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The Cauchy-Riemann Equations | Mathematics

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 + iy Then

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

Theorem 3.2. If uu and vv have continuous first partial derivatives on an open set UU and Satisfy the Cauchy-Riemann equations on UU Then 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

Section titled “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}

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 UU Then 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} = -6x So uxx+uyy=0u_{xx} + u_{yy} = 0. \checkmark

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

Section titled “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

Mistake 1: Assuming that the Cauchy-Riemann equations are sufficient for differentiability. The Cauchy-Riemann equations are necessary but not sufficient for complex differentiability. Even if uu and vv satisfy the Cauchy-Riemann equations at a point, ff may not be differentiable there if the partial derivatives are not continuous. The sufficiency condition requires continuous partial derivatives in a neighborhood.

Mistake 2: Forgetting the negative sign in the second Cauchy-Riemann equation. The Cauchy-Riemann equations are ux=vyu_x = v_y and uy=vxu_y = -v_x. The second equation has a negative sign. Forgetting this sign leads to incorrect conclusions about analyticity. Always check both equations carefully.

Mistake 3: Confusing the formula for the derivative. When the Cauchy-Riemann equations hold, the derivative is f(z)=ux+ivx=vyiuyf'(z) = u_x + iv_x = v_y - iu_y. Do not use f(z)=ux+iuyf'(z) = u_x + iu_y or other incorrect combinations. The derivative must be expressible in terms of either uxu_x and vxv_x or vyv_y and uyu_y.

Mistake 4: Assuming that a function satisfying the Cauchy-Riemann equations at a single point is analytic. Analyticity requires differentiability in a neighborhood, not just at a point. A function can satisfy the Cauchy-Riemann equations at a single point without being analytic there. For example, f(z)=z2f(z) = |z|^2 satisfies the Cauchy-Riemann equations only at z=0z = 0, but it is not analytic anywhere.

Mistake 5: Forgetting to check that partial derivatives are continuous. The sufficiency condition for the Cauchy-Riemann equations requires that the partial derivatives ux,uy,vx,vyu_x, u_y, v_x, v_y be continuous in a neighborhood. If the partial derivatives are not continuous, the function may not be differentiable even if the Cauchy-Riemann equations hold. Always verify continuity of partial derivatives.

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A[3_The Cauchy Riemann Equations] --> B[Key Concepts]
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The Cauchy-Riemann equations are the complex analogue of requiring a function to have the same derivative regardless of direction. In the real world, partial derivatives in x and y can be chosen independently, but complex differentiability forces them to be coupled. This coupling means the real part u and imaginary part v are harmonic conjugates, each satisfying Laplace’s equation. Think of a fluid flow: u might represent pressure and v the velocity potential, and the Cauchy-Riemann equations ensure the flow is irrotational and incompressible simultaneously.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.