For f(Z)=dfrac 1 Z , writing Z=re^ itheta gives f(Z)=dfrac e^ -itheta r =dfrac costheta r -idfrac sintheta r . So u=dfrac costheta r . Verifying the polar C-R equation dfrac partial u partial r =dfrac 1 r dfrac partial v partialtheta where v=-dfrac sintheta r : the left side equals:

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CSS Pure Mathematics QUESTION #4098

Question 1

For \(f(Z)=\dfrac{1}{Z}\), writing \(Z=re^{i\theta}\) gives \(f(Z)=\dfrac{e^{-i\theta}}{r}=\dfrac{\cos\theta}{r}-i\dfrac{\sin\theta}{r}\). So \(u=\dfrac{\cos\theta}{r}\). Verifying the polar C-R equation \(\dfrac{\partial u}{\partial r}=\dfrac{1}{r}\dfrac{\partial v}{\partial\theta}\) where \(v=-\dfrac{\sin\theta}{r}\): the left side equals:

\(-\dfrac{\cos\theta}{r}\)
\(\dfrac{\cos\theta}{r^2}\)
\(-\dfrac{\cos\theta}{r^2}\)✔️
\(\dfrac{\sin\theta}{r^2}\)

Correct Answer Explanation

\(\dfrac{\partial u}{\partial r}=\dfrac{\partial}{\partial r}\left(\dfrac{\cos\theta}{r}\right)=-\dfrac{\cos\theta}{r^2}\). Right side: \(\dfrac{1}{r}\dfrac{\partial}{\partial\theta}\left(-\dfrac{\sin\theta}{r}\right)=\dfrac{1}{r}\cdot\left(-\dfrac{\cos\theta}{r}\right)=-\dfrac{\cos\theta}{r^2}\). ✓ Both sides equal \(-\dfrac{\cos\theta}{r^2}\).

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