Cauchy-Riemann equations

The Cauchy–Riemann equations are the pair of partial differential equations that test whether a complex function $f(z)=u(x,y)+iv(x,y)$ is complex differentiable:

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

These are necessary conditions for $f$ to be complex differentiable at a point. Together with continuity of the partials in a neighborhood, they are also sufficient.

Derivation

Take the Complex derivative definition $f^{\prime }(z_0)={\lim }_{\Delta z\to 0}(f(z_0+\Delta z)-f(z_0))/\Delta z$ and approach $0$ along two perpendicular directions.

Along the real axis ($\Delta z=\Delta x$, real):

$f^{\prime }(z_0{)}_{\text{real approach}}={\lim }_{\Delta x\to 0}\frac{f(z_0+\Delta x)-f(z_0)}{\Delta x}=\frac{\partial f}{\partial x}{|}_{z_0}=u_x+i{v}_x.$

Along the imaginary axis ($\Delta z=i\Delta y$):

$f^{\prime }(z_0{)}_{\text{imag approach}}={\lim }_{\Delta y\to 0}\frac{f(z_0+i\Delta y)-f(z_0)}{i\Delta y}=\frac{1}{i}\frac{\partial f}{\partial y}{|}_{z_0}=-i(u_y+i{v}_y)=v_y-i{u}_y.$

If $f^{\prime }(z_0)$ exists, both expressions must agree. Equating real and imaginary parts:

$u_x=v_y,v_x=-u_y.$

These are the C–R equations.

Furthermore, the derivative can be computed two equivalent ways:

$f^{\prime }(z_0)=u_x+i{v}_x=v_y-i{u}_y.$

Verifying on known examples

$f(z)=z^2$. $u=x^2-y^2$, $v=2xy$. Partials: $u_x=2x$, $u_y=-2y$, $v_x=2y$, $v_y=2x$. C–R: $u_x=v_y$ ($2x=2x$) ✓; $u_y=-v_x$ ($-2y=-2y$) ✓. So $z^2$ is analytic. Derivative: $f^{\prime }(z)=u_x+i{v}_x=2x+i2y=2z$.

$f(z)=\bar{z}$. $u=x$, $v=-y$. Partials: $u_x=1$, $u_y=0$, $v_x=0$, $v_y=-1$. C–R fails: $u_x=1\ne -1=v_y$. Violates C–R everywhere. $\bar{z}$ is differentiable nowhere — matches the Complex derivative result.

$f(z)=e^z$. $u=e^x\cos y$, $v=e^x\sin y$. Partials: $u_x=e^x\cos y=v_y$, $u_y=-e^x\sin y=-v_x$. C–R holds everywhere. $f^{\prime }=e^x\cos y+i{e}^x\sin y=e^z$ — the complex exponential is its own derivative.

C–R alone is not sufficient

The textbook standard counterexample:

$f(z)=\{\begin{array}{ll}(\bar{z}{)}^2/z& z\ne 0\\ 0& z=0.\end{array}$

Direct computation gives $u_x(0)=1$, $v_y(0)=1$, $u_y(0)=0$, $v_x(0)=0$, so C–R holds at $z=0$. But $f$ is not complex differentiable at $0$: along $\Delta z=h$ (real) the quotient is $1$; along $\Delta z=h(1+i)$ the quotient is $-1$. The function fails the limit-along-all-directions test.

C–R plus continuity is sufficient

Theorem (sufficient condition for differentiability). If $u$ and $v$ have continuous first partial derivatives in an open neighborhood of $z_0$ and satisfy the C–R equations at $z_0$, then $f=u+iv$ is complex differentiable at $z_0$, with $f^{\prime }(z_0)=u_x+i{v}_x$.

The continuity-on-a-neighborhood condition is what makes the tangent-plane approximation work uniformly. In practice: check C–R, check that partials are continuous, conclude differentiable.

For polynomial expressions in $x,y$, partials are automatically continuous, so C–R alone is the effective test.

Polar form

For $z=r{e}^{i\theta }$ and $f(z)=u(r,\theta )+iv(r,\theta )$, the polar version of C–R:

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

Convenient when the function is naturally polar (like $\log z$, $z^n$).

C–R as conservative-and-source-free

The deepest payoff. The C–R equations $u_x=v_y$ and $u_y=-v_x$ are exactly the conditions for the 2D vector field $⟨u,-v⟩$ to be both conservative and source-free:

• Conservative: cross-partials match. For $\mathbf{F}=⟨P,Q⟩=⟨u,-v⟩$, need $P_y=Q_x$, i.e., $u_y=-v_x$.
• Source-free: $\nabla \cdot \mathbf{F}=0$. $P_x+Q_y=u_x-v_y=0$ iff $u_x=v_y$.

So a complex function $f=u+iv$ is analytic iff the underlying vector field $⟨u,-v⟩$ is both conservative and source-free.

This connects Cauchy’s theorem (closed-contour integrals of analytic functions vanish) and the Residue theorem to Green’s theorem and the Divergence theorem — they’re 2D vector calculus identities, repackaged for the field $⟨u,-v⟩$ in complex notation.

Unpacking $f(z)dz=(u+iv)(dx+idy)=(udx-vdy)+i(vdx+udy)$:

$\oint f(z)dz=\underset{\text{circulation of }⟨u,-v⟩}{\underbrace{\oint udx-vdy}}+i\underset{\text{flux of }⟨u,-v⟩\text{ in disguise}}{\underbrace{\oint vdx+udy}}.$

When $f$ is analytic, both real and imaginary parts vanish by Green’s theorem — circulation of a conservative field, flux of a source-free field. Cauchy’s theorem is two applications of Green’s theorem combined into one complex identity.

Connection to harmonicity

If $f$ is analytic, $u$ and $v$ are each harmonic: ${\nabla }^{2}u={\nabla }^{2}v=0$. The proof differentiates the C–R equations and adds:

$u_{xx}+u_{yy}=(v_y{)}_x+(-v_x{)}_y=v_{yx}-v_{xy}=0$

by Clairaut’s theorem. Same for $v$. See Harmonic function.