A uniform parallel flow with complex potential $f_0(z) = U_\infty z$ is disturbed by placing an impermeable circular cylinder of radius $a$ centered at the origin. According to the Milne-Thomson circle theorem, the resulting complex potential $w(z)$ for the flow past the cylinder is:

(A) $w(z) = U_\infty \left(z + \frac{a^2}{z}\right)$

(B) $w(z) = U_\infty \left(z - \frac{a^2}{z}\right)$

(C) $w(z) = U_\infty \left(z^2 + \frac{a^2}{z^2}\right)$

(D) $w(z) = U_\infty \ln\left(\frac{z}{a}\right)$

Answer Key:

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### 1. Strategy & Theoretical Principles The Milne-Thomson circle theorem states that if a 2D irrotational flow field has complex potential $f_0(z)$ with all singularities outside $|z| \le a$, then placing an impermeable cylinder $|z| = a$ at the origin yields the new complex potential: $$w(z) = f_0(z) + \overline{f_0\left(\frac{a^2}{\bar{z}}\right)}$$ ### 2. Step-by-Step Application Given undisturbed potential $f_0(z) = U_\infty z$ where $U_\infty$ is real: $$f_0\left(\frac{a^2}{\bar{z}}\right) = U_\infty \frac{a^2}{\bar{z}}$$ Taking the complex conjugate: $$\overline{f_0\left(\frac{a^2}{\bar{z}}\right)} = \overline{U_\infty \frac{a^2}{\bar{z}}} = U_\infty \frac{a^2}{z}$$ Adding this to $f_0(z)$: $$w(z) = U_\infty z + U_\infty \frac{a^2}{z} = U_\infty \left(z + \frac{a^2}{z}\right)$$ This reproduces the classic potential flow past a circular cylinder (uniform flow + doublet of strength $2\pi U_\infty a^2$). ### 3. Distractor Forensics - **Option A:** Correct standard theorem result. - **Option B:** Minus sign would make the cylinder boundary permeable (not a streamline). ### 4. Exam Trap & Standard Reference - **Trap Alert:** Note that taking the complex conjugate of $\bar{z}$ in the denominator returns $z$ in the denominator. - **Standard Reference:** Milne-Thomson, L.M., *Theoretical Hydrodynamics*, 5th Edition, Dover Publications.

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