3.1 Statement
Theorem 3.1 (Cauchy-Riemann Equations). If f(z)=u(x,y)+iv(x,y) is differentiable at z=x+iyThen
∂x∂u=∂y∂v,∂y∂u=−∂x∂v
Proof. Compute the limit along the real axis (h∈R, h→0):
f"(z)=limh→0hu(x+h,y)−u(x,y)+ilimh→0hv(x+h,y)−v(x,y)=∂x∂u+i∂x∂v
Compute along the imaginary axis (h=ik, k∈R, k→0):
f′(z)=limk→0iku(x,y+k)−u(x,y)+ilimk→0ikv(x,y+k)−v(x,y)=−i∂y∂u+∂y∂v
Equating real and imaginary parts: ∂x∂u=∂y∂v And ∂x∂v=−∂y∂u. ■
3.2 Sufficiency Condition
Theorem 3.2. If u and v have continuous first partial derivatives on an open set U and Satisfy the Cauchy-Riemann equations on UThen f=u+iv is analytic on U.
Proof. Since ux,uy,vx,vy are continuous on U, u and v are (real) differentiable. Let Δz=Δx+iΔy. By real differentiability:
u(x+Δx,y+Δy)−u(x,y)=uxΔx+uyΔy+ε1 v(x+Δx,y+Δy)−v(x,y)=vxΔx+vyΔy+ε2
Where ε1,ε2=o(∣Δz∣). Therefore
Δzf(z+Δz)−f(z)=Δx+iΔy(ux+ivx)Δx+(uy+ivy)Δy+ε1+iε2
By CR: uy+ivy=−vx+iux=i(ux+ivx). Substituting:
=(ux+ivx)Δx+iΔyΔx+iΔy+Δzo(∣Δz∣)→ux+ivx
As Δz→0. ■
3.3 The Derivative in Terms of Partial Derivatives
When the Cauchy-Riemann equations hold:
f′(z)=∂x∂u+i∂x∂v=∂y∂v−i∂y∂u
3.4 Harmonic Functions
Definition. A real-valued function ϕ(x,y) is harmonic if ϕxx+ϕyy=0 (Laplace’s equation).
Proposition 3.3. If f=u+iv is analytic, then u and v are harmonic.
Proof. From the Cauchy-Riemann equations: ux=vy and uy=−vx. Differentiating: uxx=vyx and uyy=−vxy. By equality of mixed partials, uxx+uyy=vyx−vxy=0. Similarly for v. ■
Definition. If u and v are harmonic on U and satisfy the Cauchy-Riemann equations, then v is the harmonic conjugate of u.
Proposition 3.4. If U is a connected domain and u is harmonic on UThen u has A harmonic conjugate on UUnique up to an additive constant.
Proof. Define v(x,y)=∫(x0,y0)(x,y)(−uydx+uxdy). The integrand is closed (since (−uy)y=−uyy=uxx=(ux)x) and since U is Connected, v is well-defined (path-independent) by Green’s theorem. Then vx=−uy and vy=uxWhich are the CR equations. ■
Solution
Problem. Find the harmonic conjugate of u(x,y)=x3−3xy2.
Verify u is harmonic: uxx=6x, uyy=−6xSo uxx+uyy=0. ✓
By CR: vy=ux=3x2−3y2So v=3x2y−y3+g(x). Also vx=−uy=6xySo 6xy=6xy+g′(x)Giving g′(x)=0So g(x)=C.
Harmonic conjugate: v(x,y)=3x2y−y3+C.
Note: f(z)=u+iv=x3−3xy2+i(3x2y−y3)=(x+iy)3=z3.
Problem. Show that u(x,y)=ln(x2+y2) is harmonic on R2∖{0} but Has no harmonic conjugate on R2∖{0}.
ux=x2+y22x, uxx=(x2+y2)22(y2−x2). uy=x2+y22y, uyy=(x2+y2)22(x2−y2). Δu=0. ✓
However, ∮∣z∣=1(−uydx+uxdy)=∮∣z∣=1x2+y2−ydx+xdy=∫02π1dθ=2π=0.
Since R2∖{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)=ez satisfies the Cauchy-Riemann equations and find f′(z).
Solution. ez=ex+iy=ex(cosy+isiny). So u=excosy and v=exsiny.
ux=excosy, uy=−exsiny, vx=exsiny, vy=excosy.
Cauchy-Riemann: ux=excosy=vy and uy=−exsiny=−vx. Both satisfied.
f′(z)=ux+ivx=excosy+iexsiny=ez. ■
Problem. Verify CR for f(z)=sinz and find f′(z).
sinz=sin(x+iy)=sinxcoshy+icosxsinhy.
u=sinxcoshy, v=cosxsinhy.
ux=cosxcoshy, uy=sinxsinhy. vx=−sinxsinhy, vy=cosxcoshy.
CR: ux=cosxcoshy=vy ✓ and uy=sinxsinhy=−vx ✓.
f′(z)=ux+ivx=cosxcoshy−isinxsinhy=cosz. ■
Problem. Show f(z)=z1 satisfies CR on C∖{0}.
z1=∣z∣2zˉ=x2+y2x−iy.
u=x2+y2x, v=x2+y2−y.
ux=(x2+y2)2y2−x2, vy=(x2+y2)2y2−x2. So ux=vy. ✓
uy=(x2+y2)2−2xy, vx=(x2+y2)22xy. So uy=−vx. ✓
f′(z)=ux+ivx=(x2+y2)2−(x2−y2+2ixy)=z2−1. ■