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The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.

On $\zeta =a{e}^{i\theta }$, the Joukowski map gives $z=2a\cos \theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidatesOn $D=\mathbb{C}\setminus [-2a,2a]$, choose the analytic square root $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to 1$ at infinity. One explicit definition iswhere the square root on the right is the principal square root. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required holomorphic function. Then $g(z)=(z+s(z))/2\sim z$.

The two quadratic roots have product $a^2$. For $z\notin S$, neither root lies on $|\zeta |=a$, since that circle maps into $S$. Consequently exactly one root lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, thenwhose second factor cannot vanish. Thus the Joukowski map is a conformal map from $|\zeta |>a$ onto $D$, and $g$ is conformal throughout $D$. The two endpoints are branch points and are excluded.

The complex potential in the circle plane is the sum of the uniform flow term $U{e}^{-i\alpha }\zeta$, its circle-image term $U{a}^2{e}^{i\alpha }/\zeta$, and the point vortex term $-i\Gamma \log \zeta /(2\pi )$. On $\zeta =a{e}^{i\theta }$, the first two sum to $2Ua\cos (\theta -\alpha )$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma \log a/(2\pi )$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required uniform flow; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.

Conformal transport therefore givesIn the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The complex velocity isAt the trailing edge $z=2a$, corresponding to $\zeta =a$, the denominator has a simple zero. Finite velocity requiresso the Kutta condition isWith this choice, cancellation of the common zero givesThus the limiting velocity has zero transverse component and is tangent to the plate.

Orient $\partial B$ counterclockwise, and define positive circulation by that orientation. The outward normal from the body satisfies $nds=(dy,-dx)$. Pressure acts inward, soThe steady Bernoulli equation for this constant-density irrotational flow is $p+\rho |v{|}^2/2=C$. The constant has zero integral around a closed curve. With complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, let $W=w^{\prime }=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline{W}d\overline{z}$, and $|W{|}^{2}d\overline{z}=W^{2}dz$. Substitution proves Blasius force theorem:At infinity the Laurent expansion has the formThere is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma (U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the residue theorem gives $\oint W^{2}dz=2\Gamma (U-iV)$. HenceIts component along $(U,V)$ is zero: the drag is zero. The lift magnitude is $\boxed{\rho |\Gamma |\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho \Gamma \sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the Kutta–Joukowski theorem with its orientation made explicit.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$, and the complex velocity as $q=w^{\prime }=u-iv$. The lower solid wall has $\psi =0$. The upper solid wall and its continuing free streamline have constant $\psi$ equal to the far-field volume flux per unit depth, namely $a$. Consequently the potential domain is the infinite stripThe points $A,E$ correspond to $\varphi \to -\infty$, $B,C$ to $\varphi \to +\infty$, and $D$ is $w=ia$ by the chosen potential normalization.

In the hodograph plane, the horizontal wall has $q$ real positive, the vertical wall has $q$ imaginary positive because the physical velocity is downward, and the fluid free boundary has $|q|=1$. The upstream reservoir velocity tends to zero, the outgoing velocity is one, and the velocity at $D$ is $i$. Thus the hodograph is the first-quadrant unit quarter disk, with $A,E$ at $q=0$, $B,C$ at $q=1$, and $D$ at $q=i$.

The complex potential of a two-dimensional potential flow maps the fluid region to the strip $0<\mathrm{Im}w<1$. Its lower boundary is the wall, with $B$ at $w=0$; upstream is $\mathrm{Re}w\to -\infty$ and downstream is $\mathrm{Re}w\to +\infty$. The free streamline is the upper strip boundary.

Write $q=w^{\prime }=u-iv$. The horizontal wall gives positive real velocities running from $q=1$ to the stagnation value $q=0$ at $B$. The vertical wall gives negative imaginary velocities from zero to $-i$. On the free surface $|q|=1$, with the angle turning continuously from zero to $-\pi /2$. Thus the hodograph plane domain is the fourth-quadrant quarter of the unit disk, bounded by those two radii and the unit-circle arc.

The following successive conformal maps straighten it. First $r=q^2$ maps to the lower half unit disk. Next $s=(r-1)/(r+1)$ maps to the third quadrant, since $\mathrm{Re}s=(|r{|}^2-1)/|r+1{|}^2<0$ and $\mathrm{Im}s=2\mathrm{Im}r/|r+1{|}^2<0$. Finally $s^2$ maps onto the upper half-plane. The three wall landmarks $q=1,0,-i$ go to $0,1,\infty$ respectively.

The strip map $e^{\pi w}$ also maps onto the upper half-plane, with these same landmarks $0,1,\infty$. The unique half-plane automorphism fixing all three is the identity. ThereforeThe free surface maps to the negative real axis in either construction, as required by its streamfunction value one. This completes the right-angle free-streamline hodograph map construction.

Milne-Thomson's circle theorem states that if $f(z)$ is a complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenhas the circle as a streamline, with the original exterior singularities unchanged. Additive constants in the potential are immaterial.

Take $f(z)=\Gamma (2\pi i{)}^{-1}\log (z-z_1)$, using the convention that positive $\Gamma$ is positive circulation. Reflection in the circle gives, up to a constant,The reflected vortex at $a^2/{\bar{z}}_1$ has strength $-\Gamma$, while the central term has strength $+\Gamma$. Their combined circulation around the cylinder is zero; this is the circulation normalization implicit in the requested equation.

The vortex moves with the velocity induced by the other singularities, excluding its own singular term. Since complex velocity is $dw/dz=u-iv$, its value at the vortex is $d{\bar{z}}_1/dt$. ThusWriting $z_1=r{e}^{i\theta }$ and multiplying this equation by $z_1$ givesIts real part implies $\dot{r}=0$, and its imaginary part givesThe radius is constant and the angular velocity is constant on that orbit. For positive $\Gamma$ the orbit is clockwise with this convention. Retaining the conjugate on the time derivative is essential for the sign.

For incompressible planar flow define the streamfunction by $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ when the flow is irrotational; the differential condition is $v_x-u_y=0$. The complex potential of a two-dimensional potential flow is $w=\varphi +i\psi$, with derivative $w^{\prime }=u-iv$.

For $w=m\log (z-z_0)/(2\pi )$, its derivative is $m/[2\pi (z-z_0)]$. This gives outward radial velocity $m/(2\pi r)$, whose flux across a circle is $2\pi r\cdot m/(2\pi r)=m$. It is therefore the stated line source.

Superposition givesFor positive $U,m$, the stagnation points areThey lie on the real axis, one upstream of the source and the other downstream of the sink. Streamlines are level sets of $\psi =Uy+(m/2\pi )[\arg z-\arg (z-1)]$. The dividing streamline passes through both stagnation points and forms a Rankine oval enclosing the source and sink; outside, streamlines tend to uniform horizontal flow.

Use the convention that positive vortex strength is counterclockwise and complex velocity is $u-iv$. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices isLocal choices of logarithm branch change no velocity. Differentiation givesFor nonzero vortex strength and $U>0$, its two zeros areThe number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity $-\Gamma /(4\pi a)$, while the background contributes $(U,0)$. Their velocities are therefore equal. Their separation stays $(2a,0)$, making this induced velocity constant for all time. Both travel along parallel straight lines with

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.

The circle theorem says that if a complex potential $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), thenadds image singularities inside the circle and makes it a streamline. On the circle $a^2/\bar{z}=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its power series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed circulation handled separately.

For the fluid vortex use $f(z)=-i\Gamma \log (z-b)/(2\pi )$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma \log z/(2\pi )$, up to constants. The zero-circulation circle-theorem construction has both these terms. To impose the specified circulation $-\Gamma$ about the obstacle, add $+i\Gamma \log z/(2\pi )$, canceling the central term. ThusIt has the stated circulation and the same impermeable boundary, since the added central-fluid vortex streamfunction is constant on the circle.

For the given Blasius integral take the obstacle contour counterclockwise. Inside it the sole singularity ofis at $d$. Its residue is ${\Gamma }^2/[2{\pi }^2(d-b)]$, from the cross term. The residue theorem then givesTherefore $\boxed{F_x=\rho {\Gamma }^{2}b/[2\pi (b^2-a^2)],F_y=0}$. The force on the obstacle points toward the external fluid vortex.

A planar potential dipole at $a$ with complex moment $m$ has complex potential of a two-dimensional potential flow $w(z)=m/(z-a)$. Its complex velocity is $w^{\prime }(z)=-m/(z-a{)}^2$, interpreted as $v_x-i{v}_y$. Unlike the three-dimensional potential dipole, this is a two-dimensional singularity of the logarithmic-source potential. For example, two opposite source coefficients $s$ at $a\pm \delta /2$ have potential $s[\log (z-a-\delta /2)-\log (z-a+\delta /2)]$, whose leading term is $-s\delta /(z-a)$ as the separation tends to zero with $s\delta$ fixed. This gives the dipole as a limiting source–sink pair, with the complex sign and orientation included in $m$.

Let $\zeta =g(z)$ be a conformal map, and let a point vortex of circulation $\Gamma$ lie at $z=c$. Suppose the mapped complex potential of a two-dimensional potential flow iswhere $R$ is a holomorphic function near $g(c)$ and represents the other point vortices, boundaries, and any imposed flow. The physical complex velocity is $w^{\prime }=g^{\prime }{W}^{\prime }$. The point vortex follows its regular velocity, found by removing $\Gamma /[2\pi i(z-c)]$ before taking the limit. A Taylor expansion proveswhich gives the title formula. The complex conjugate occurs because the complex velocity convention is $w^{\prime }=u-iv$, whereas $\dot{c}=u+iv$.

For a mapped upper half-plane with no imposed flow, the method of images gives $R(\zeta )=-\Gamma \log (\zeta -\overline{g(c)})/(2\pi i)$, henceThe first term is essential: excluding a mapped singularity and then transforming its regular velocity does not in general equal subtracting the physical singularity after transformation.

For a point vortex of circulation $\Gamma$ in the wedge $0<\theta <\pi /n$, with positive integer $n$, the conformal map $z\mapsto z^n$ sends the domain to the upper half-plane. An opposite image vortex gives the complex potentialRemove the singular self-induced velocity before setting $z=c$. The resulting complex velocity isThe finite first term follows from the expansion of $(z^n-c^n)/(z-c)$ at $c$. In polar coordinates this becomes $\dot{r}=n\Gamma \cot (n\theta )/(4\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Therefore $r\sin (n\theta )$ is constant along a vortex trajectory. The case $n=1$ is motion parallel to a plane wall, while $n=3$ describes a sixty-degree wedge.

For a complex potential $w(z)=m/(z-a)$, reflection in the straight wall through zero at angle $\theta$ is $Rz=e^{2i\theta }\bar{z}$. The image construction $W(z)=w(z)+\overline{w(Rz)}$ givesOn the wall $Rz=z$, this is real, so its stream function is constant there. This proves the impermeability condition. Thus the image position is $a^{\prime }=e^{2i\theta }\bar{a}$ and its moment is $m^{\prime }=e^{2i\theta }\bar{m}$. If the point dipole is advected by the regular image flow, the complex velocity convention $W^{\prime }=u-iv$ giveswhere $d=\mathrm{Im}(e^{-i\theta }a)$ is its signed distance from the wall. The last equality follows from $a-a^{\prime }=2i{e}^{i\theta }d$. It applies when $d\ne 0$.

For an inviscid irrotational unit-speed layer turning through a right angle, the complex potential of a two-dimensional potential flow occupies the strip $0<\mathrm{Im}w<1$ and its complex velocity $q$ occupies a fourth-quadrant quarter disk. The maps $r=q^2$, $s=(r-1)/(r+1)$ and $s^2$ take this disk to the upper half-plane, sending upstream velocity one, corner stagnation zero, and downstream velocity $-i$ to $0,1,\infty$. The map $e^{\pi w}$ has the same normalization, proving the displayed relation.

For complex potential of a two-dimensional potential flow $w=U(z+a^2/z)-i\Gamma \log z/(2\pi )$, zeros of the complex velocity solve a quadratic. If $0<\gamma <1$, the displayed two roots are distinct and both have modulus $a$; positive circulation places them above the centreline. Their polar angles satisfy $\sin \theta =\gamma$. At the threshold $\gamma =1$ they merge at the top of the cylinder. This follows also from the surface tangential velocity $-2U\sin \theta +\Gamma /(2\pi a)$. The shifted stream function $U(r-a^2/r)\sin \theta -\Gamma \log (r/a)/(2\pi )$ is constant on the cylinder.

A two-dimensional point sink of strength $Q>0$ removes volume flux $Q$ per unit depth at $z_0$. Its complex potential of a two-dimensional potential flow is the displayed expression, so $u-iv=-Q/[2\pi (z-z_0)]$. The radial velocity is $u_r=-Q/(2\pi r)$: integrating over a circle gives outward flux $2\pi r{u}_r=-Q$. This is the planar cross-section of a line sink. A passive particle satisfies $d(r^2)/dt=-Q/\pi$ and reaches the ideal point singularity in finite time. This is a flow singularity, distinct from a stable sink equilibrium of a regular dynamical system.

A two-dimensional potential doublet is a localized potential flow singularity with complex potential of a two-dimensional potential flow $W=M/(z-b)$, where $M=m{e}^{i\alpha }$ specifies strength and orientation. It is the limiting field of an equal point source and sink as their separation tends to zero with strength times separation fixed, with the sign absorbed in $M$. Its complex velocity is $-M/(z-b{)}^2$, so the velocity decays as inverse distance squared. The total volumetric flow rate around the singularity is zero. Its two-dimensional decay differs from that of the three-dimensional source dipole.