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proof to Cauchy-Riemann equations (polar coordinates)

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proof to Cauchy-Riemann equations (polar coordinates)

If f⁢(z)fzf(z) is differentialble at z0subscriptz0z_{0} then the following limit

f′⁢(z0)superscriptfnormal-′subscriptz0\displaystyle f^{{\prime}}(z_{0}) =\displaystyle= limξ→0⁡f⁢(z0+ξ)-f⁢(z0)ξsubscriptnormal-→ξ0fsubscriptz0ξfsubscriptz0ξ\displaystyle\lim_{{\xi\to 0}}\frac{f(z_{0}+\xi)-f(z_{0})}{\xi}

will remain the same approaching from any direction. First we fix θθ\theta as θ0subscriptθ0\theta_{0} then we take the limit along the ray where the argument is equal to θ0subscriptθ0\theta_{0}. Then

f′⁢(z0)superscriptfnormal-′subscriptz0\displaystyle f^{{\prime}}(z_{0}) =\displaystyle= limh→0⁡f⁢(r0⁢ei⁢θ0+h⁢ei⁢θ0)-f⁢(r0⁢ei⁢θ0)h⁢ei⁢θ0subscriptnormal-→h0fsubscriptr0superscripteisubscriptθ0hsuperscripteisubscriptθ0fsubscriptr0superscripteisubscriptθ0hsuperscripteisubscriptθ0\displaystyle\lim_{{h\to 0}}\frac{f(r_{0}e^{{i\theta_{0}}}+he^{{i\theta_{0}}})% -f(r_{0}e^{{i\theta_{0}}})}{he^{{i\theta_{0}}}}
=\displaystyle= limh→0⁡f⁢((r0+h)⁢ei⁢θ0)-f⁢(r0⁢ei⁢θ0)h⁢ei⁢θ0subscriptnormal-→h0fsubscriptr0hsuperscripteisubscriptθ0fsubscriptr0superscripteisubscriptθ0hsuperscripteisubscriptθ0\displaystyle\lim_{{h\to 0}}\frac{f((r_{0}+h)e^{{i\theta_{0}}})-f(r_{0}e^{{i% \theta_{0}}})}{he^{{i\theta_{0}}}}
=\displaystyle= limh→0⁡u⁢(r0+h,θ0)+i⁢v⁢(r0+h,θ0)-u⁢(r0,θ0)-i⁢v⁢(r0,θ0)h⁢ei⁢θ0subscriptnormal-→h0usubscriptr0hsubscriptθ0ivsubscriptr0hsubscriptθ0usubscriptr0subscriptθ0ivsubscriptr0subscriptθ0hsuperscripteisubscriptθ0\displaystyle\lim_{{h\to 0}}\frac{u(r_{0}+h,\theta_{0})+iv(r_{0}+h,\theta_{0})% -u(r_{0},\theta_{0})-iv(r_{0},\theta_{0})}{he^{{i\theta_{0}}}}
=\displaystyle= 1ei⁢θ0⁢[limh→0⁡u⁢(r0+h,θ0)-u⁢(r0,θ0)h+i⁢limh→0⁡v⁢(r0+h,θ0)-v⁢(r0,θ0)h]1superscripteisubscriptθ0subscriptnormal-→h0usubscriptr0hsubscriptθ0usubscriptr0subscriptθ0hisubscriptnormal-→h0vsubscriptr0hsubscriptθ0vsubscriptr0subscriptθ0h\displaystyle\frac{1}{e^{{i\theta_{0}}}}\Bigg[\lim_{{h\to 0}}\frac{u(r_{0}+h,% \theta_{0})-u(r_{0},\theta_{0})}{h}+i\lim_{{h\to 0}}\frac{v(r_{0}+h,\theta_{0}% )-v(r_{0},\theta_{0})}{h}\Bigg]
=\displaystyle= 1ei⁢θ0⁢[∂⁡u∂⁡r⁢(r0,θ0)+i⁢∂⁡v∂⁡r⁢(r0,θ0)]1superscripteisubscriptθ0ursubscriptr0subscriptθ0ivrsubscriptr0subscriptθ0\displaystyle\frac{1}{e^{{i\theta_{0}}}}\Bigg[\frac{\partial u}{\partial r}(r_% {0},\theta_{0})+i\frac{\partial v}{\partial r}(r_{0},\theta_{0})\Bigg]

Similarly, if we take the limit along the circle with fixed rrr equals r0subscriptr0r_{0}. Then

f′⁢(z0)superscriptfnormal-′subscriptz0\displaystyle f^{{\prime}}(z_{0}) =\displaystyle= limh→0⁡f⁢(r0⁢ei⁢θ0+r0⁢ei⁢(θ0+h))-f⁢(r0⁢ei⁢θ0)r0⁢ei⁢θ0⁢(ei⁢h-1)subscriptnormal-→h0fsubscriptr0superscripteisubscriptθ0subscriptr0superscripteisubscriptθ0hfsubscriptr0superscripteisubscriptθ0subscriptr0superscripteisubscriptθ0superscripteih1\displaystyle\lim_{{h\to 0}}\frac{f(r_{0}e^{{i\theta_{0}}}+r_{0}e^{{i(\theta_{% 0}+h)}})-f(r_{0}e^{{i\theta_{0}}})}{r_{0}e^{{i\theta_{0}}}(e^{{ih}}-1)}
=\displaystyle= limh→0⁡f⁢(r0⁢ei⁢(θ0+h))-f⁢(r0⁢ei⁢θ0)h⁢ei⁢θ0subscriptnormal-→h0fsubscriptr0superscripteisubscriptθ0hfsubscriptr0superscripteisubscriptθ0hsuperscripteisubscriptθ0\displaystyle\lim_{{h\to 0}}\frac{f(r_{0}e^{{i(\theta_{0}+h)}})-f(r_{0}e^{{i% \theta_{0}}})}{he^{{i\theta_{0}}}}
=\displaystyle= limh→0⁡u⁢(r0,θ0+h)+i⁢v⁢(r0,θ0+h)-u⁢(r0,θ0)-i⁢v⁢(r0,θ0)h⁢ei⁢θ0subscriptnormal-→h0usubscriptr0subscriptθ0hivsubscriptr0subscriptθ0husubscriptr0subscriptθ0ivsubscriptr0subscriptθ0hsuperscripteisubscriptθ0\displaystyle\lim_{{h\to 0}}\frac{u(r_{0},\theta_{0}+h)+iv(r_{0},\theta_{0}+h)% -u(r_{0},\theta_{0})-iv(r_{0},\theta_{0})}{he^{{i\theta_{0}}}}
=\displaystyle= 1r0⁢ei⁢θ0⁢[limh→0⁡u⁢(r0+h,θ0)-u⁢(r0,θ0)h⋅hei⁢h-1+i⁢limh→0⁡v⁢(r0+h,θ0)-v⁢(r0,θ0)h⋅hei⁢h-1]1subscriptr0superscripteisubscriptθ0subscriptnormal-→h0normal-⋅usubscriptr0hsubscriptθ0usubscriptr0subscriptθ0hhsuperscripteih1isubscriptnormal-→h0normal-⋅vsubscriptr0hsubscriptθ0vsubscriptr0subscriptθ0hhsuperscripteih1\displaystyle\frac{1}{r_{0}e^{{i\theta_{0}}}}\Bigg[\lim_{{h\to 0}}\frac{u(r_{0% }+h,\theta_{0})-u(r_{0},\theta_{0})}{h}\cdot\frac{h}{e^{{ih}}-1}+i\lim_{{h\to 0% }}\frac{v(r_{0}+h,\theta_{0})-v(r_{0},\theta_{0})}{h}\cdot\frac{h}{e^{{ih}}-1}\Bigg]
=\displaystyle= 1r0⁢ei⁢θ0⁢[limh→0⁡u⁢(r0+h,θ0)-u⁢(r0,θ0)h⋅limh→0⁡hei⁢h-1+i⁢limh→0⁡v⁢(r0+h,θ0)-v⁢(r0,θ0)h⋅limh→0⁡hei⁢h-1]1subscriptr0superscripteisubscriptθ0subscriptnormal-→h0normal-⋅usubscriptr0hsubscriptθ0usubscriptr0subscriptθ0hsubscriptnormal-→h0hsuperscripteih1isubscriptnormal-→h0normal-⋅vsubscriptr0hsubscriptθ0vsubscriptr0subscriptθ0hsubscriptnormal-→h0hsuperscripteih1\displaystyle\frac{1}{r_{0}e^{{i\theta_{0}}}}\Bigg[\lim_{{h\to 0}}\frac{u(r_{0% }+h,\theta_{0})-u(r_{0},\theta_{0})}{h}\cdot\lim_{{h\to 0}}\frac{h}{e^{{ih}}-1% }+i\lim_{{h\to 0}}\frac{v(r_{0}+h,\theta_{0})-v(r_{0},\theta_{0})}{h}\cdot\lim% _{{h\to 0}}\frac{h}{e^{{ih}}-1}\Bigg]
=\displaystyle= 1r0⁢ei⁢θ0⁢[∂⁡u∂⁡θ⁢(r0,θ0)⁢1i+∂⁡v∂⁡θ⁢(r0,θ0)]1subscriptr0superscripteisubscriptθ0uθsubscriptr0subscriptθ01ivθsubscriptr0subscriptθ0\displaystyle\frac{1}{r_{0}e^{{i\theta_{0}}}}\Bigg[\frac{\partial u}{\partial% \theta}(r_{0},\theta_{0})\frac{1}{i}+\frac{\partial v}{\partial\theta}(r_{0},% \theta_{0})\Bigg]
=\displaystyle= 1r0⁢ei⁢θ0⁢[∂⁡v∂⁡θ⁢(r0,θ0)-i⁢∂⁡u∂⁡θ⁢(r0,θ0)]1subscriptr0superscripteisubscriptθ0vθsubscriptr0subscriptθ0iuθsubscriptr0subscriptθ0\displaystyle\frac{1}{r_{0}e^{{i\theta_{0}}}}\Bigg[\frac{\partial v}{\partial% \theta}(r_{0},\theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})\Bigg]

Note: We use l’Hôpital’s rule to obtain the following result used above limh→0⁡hei⁢h-1=1isubscriptnormal-→h0hsuperscripteih11i\lim_{{h\to 0}}\frac{h}{e^{{ih}}-1}=\frac{1}{i}.

Now, since the limit is the same along the circle and the ray then they are equal:

1ei⁢θ0⁢[∂⁡u∂⁡r⁢(r0,θ0)+i⁢∂⁡v∂⁡r⁢(r0,θ0)]1superscripteisubscriptθ0ursubscriptr0subscriptθ0ivrsubscriptr0subscriptθ0\displaystyle\frac{1}{e^{{i\theta_{0}}}}\Bigg[\frac{\partial u}{\partial r}(r_% {0},\theta_{0})+i\frac{\partial v}{\partial r}(r_{0},\theta_{0})\Bigg] =\displaystyle= 1r0⁢ei⁢θ0⁢[∂⁡v∂⁡θ⁢(r0,θ0)-i⁢∂⁡u∂⁡θ⁢(r0,θ0)]1subscriptr0superscripteisubscriptθ0vθsubscriptr0subscriptθ0iuθsubscriptr0subscriptθ0\displaystyle\frac{1}{r_{0}e^{{i\theta_{0}}}}\Bigg[\frac{\partial v}{\partial% \theta}(r_{0},\theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})\Bigg]
[∂⁡u∂⁡r⁢(r0,θ0)+i⁢∂⁡v∂⁡r⁢(r0,θ0)]ursubscriptr0subscriptθ0ivrsubscriptr0subscriptθ0\displaystyle\Bigg[\frac{\partial u}{\partial r}(r_{0},\theta_{0})+i\frac{% \partial v}{\partial r}(r_{0},\theta_{0})\Bigg] =\displaystyle= 1r0⁢[∂⁡v∂⁡θ⁢(r0,θ0)-i⁢∂⁡u∂⁡θ⁢(r0,θ0)]1subscriptr0vθsubscriptr0subscriptθ0iuθsubscriptr0subscriptθ0\displaystyle\frac{1}{r_{0}}\Bigg[\frac{\partial v}{\partial\theta}(r_{0},% \theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})\Bigg]

which implies that

∂⁡u∂⁡rur\displaystyle\frac{\partial u}{\partial r} =\displaystyle= 1r⁢∂⁡v∂⁡θ1rvθ\displaystyle\frac{1}{r}\frac{\partial v}{\partial\theta}
∂⁡v∂⁡rvr\displaystyle\frac{\partial v}{\partial r} =\displaystyle= -1r⁢∂⁡u∂⁡θ1ruθ\displaystyle-\frac{1}{r}\frac{\partial u}{\partial\theta}

QED

Type of Math Object: 
Proof
Major Section: 
Reference
Parent: 
Cauchy-Riemann equations (polar coordinates)
Groups audience: 
Editing Group for ProofToCauchyRiemannEquationsPolarCoordinates

Mathematics Subject Classification

30E99 None of the above, but in MSC2010 section 30Exx

Comments

Submitted by zbeekman on Fri, 02/15/2008 - 19:05

The second part, for varying theta is not correct.

Submitted by azdbacks4234 on Fri, 02/15/2008 - 21:07

If you feel that something should be corrected it's best to file a formal correction specifying exactly what you think is wrong as opposed to attaching a vague assertion to the object itself.

Subscribe to Comments for "proof to Cauchy-Riemann equations (polar coordinates)"

Info

Owner: Daume
Added: 2004-01-06 - 21:17
Author(s): Daume

Corrections

Variable flip by jacou

Versions

(v5) by Daume 2013-03-22
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