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The Cauchy-Riemann Equations

3.1 Statement

Theorem 3.1 (Cauchy-Riemann Equations). If f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y) is differentiable at z=x+iyz = x + iyThen

ux=vy,uy=vx\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

Proof. Compute the limit along the real axis (hRh \in \mathbb{R}, h0h \to 0):

f"(z)=limh0u(x+h,y)u(x,y)h+ilimh0v(x+h,y)v(x,y)h=ux+ivxf"(z) = \lim_{h \to 0} \frac{u(x+h, y) - u(x, y)}{h} + i\lim_{h \to 0} \frac{v(x+h, y) - v(x, y)}{h} = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}

Compute along the imaginary axis (h=ikh = ik, kRk \in \mathbb{R}, k0k \to 0):

f(z)=limk0u(x,y+k)u(x,y)ik+ilimk0v(x,y+k)v(x,y)ik=iuy+vyf'(z) = \lim_{k \to 0} \frac{u(x, y+k) - u(x, y)}{ik} + i\lim_{k \to 0} \frac{v(x, y+k) - v(x, y)}{ik} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}

Equating real and imaginary parts: ux=vy\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} And vx=uy\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}. \blacksquare

3.2 Sufficiency Condition

Theorem 3.2. If uu and vv have continuous first partial derivatives on an open set UU and Satisfy the Cauchy-Riemann equations on UUThen f=u+ivf = u + iv is analytic on UU.

Proof. Since ux,uy,vx,vyu_x, u_y, v_x, v_y are continuous on UU, uu and vv are (real) differentiable. Let Δz=Δx+iΔy\Delta z = \Delta x + i\Delta y. By real differentiability:

u(x+Δx,y+Δy)u(x,y)=uxΔx+uyΔy+ε1u(x + \Delta x, y + \Delta y) - u(x, y) = u_x\,\Delta x + u_y\,\Delta y + \varepsilon_1 v(x+Δx,y+Δy)v(x,y)=vxΔx+vyΔy+ε2v(x + \Delta x, y + \Delta y) - v(x, y) = v_x\,\Delta x + v_y\,\Delta y + \varepsilon_2

Where ε1,ε2=o(Δz)\varepsilon_1, \varepsilon_2 = o(|\Delta z|). Therefore

f(z+Δz)f(z)Δz=(ux+ivx)Δx+(uy+ivy)Δy+ε1+iε2Δx+iΔy\frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{(u_x + iv_x)\Delta x + (u_y + iv_y)\Delta y + \varepsilon_1 + i\varepsilon_2}{\Delta x + i\Delta y}

By CR: uy+ivy=vx+iux=i(ux+ivx)u_y + iv_y = -v_x + iu_x = i(u_x + iv_x). Substituting:

=(ux+ivx)Δx+iΔyΔx+iΔy+o(Δz)Δzux+ivx= (u_x + iv_x)\frac{\Delta x + i\Delta y}{\Delta x + i\Delta y} + \frac{o(|\Delta z|)}{\Delta z} \to u_x + iv_x

As Δz0\Delta z \to 0. \blacksquare

3.3 The Derivative in Terms of Partial Derivatives

When the Cauchy-Riemann equations hold:

f(z)=ux+ivx=vyiuyf'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}

3.4 Harmonic Functions

Definition. A real-valued function ϕ(x,y)\phi(x, y) is harmonic if ϕxx+ϕyy=0\phi_{xx} + \phi_{yy} = 0 (Laplace’s equation).

Proposition 3.3. If f=u+ivf = u + iv is analytic, then uu and vv are harmonic.

Proof. From the Cauchy-Riemann equations: ux=vyu_x = v_y and uy=vxu_y = -v_x. Differentiating: uxx=vyxu_{xx} = v_{yx} and uyy=vxyu_{yy} = -v_{xy}. By equality of mixed partials, uxx+uyy=vyxvxy=0u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0. Similarly for vv. \blacksquare

Definition. If uu and vv are harmonic on UU and satisfy the Cauchy-Riemann equations, then vv is the harmonic conjugate of uu.

Proposition 3.4. If UU is a connected domain and uu is harmonic on UUThen uu has A harmonic conjugate on UUUnique up to an additive constant.

Proof. Define v(x,y)=(x0,y0)(x,y)(uydx+uxdy)v(x, y) = \int_{(x_0, y_0)}^{(x, y)} (-u_y\, dx + u_x\, dy). The integrand is closed (since (uy)y=uyy=uxx=(ux)x(-u_y)_y = -u_{yy} = u_{xx} = (u_x)_x) and since UU is Connected, vv is well-defined (path-independent) by Green’s theorem. Then vx=uyv_x = -u_y and vy=uxv_y = u_xWhich are the CR equations. \blacksquare

Solution

Problem. Find the harmonic conjugate of u(x,y)=x33xy2u(x, y) = x^3 - 3xy^2.

Verify uu is harmonic: uxx=6xu_{xx} = 6x, uyy=6xu_{yy} = -6xSo uxx+uyy=0u_{xx} + u_{yy} = 0. \checkmark

By CR: vy=ux=3x23y2v_y = u_x = 3x^2 - 3y^2So v=3x2yy3+g(x)v = 3x^2 y - y^3 + g(x). Also vx=uy=6xyv_x = -u_y = 6xySo 6xy=6xy+g(x)6xy = 6xy + g'(x)Giving g(x)=0g'(x) = 0So g(x)=Cg(x) = C.

Harmonic conjugate: v(x,y)=3x2yy3+Cv(x, y) = 3x^2 y - y^3 + C.

Note: f(z)=u+iv=x33xy2+i(3x2yy3)=(x+iy)3=z3f(z) = u + iv = x^3 - 3xy^2 + i(3x^2 y - y^3) = (x + iy)^3 = z^3.

Problem. Show that u(x,y)=ln(x2+y2)u(x, y) = \ln(x^2 + y^2) is harmonic on R2{0}\mathbb{R}^2 \setminus \{0\} but Has no harmonic conjugate on R2{0}\mathbb{R}^2 \setminus \{0\}.

ux=2xx2+y2u_x = \frac{2x}{x^2 + y^2}, uxx=2(y2x2)(x2+y2)2u_{xx} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2}. uy=2yx2+y2u_y = \frac{2y}{x^2 + y^2}, uyy=2(x2y2)(x2+y2)2u_{yy} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}. Δu=0\Delta u = 0. \checkmark

However, z=1(uydx+uxdy)=z=1ydx+xdyx2+y2=02π1dθ=2π0\oint_{|z|=1} (-u_y\, dx + u_x\, dy) = \oint_{|z|=1} \frac{-y\, dx + x\, dy}{x^2 + y^2} = \int_0^{2\pi} 1\, d\theta = 2\pi \neq 0.

Since R2{0}\mathbb{R}^2 \setminus \{0\} is not connected and this integral is non-zero, no Harmonic conjugate exists on this domain.

3.5 Worked Examples: Verifying CR Equations

Solution

Problem. Verify that f(z)=ezf(z) = e^z satisfies the Cauchy-Riemann equations and find f(z)f'(z).

Solution. ez=ex+iy=ex(cosy+isiny)e^z = e^{x+iy} = e^x(\cos y + i\sin y). So u=excosyu = e^x \cos y and v=exsinyv = e^x \sin y.

ux=excosyu_x = e^x \cos y, uy=exsinyu_y = -e^x \sin y, vx=exsinyv_x = e^x \sin y, vy=excosyv_y = e^x \cos y.

Cauchy-Riemann: ux=excosy=vyu_x = e^x \cos y = v_y and uy=exsiny=vxu_y = -e^x \sin y = -v_x. Both satisfied.

f(z)=ux+ivx=excosy+iexsiny=ezf'(z) = u_x + iv_x = e^x \cos y + ie^x \sin y = e^z. \blacksquare

Problem. Verify CR for f(z)=sinzf(z) = \sin z and find f(z)f'(z).

sinz=sin(x+iy)=sinxcoshy+icosxsinhy\sin z = \sin(x + iy) = \sin x \cosh y + i\cos x \sinh y.

u=sinxcoshyu = \sin x \cosh y, v=cosxsinhyv = \cos x \sinh y.

ux=cosxcoshyu_x = \cos x \cosh y, uy=sinxsinhyu_y = \sin x \sinh y. vx=sinxsinhyv_x = -\sin x \sinh y, vy=cosxcoshyv_y = \cos x \cosh y.

CR: ux=cosxcoshy=vyu_x = \cos x \cosh y = v_y \checkmark and uy=sinxsinhy=vxu_y = \sin x \sinh y = -v_x \checkmark.

f(z)=ux+ivx=cosxcoshyisinxsinhy=coszf'(z) = u_x + iv_x = \cos x \cosh y - i\sin x \sinh y = \cos z. \blacksquare

Problem. Show f(z)=1zf(z) = \frac{1}{z} satisfies CR on C{0}\mathbb{C} \setminus \{0\}.

1z=zˉz2=xiyx2+y2\frac{1}{z} = \frac{\bar{z}}{|z|^2} = \frac{x - iy}{x^2 + y^2}.

u=xx2+y2u = \frac{x}{x^2 + y^2}, v=yx2+y2v = \frac{-y}{x^2 + y^2}.

ux=y2x2(x2+y2)2u_x = \frac{y^2 - x^2}{(x^2 + y^2)^2}, vy=y2x2(x2+y2)2v_y = \frac{y^2 - x^2}{(x^2 + y^2)^2}. So ux=vyu_x = v_y. \checkmark

uy=2xy(x2+y2)2u_y = \frac{-2xy}{(x^2 + y^2)^2}, vx=2xy(x2+y2)2v_x = \frac{2xy}{(x^2 + y^2)^2}. So uy=vxu_y = -v_x. \checkmark

f(z)=ux+ivx=(x2y2+2ixy)(x2+y2)2=1z2f'(z) = u_x + iv_x = \frac{-(x^2 - y^2 + 2ixy)}{(x^2 + y^2)^2} = \frac{-1}{z^2}. \blacksquare