \]The equations require
\[ 2x=-1 \]and
\[ 2y=-1. \]Therefore they hold only at
\[ \boxed{ \left(-\frac12,-\frac12\right). } \]Practice 5
\[ u_x=3x^2-3y^2, \] \[ u_{xx}=6x. \]Also,
\[ u_y=-6xy, \] \[ u_{yy}=-6x. \]Therefore,
\[ u_{xx}+u_{yy}=0. \]Hence,
\[ \boxed{ u\text{ is harmonic}. } \]Practice 6
Since
\[ u_x=2x=v_y, \]integrate with respect to \(y\):
\[ v=2xy+\phi(x). \]The second equation forces
\[ \phi'(x)=0. \]Therefore,
\[ \boxed{ v=2xy+C. } \]Practice 7
Combining the components,
\[ f(z) = x^2-y^2+i2xy+iC. \]Therefore,
\[ \boxed{ f(z)=z^2+iC. } \]Practice 8
\[ u_x=e^x\cos y, \qquad v_y=e^x\cos y, \]and
\[ u_y=-e^x\sin y, \qquad -v_x=-e^x\sin y. \]Therefore,
\[ \boxed{ \text{both Cauchy–Riemann relationships hold everywhere}. } \]Practice 9
The local scale factor is
\[ |2+2i| = \sqrt{4+4} = 2\sqrt2. \]The argument is
\[ \frac{\pi}{4}. \]Therefore,
\[ \boxed{ \text{scale by }2\sqrt2 \text{ and rotate by }\frac{\pi}{4}. } \]Practice 10
\[ \boxed{ u_r=\frac1r v_\theta, \qquad v_r=-\frac1r u_\theta, \qquad r>0. } \]Checking Strategies
- Write the two equations side by side before substituting derivatives.
- Check the negative sign in \(u_y=-v_x\).
- State where the equations hold, not merely whether they hold somewhere.
- Distinguish pointwise satisfaction from analyticity on a neighborhood.
- When finding a harmonic conjugate, include the extra function of the variable held constant during integration.
- Verify a recovered analytic function by differentiating or rechecking both equations.
Connections to Earlier Mathematics
The Cauchy–Riemann relationships connect several major areas:
- Complex differentiation: they express directional consistency of the derivative.
- Multivariable calculus: they constrain the Jacobian matrix.
- Linear algebra: the Jacobian takes the matrix form of complex multiplication.
- Partial differential equations: analytic components become harmonic and satisfy Laplace's equation.
- Geometry: gradients of harmonic conjugates are orthogonal and equal in magnitude.
- Conformal mapping: the local differential becomes a scale-and-rotation map.
What to Remember
For
\[ f(z)=u(x,y)+iv(x,y), \]the Cartesian Cauchy–Riemann relationships are
\[ \boxed{ u_x=v_y, \qquad u_y=-v_x. } \]When the function is complex differentiable,
\[ \boxed{ f'(z)=u_x+iv_x=v_y-iu_y. } \]Under suitable smoothness assumptions, these equations provide a practical test for analyticity.
They also imply that the real and imaginary components of an analytic function are harmonic:
\[ \boxed{ u_{xx}+u_{yy}=0, \qquad v_{xx}+v_{yy}=0. } \]The essential structural idea is: complex differentiability forces two real component functions to behave like one coherent rotation-and-scaling transformation.
Where We Go Next
We now have a practical differential test for analytic behavior. The next major step is to integrate complex functions along curves in the complex plane.
The next lesson, Complex Integration, introduces parametrized contours, line integrals of complex functions, orientation, antiderivatives, and the basic computational rules that lead toward Cauchy's theorem and Cauchy's integral formula.