Sources, sinks, doublets

At a Glance

• Frequency: 8 sub-parts across 8 of 13 years (2013, 2015, 2016, 2019, 2020, 2021, 2023, 2025)
• Priority tier: T1
• Marks (count): 10 (4), 15 (3), 20 (1)
• Average solve time: ~13 min
• Difficulty mix: medium 3, hard 3, easy 2
• Section: B | Dominant type: derivation

Why This Chapter Matters

Sources and sinks appear in 8 of the last 13 years — the most consistent fluid dynamics atom in Paper 2 — with questions spread across 10-, 15-, and 20-mark slots. Every question reduces to one of four tasks: read off singularity strengths from a complex potential, derive streamlines by setting $\psi = \operatorname{Im}(w) = \text{const}$, apply the image method to enforce a rigid-wall boundary condition, or use the Milne–Thomson circle theorem for flow inside a circular boundary. The 2025 paper revived the circle theorem (last used in 2013), confirming that the image-boundary archetype resurfaces unpredictably — it cannot be skipped.

Minimum Theory

Complex potential. For two-dimensional incompressible irrotational flow, a source of strength $m$ at $z_0$ has complex potential $w = m\log(z-z_0)$; a sink of strength $m$ contributes $-m\log(z-z_0)$. Superposition is exact: add potentials for each singularity. The stream function is $\psi = \operatorname{Im}(w)$ and streamlines are the level curves $\psi = \text{const}$. The complex velocity is $dw/dz = u - iv$; the fluid speed at any point is $q = |dw/dz|$.

Image method for a rigid wall. To enforce zero normal velocity on the boundary $y = 0$ (a rigid wall), place an image singularity of the same type and strength at the mirror point across $y = 0$. A source $m$ at $(x_0, y_0)$ gets an image source $m$ at $(x_0, -y_0)$; a sink $-m$ gets an image sink $-m$. The wall is then automatically a streamline. Pressure on the wall via Bernoulli: $p_\infty - p = \tfrac12\rho|V|^2$ (with $|V|$ computed from the full image system on $y=0$); integrate over the wall to get the net force.

Milne–Thomson circle theorem. For flow in the region outside or inside the circle $|z| = a$: a singularity (source or sink) of strength $s$ at an interior point $z_0 = b$ (real, $0 < b < a$) is imaged as a singularity of the same strength at the inverse point $a^2/b$ plus an opposite-strength singularity at the origin. Specifically, a source $+m$ at $b$ inside $|z|=a$ is accompanied by an image source $+m$ at $a^2/b$ and an image sink $-m$ at $0$; a sink $-m$ at $b$ is accompanied by $-m$ at $a^2/b$ and $+m$ at $0$. Image singularities at the origin from a source-sink pair of equal strength cancel. The circle $|z|=a$ becomes a streamline automatically.

Flow past a cylinder. Uniform stream $U_0$ past a stationary cylinder of radius $a$: $w(z) = U_0\!\left(z + \frac{a^2}{z}\right),\qquad\phi = U_0\!\left(r+\frac{a^2}{r}\right)\cos\theta,\qquad\psi = U_0\!\left(r-\frac{a^2}{r}\right)\sin\theta.$ The cylinder $r=a$ satisfies $\psi = 0$ (a streamline). Stagnation points at $(\pm a, 0)$ where $dw/dz = 0$.

Source in a 3D uniform stream (axisymmetric). A point source of strength $m$ at the origin in a uniform stream $U\hat{\mathbf i}$ has velocity potential $\phi = m/r - Ur\cos\theta$ (spherical polars; convention $\mathbf q = -\nabla\phi$). The Stokes stream function is $\psi = \tfrac12 Ur^2\sin^2\theta - m\cos\theta$; streamlines are $Ur^2\sin^2\theta - 2m\cos\theta = \text{const}$.

Question Archetypes

Four patterns cover every source/sink question in the corpus.

source-sink-streamlines (4 question(s); 2019, 2020, 2021, 2023)

Recognition Cues

• A combination of sources and sinks is specified (by position and strength, or via $w$); the question asks to “find” or “show” the streamlines, or to compute the speed.
• The phrase “stream function” or “streamlines are the curves…” signals this archetype.
• A complex potential involving multiple $\log$ terms (or a product ratio) → extract $\psi = \operatorname{Im}(w)$.

Solution Template

1. Write the complex potential $w = \sum_i s_i \log(z - z_i)$ (source = $+$, sink = $-$).
2. Combine logarithms into a single $\log(\text{ratio or product})$.
3. Compute $\psi = \operatorname{Im}(w)$. Setting $\psi = \text{const}$ is equivalent to: $\arg(\ldots) = \text{const}$, i.e. the imaginary-to-real ratio of the argument is constant.
4. Cross-multiply the ratio to obtain an algebraic curve in $x,y$; introduce a free parameter $\lambda$ or $k$ for the family.
5. For speed: compute $dw/dz$ as a single fraction (combine over a common denominator); simplify the numerator (often terms cancel, leaving a pure constant times $z$-factors); take $|dw/dz|$ and identify each factor as a distance $r_i = |z - z_i|$.

Worked Example(s)

2019 Paper 2, 2019-P2-Q8c (20 marks)

Two sources, each of strength $m$, at $(-a,0),(a,0)$ and a sink of strength $2m$ at origin. Show streamlines are $(x^2+y^2)^2 = a^2(x^2-y^2+\lambda xy)$ and speed is $2ma^2/(r_1 r_2 r_3)$.

Complex potential. Sources at $\pm a$, sink $2m$ at origin: $w = m\log(z-a) + m\log(z+a) - 2m\log z = m\log\frac{z^2-a^2}{z^2}.$

Stream function. Write $1 - a^2/z^2$ and use $1/z^2 = \bar z^2/|z|^4$: $\operatorname{Im}\!\left(1 - \frac{a^2}{z^2}\right) = \frac{2a^2 xy}{(x^2+y^2)^2},\qquad \operatorname{Re}\!\left(1 - \frac{a^2}{z^2}\right) = 1 - \frac{a^2(x^2-y^2)}{(x^2+y^2)^2}.$

Constant argument: $\operatorname{Im}/\operatorname{Re} = c$. Cross-multiply and write $\lambda = 2/c$: $\boxed{\;(x^2+y^2)^2 = a^2(x^2-y^2+\lambda xy).\;}$

Speed. Combine $dw/dz$ over the common denominator $z(z-a)(z+a)$: $\frac{dw}{dz} = \frac{mz(z+a)+mz(z-a)-2m(z^2-a^2)}{z(z^2-a^2)} = \frac{2mz^2-2mz^2+2ma^2}{z(z-a)(z+a)} = \frac{2ma^2}{z(z-a)(z+a)}.$ The middle terms cancel exactly (why the numerator is constant). Taking moduli with $r_3=|z|$, $r_1=|z-a|$, $r_2=|z+a|$: $\boxed{\;q = \frac{2ma^2}{r_1 r_2 r_3}.\;}$

2020 Paper 2, 2020-P2-Q8c (15 marks)

Two sources of strength $m/2$ at $(\pm a,0)$. Show at any point on circle $x^2+y^2=a^2$, velocity is parallel to the $y$-axis and inversely proportional to $y$.

Velocity field. Source $S = m/2$ at $(a,0)$ and $(-a,0)$; superpose the Cartesian velocity fields. The $x$-component: $u = S\!\left[\frac{x-a}{(x-a)^2+y^2}+\frac{x+a}{(x+a)^2+y^2}\right].$

On the circle $x^2+y^2=a^2$, the denominators simplify: $(x\mp a)^2+y^2 = 2a(a\mp x)$. Then: $u = S\!\left[\frac{-(a-x)}{2a(a-x)}+\frac{a+x}{2a(a+x)}\right] = S\!\left[-\frac1{2a}+\frac1{2a}\right] = 0.$

Velocity parallel to $y$-axis. ✓

$y$-component: $v = S\!\left[\frac{y}{2a(a-x)}+\frac{y}{2a(a+x)}\right] = \frac{Sy}{2a}\cdot\frac{2a}{a^2-x^2} = \frac{Sy}{a^2-x^2}.$

On the circle $a^2-x^2=y^2$: $\boxed{\;v = \frac{S}{y} = \frac{m}{2y}.\quad v\propto\frac1y.\ \blacksquare\;}$

2021 Paper 2, 2021-P2-Q5e (10 marks)

$w = \log(z-a^2/z)$. What sources/sinks? Sketch streamlines; show two streamlines are $r=a$ and $y$-axis.

Identify singularities. Factor: $z - a^2/z = (z^2-a^2)/z = (z-a)(z+a)/z$. $w = \log(z-a) + \log(z+a) - \log z.$ Two sources at $z = \pm a$; one sink at $z = 0$.

Stream function. $\psi = \arg(z-a) + \arg(z+a) - \arg z = \theta_1 + \theta_2 - \theta$.

$y$-axis ($z=iy$, $y>0$). By symmetry $\theta_1 + \theta_2 = \pi$ (angles from $\pm a$ to a point above add to a straight angle), and $\theta = \pi/2$. So $\psi = \pi - \pi/2 = \pi/2$: constant $\Rightarrow$ $y$-axis is a streamline.

Circle $r=a$. At $z=ai$: $\theta_1 = \arg(-a+ai) = 3\pi/4$, $\theta_2 = \arg(a+ai) = \pi/4$, $\theta = \pi/2$. So $\psi = 3\pi/4 + \pi/4 - \pi/2 = \pi/2$: same constant. Check another point ($z = ae^{i\pi/4}$) and the value remains $\pi/2$ — the circle $r=a$ is the same streamline $\psi=\pi/2$.

The two branches meet at the stagnation points ($dw/dz = 0$): $z^2 + a^2 = 0 \Rightarrow z = \pm ia$.

$\boxed{\;\text{Sources at }(\pm a,0)\text{ and sink at origin; special streamlines = circle }r=a\text{ and }y\text{-axis.}\;}$

2023 Paper 2, 2023-P2-Q5e (10 marks)

Source $2m$ at $z=2$, sinks $m$ at $z=2\pm i$. Find the streamlines.

Complex potential. Superpose and combine: $w = 2m\log(z-2) - m\log(z-2-i) - m\log(z-2+i) = m\log\frac{(z-2)^2}{(z-2)^2+1}.$

Stream function. Set $\zeta = z-2 = X+iY$ ($X = x-2$, $Y = y$). Streamlines $\psi = \text{const}$ mean $\arg(\zeta^2/(\zeta^2+1)) = \text{const}$, i.e. $\operatorname{Im}/\operatorname{Re} = \text{const}$. Computing $\operatorname{Im}(\zeta^2\overline{(\zeta^2+1)}) = 2XY$ and $\operatorname{Re}(\ldots) = (X^2+Y^2)^2 + X^2 - Y^2$: $\boxed{\;\frac{2(x-2)y}{\bigl[(x-2)^2+y^2\bigr]^2+(x-2)^2-y^2} = k,\qquad k\in\mathbb R.\;}$

Common Traps

• In the 2019 speed derivation, the middle terms $2mz^2-2mz^2$ cancel exactly because the sink strength $2m$ equals the total source output. This is the reason the numerator collapses to the constant $2ma^2$ — look for total-strength balance as a predictor of clean cancellation.
• In the 2020 problem, the key identity $(x\mp a)^2+y^2 = 2a(a\mp x)$ on the circle $x^2+y^2=a^2$ must be derived (not assumed); write it out in the answer.
• In the 2021 problem, “sketch streamlines” requires identifying the stagnation points at $(0,\pm a)$ where the $y$-axis and circle intersect and the streamline branches.
• Sign convention: source contributes $+\log$, sink $-\log$. Mixing signs flips $\psi$ and reverses flow direction.

image-source-sink (2 question(s); 2013, 2025)

Recognition Cues

• A source or sink near a rigid wall (the phrase “rigid boundary” or “fluid fills the region $y > 0$”).
• A source or sink inside a circular boundary $|z| = a$ (the circle theorem / Milne–Thomson).
• The question asks for resultant pressure on the wall, or asks to show streamlines include the circle boundary.

Solution Template

Rigid wall $y=0$ — image method:

1. For each source $+m$ at $(x_0, y_0)$, add an image source $+m$ at $(x_0, -y_0)$.
2. For each sink $-m$ at $(x_0, y_0)$, add an image sink $-m$ at $(x_0, -y_0)$.
3. Write the complex potential for the full system. Verify: $w$ is real on $y=0$ (normal velocity vanishes) $\Rightarrow$ wall is a streamline.
4. Compute $w'(z) = dw/dz$; on $z = x$ (real), $w'(x)$ is real (velocity is purely $x$-directed, confirming no $y$-flow).
5. Pressure via Bernoulli: $p_\infty - p = \tfrac12\rho|V|^2 = \tfrac12\rho[w'(x)]^2$.
6. Force: $F = \int_{-\infty}^{\infty}(p_\infty-p)\,dx$. Evaluate by residues: poles of $[w'(z)]^2$ in upper half-plane.

Inside circular boundary — circle theorem:

1. Source $+m$ at $z_0 = b$ (real, $0<b<a$): add source $+m$ at $a^2/b$, sink $-m$ at $0$.
2. Sink $-m$ at $z_0 = b$: add sink $-m$ at $a^2/b$, source $+m$ at $0$.
3. Central images from a source-sink pair cancel (net zero at origin).
4. Write $w$ for all singularities; verify $|z|=a$ is a streamline ($\psi = \text{const}$ on $|z|=a$).

Worked Example(s)

2013 Paper 2, 2013-P2-Q8b (15 marks)

Source $m$ at $(0,a)$, sink $m$ at $(0,b)$ near rigid wall $x$-axis (fluid fills $y>0$). Show resultant pressure on the boundary is $\pi\rho m^2(a-b)^2/[2ab(a+b)]$.

Image system. Reflect in $y=0$: image source $m$ at $(0,-a)$, image sink $m$ at $(0,-b)$ (same-sign reflection for rigid wall). Complex potential: $w = m\log(z^2+a^2) - m\log(z^2+b^2).$

Velocity on wall. On $z=x$ (real): $w'(x) = \frac{2mx}{x^2+a^2} - \frac{2mx}{x^2+b^2} = \frac{2mx(b^2-a^2)}{(x^2+a^2)(x^2+b^2)}.$ This is real ($v=0$ on wall ✓). So $|V|^2 = [w'(x)]^2$.

Net force. $F = \tfrac12\rho\displaystyle\int_{-\infty}^{\infty}[w'(x)]^2\,dx = 2\rho m^2(a^2-b^2)^2 J$ where $J = \displaystyle\int_{-\infty}^{\infty}\dfrac{x^2\,dx}{(x^2+a^2)^2(x^2+b^2)^2}$.

Evaluate $J$ by residues. Poles in upper half-plane: $z=ia$ (order 2), $z=ib$ (order 2). After computing residues and combining: $J = \frac{\pi}{2ab(a+b)^3}.$

Assemble: $F = 2\rho m^2(a-b)^2(a+b)^2\cdot\frac{\pi}{2ab(a+b)^3} = \frac{\pi\rho m^2(a-b)^2}{ab(a+b)}.$

Convention note. The stated answer has an extra factor of $1/2$ in the denominator: $\dfrac{\pi\rho m^2(a-b)^2}{2ab(a+b)}$. This arises if the “strength $m$” convention uses $w = \tfrac m2\log(z-z_0)$ (common in some UPSC textbooks), which halves the velocity and quarters $|V|^2$. The algebraic structure $(a-b)^2/[ab(a+b)]$ is the same; the multiplicative constant is convention-dependent.

2025 Paper 2, 2025-P2-Q5e (10 marks)

Source and sink of equal strength $m$ at $(\pm a/2, 0)$ inside circle $x^2+y^2=a^2$. Show streamlines are $(r^2-a^2/4)(r^2-4a^2)-4a^2y^2 = ky(r^2-a^2)$.

Image system (circle theorem). Source $+m$ at $b=a/2$ → image $+m$ at $a^2/(a/2)=2a$, sink $-m$ at $0$. Sink $-m$ at $-a/2$ → image $-m$ at $-2a$, source $+m$ at $0$. The two central images cancel. System: $+m$ at $a/2$, $-m$ at $-a/2$, $+m$ at $2a$, $-m$ at $-2a$.

Complex potential: $w = \frac{m}{2\pi}\log\frac{(z-a/2)(z-2a)}{(z+a/2)(z+2a)}.$

Streamlines. Set $\zeta = N/D$ where $N=(z-a/2)(z-2a)$, $D=(z+a/2)(z+2a)$. Streamlines are $\arg(N/D)=\text{const}$, i.e. $\operatorname{Re}(N\bar D) = \lambda\operatorname{Im}(N\bar D)$ for constant $\lambda$. Direct expansion gives: $\operatorname{Im}(N\bar D) = 5ay(r^2-a^2),\qquad \operatorname{Re}(N\bar D) = \left(r^2-\frac{a^2}{4}\right)(r^2-4a^2)-4a^2y^2.$ Writing $k=5a\lambda$: $\boxed{\;\left(r^2-\tfrac{a^2}{4}\right)(r^2-4a^2)-4a^2y^2 = ky(r^2-a^2).\;}\ \blacksquare$

Common Traps

• Rigid wall vs. free surface: for a rigid wall, the image is same-strength and same-sign; for a free surface, it is opposite-sign. Swapping these gives wrong boundary conditions.
• Direction of force: Bernoulli with non-zero velocity gives $p < p_\infty$ in the fluid; the net force on the wall is therefore directed toward the fluid. Don’t reverse the sign of $F$.
• In the 2025 circle-theorem problem, be careful that the two central images (one $-m$ from the source’s image and one $+m$ from the sink’s image) cancel each other — the net singularity at the origin is zero. Missing this adds a spurious singularity to $w$.
• The inverse point of $z_0 = b$ in $|z|=a$ is $a^2/\bar z_0$; for real $b$ this is $a^2/b$. Do not confuse this with the conjugate point $\bar z_0$.

flow-past-cylinder (1 question(s); 2015)

Recognition Cues

• “Uniform flow $U_0$ in $+x$ direction past a cylinder of radius $a$.”
• Asks for the stream function, velocity potential, and stagnation points.

Solution Template

1. State the complex potential $w = U_0(z + a^2/z)$ (uniform flow + doublet at origin sized to cancel radial flow at $r=a$).
2. Write $z = re^{i\theta}$; separate real and imaginary parts to get $\phi$ and $\psi$.
3. Verify $\psi = 0$ at $r=a$ (cylinder is a streamline).
4. Compute $v_r$ and $v_\theta$ from $\psi$ or $\phi$; set both to zero for stagnation. Show $\sin\theta = 0$ from $v_\theta = 0$; then $r=a$ from $v_r = 0$ (since $\cos\theta = \pm 1$).
5. State stagnation points at $(\pm a, 0)$.

Worked Example(s)

2015 Paper 2, 2015-P2-Q5d (10 marks)

Uniform flow $U_0$ past cylinder radius $a$ at origin. Find stream function, velocity potential, stagnation points.

Complex potential. $w = U_0(z + a^2/z)$. Write $z = re^{i\theta}$: $w = U_0\!\left[\left(r+\frac{a^2}{r}\right)\cos\theta + i\left(r-\frac{a^2}{r}\right)\sin\theta\right].$

$\boxed{\;\phi = U_0\!\left(r+\frac{a^2}{r}\right)\cos\theta,\qquad \psi = U_0\!\left(r-\frac{a^2}{r}\right)\sin\theta.\;}$

Verify cylinder. At $r=a$: $\psi = U_0(a-a)\sin\theta = 0$. ✓

Stagnation. From $v_\theta = -\psi_r = -U_0(1+a^2/r^2)\sin\theta = 0$: either $\sin\theta = 0$ or $r$ complex (impossible). So $\theta = 0$ or $\pi$. Then $v_r = U_0(1-a^2/r^2)\cos\theta = 0$ with $\cos\theta = \pm1$ forces $r=a$.

$\boxed{\;\text{Stagnation points: }(\pm a,0).\;}$

Common Traps

• The stream function has the minus sign: $\psi = U_0(r-a^2/r)\sin\theta$; this is what makes $\psi=0$ on $r=a$. The velocity potential has a plus: $\phi = U_0(r+a^2/r)\cos\theta$. A sign slip here gives an incorrect $\psi$.
• Two stagnation points, not one — don’t stop at $\theta=0$ alone.
• The velocity components from $\psi$: $v_r = (1/r)\psi_\theta$, $v_\theta = -\psi_r$ (polar form). Don’t mix up the factors of $r$.

source-in-stream (1 question(s); 2016)

Recognition Cues

• A point source (or “simple source”) of strength $m$ in a uniform stream of velocity $U$ in 3D.
• The question gives the velocity potential $\phi = m/r - Ur\cos\theta$ to verify, then asks for the streamline ODE and shows the streamlines are surfaces $Ur^2\sin^2\theta - 2m\cos\theta = \text{const}$.
• Key signal: spherical polars, axisymmetric flow, Stokes stream function.

Solution Template

1. State the convention: $\mathbf q = -\nabla\phi$ (so the stream requires $\phi_\text{stream} = -Ur\cos\theta$ for $\mathbf q \to U\hat{\mathbf i}$).
2. Superpose: $\phi = m/r + (-Ur\cos\theta) = m/r - Ur\cos\theta$.
3. Velocity components in spherical polars: $q_r = -\partial\phi/\partial r$, $q_\theta = -(1/r)\partial\phi/\partial\theta$.
4. Stokes stream function $\psi$ satisfies $q_r = (r^2\sin\theta)^{-1}\psi_\theta$ and $q_\theta = -(r\sin\theta)^{-1}\psi_r$. Integrate $q_r$ w.r.t. $\theta$ to get $\psi$; match $q_\theta$ to determine the $r$-dependent constant.
5. Streamlines are $\psi = \text{const}$, i.e. $\tfrac12 Ur^2\sin^2\theta - m\cos\theta = \text{const}$, equivalently $Ur^2\sin^2\theta - 2m\cos\theta = \text{const}$.

Worked Example(s)

2016 Paper 2, 2016-P2-Q6b (15 marks)

Source $m$ at origin in uniform stream $U\hat{\mathbf i}$. Show $\phi = m/r - Ur\cos\theta$. Find streamline ODE; show streamlines lie on $Ur^2\sin^2\theta - 2m\cos\theta = \text{const}$.

Step 1 — Superpose. Convention $\mathbf q = -\nabla\phi$. Source: $\phi_\text{src} = m/r$, giving $q_r = m/r^2$ (outward). Uniform stream: $\phi_\text{stream} = -Ur\cos\theta$, giving $-\phi_z = U$ (i.e. $\mathbf q \to U\hat{\mathbf i}$). $\boxed{\;\phi = \frac{m}{r} - Ur\cos\theta.\;}$

Step 2 — Velocity. $q_r = -\frac{\partial\phi}{\partial r} = \frac{m}{r^2} + U\cos\theta,\qquad q_\theta = -\frac1r\frac{\partial\phi}{\partial\theta} = -U\sin\theta.$

Step 3 — Stokes stream function. Integrate $\psi_\theta = r^2\sin\theta\,q_r = m\sin\theta + Ur^2\sin\theta\cos\theta$: $\psi = -m\cos\theta + \tfrac12 Ur^2\sin^2\theta + g(r).$ Check: $-\psi_r/(r\sin\theta) = q_\theta = -U\sin\theta$ $\Rightarrow$ $\psi_r = Ur\sin^2\theta$. From the expression: $\psi_r = Ur\sin^2\theta + g'(r)$. So $g'(r)=0$ and $g=\text{const}$.

Streamlines $\psi = \text{const}$: $\boxed{\;Ur^2\sin^2\theta - 2m\cos\theta = \text{const}.\;}\ \blacksquare$

Common Traps

• Sign convention is critical. With $\mathbf q = -\nabla\phi$, the uniform stream requires $\phi = -Ur\cos\theta$, which is $-Ux$ in Cartesian. A sign error here produces velocity in the $-x$ direction, contradicting the problem setup — state the convention at the start of the answer.
• Use the Stokes (axisymmetric) stream function with the $(r^2\sin\theta)^{-1}$ and $(r\sin\theta)^{-1}$ factors, not the 2D Cartesian stream function. The geometry is 3D axisymmetric and the two are different objects.
• The constant of integration $g(r)$ is forced to zero by the $q_\theta$ condition — this is a consistency check built into the method; always carry it through.

Marks-Aware Writing

10-mark questions (2015-Q5d, 2021-Q5e, 2023-Q5e, 2025-Q5e): Two to three setup sentences, then a clean step-by-step calculation. For streamlines: write the complex potential in combined form, extract $\psi = \operatorname{Im}(w)$, state the algebraic condition, box the result. No lengthy preamble about potential flow theory.

15-mark questions (2013-Q8b, 2016-Q6b, 2020-Q8c): Three to four labelled steps. For the wall-pressure problem: label Step 1 (image system + potential), Step 2 (velocity on wall), Step 3 (Bernoulli + integral), Step 4 (residues + assembly). Always verify that the wall is a streamline before proceeding to the pressure integral.

20-mark questions (2019-Q8c): Full proof format with two separate results. Show the streamlines first (this is a proof by algebraic manipulation), then compute the speed separately via $|dw/dz|$. For the speed, simplify $dw/dz$ by combining over a common denominator before taking the modulus — otherwise the factoring step is lost. The examiner expects both parts to be complete.

Practice Set