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Lecture 14: Cauchy-Riemann Equations:Polar Form
Dan Sloughter FurmanUniversity Mathematics39
March 31, 2004
14.1 Polarform of the Cauchy-Riemann Equations Theorem 14.1.Supposefis defined on anneighborhoodUof a point iθ0 z0=r0e, iθ f(re) =u(r, θ) +iv(r, θ), andur,uθ,vr, andvθexist onUand are continuous at (r0, θ0). Iffis differentiable atz0, then r0ur(r0, θ0) =vθ(r0, θ0) anduθ(r0, θ0) =−r0vr(r0, θ0) and 0 −iθ0 f(z0) =e(ur(r0, θ0) +ivr(r0, θ0)). Proof.By the chain rule from multi-variable calculus, ∂u ∂u∂x ∂u∂y = + ∂r ∂x∂r ∂y∂r and ∂u ∂u∂x ∂u∂y = +. ∂θ ∂x∂θ ∂y∂θ Sincex=rcos(θ) andy=rsin(θ), we have ur=uxcos(θ) +uysin(θ)
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and uθ=−uxrsin(θ) +uyrcos(θ). Similarly, vr=vxcos(θ) +vysin(θ) and vθ=−vxrsin(θ) +vyrcos(θ). Nowfis differentiable atz0, and so satisfies the Cauchy-Riemann equations atz0. Thatis, ux(r0, θ0) =vy(r0, θ0) anduy(r0, θ0) =−vx(r0, θ0). Hence vr(r0, θ0) =−uy(r0, θ0) cos(θ) +ux(r0, θ0) sin(θ) and vθ(r0, θ0) =uy(r0, θ0)rsin(θ) +ux(r0, θ0)rcos(θ). It follows that r0ur(r0, θ0) =vθ(r0, θ0) anduθ(r0, θ0) =−r0vr(r0, θ0) The final statement of the theorem is left as an exercise. Theorem 14.2.Supposefis defined on anneighborhoodUof a point iθ0 z0=r0e, iθ f(re) =u(r, θ) +iv(r, θ), andur,uθ,vr, andvθexist onUand are continuous at (r0, θ0). If r0ur(r0, θ0) =vθ(r0, θ0) anduθ(r0, θ0) =−r0vr(r0, θ0), thenfis differentiable atz0. Proof.The proof is left as a homework exercise. Example 14.1.Forz6= 0, let 1 f(z) =. 2 z iθ If we writez=re, then 1 1 f(z= (cos(2) =θ)−isin(2θ)). 2 2iθ2 r er 2
Hence, in the notation of the above theorems, 1 u(r, θcos(2) =θ) 2 r and 1 v(r, θ) =−sin(2θ). 2 r It follows that 2 2 ur(r, θ) =−cos(2θ) anduθ(r, θ) =−sin(2θ) 3 2 r r and 2 2 vr(r, θsin(2) =θ) andvθ(r, θ) =−cos(2θ). 3 2 r r Thus rur(r, θ) =vθ(r, θ) anduθ(r, θ) =−rvr(r, θ), and sofis differentiable at allz6= 0.Moreover, 2 2 0 −iθ f(z) =e−cos(2θ) +isin(2θ) 3 3 r r 2 −iθ−2iθ =−e e 3 r 2 −3iθ =−e 3 r 2 =−. 3 z
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