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If a nonconstant holomorphic function has its first nonzero local Taylor coefficient at order $m$, small rays from $z_0$ have image arguments $\arg a+m\arg (z-z_0)$ to leading order. Their oriented angles are therefore multiplied by $m$ modulo $2\pi$. The case $m=1$ is locally conformal; $m>1$ gives a critical point, and the local inverse has $m$ branches. For the Joukowski map at $z_0=1$, $f(1+\zeta )=2+{\zeta }^2+O({\zeta }^3)$, so $m=2$.

For $b>0$, the conformal map $\cosh (\pi z/b)$ takes $\{\mathrm{Re}z>0,0<\mathrm{Im}z<b\}$ bijectively onto the upper half-plane. The exponential maps this half-strip onto the upper half-plane exterior to the unit circle; then $\cosh (\pi z/b)=(w+w^{-1})/2$ is a scaled Joukowski map. All critical points of the entire map are $z=inb$, mapping alternately to $1$ and $-1$.

For $\rho >0$, substitution in the Joukowski map gives the title parametrization. If $\rho \ne 1$, its image is an ellipse with semiaxes $\rho +{\rho }^{-1}$ and $|\rho -{\rho }^{-1}|$ and foci $\pm 2$. Reciprocal radii give the same ellipse in opposite orientations. At $\rho =1$ the image degenerates to $[-2,2]$, each interior point covered once by each unit semicircle.

The Joukowski map restricts to a conformal map from $\{\mathrm{Im}z>0,|z|>1\}$ onto the upper half-plane. Indeed, $\mathrm{Im}g(z)=\mathrm{Im}z(1-|z{|}^{-2})>0$ there. For any $\mathrm{Im}\zeta >0$, the roots of $z^2-\zeta z+1=0$ have product one and cannot lie on the unit circle; exactly one has modulus greater than one, and that root has positive imaginary part. This proves that the map is a bijection.

Its inverse function iswith each factor taken as the principal square root. This product is the holomorphic square root of ${\zeta }^2-4$ that behaves as $\zeta$ at infinity in the upper half-plane. The upper unit semicircle maps to $[-2,2]$; the exterior real rays map to the remaining real rays. The derivative $g^{\prime }=1-z^{-2}$ vanishes only at the boundary endpoints $z=\pm 1$.

The Kutta condition selects the circulation of an idealized potential flow past a body with a sharp trailing edge by requiring finite limiting velocity at that edge. For the Joukowski map $z=\zeta +a^2/\zeta$, the trailing edge $z=2a$ corresponds to a simple zero of $dz/d\zeta$ at $\zeta =a$. If $W$ is the complex potential in the circle plane, then $dw/dz=W^{\prime }(\zeta )/(1-a^2/{\zeta }^2)$ is finite there only if $W^{\prime }(a)=0$. For a uniform flow of speed $U$ at angle $\alpha$ and a point vortex of circulation $\Gamma$, this gives $\Gamma =-4\pi Ua\sin \alpha$, and cancellation of the simple zeros gives the tangential trailing-edge velocity $U\cos \alpha$. The opposite endpoint can retain a leading-edge singularity.

A conformal map preserves oriented angles between intersecting smooth curves at each point, with a nonzero local scale factor. A sufficient criterion is that $f$ be a holomorphic function on the domain and that $f^{\prime }(z)\ne 0$ everywhere there. Its first-order action then multiplies every tangent direction by the same nonzero complex number, combining a rotation and a dilation. A conformal bijection additionally requires injectivity and surjectivity onto the target.

For $z=x+iy$ in the given half-strip,Thus $0<|w|<1$ and $0<\arg w<\pi$. Conversely every such $w$ has a unique argument in $(0,\pi )$ and unique radius in $(0,1)$, giving $x=\arg w$, $y=-\log |w|$. HenceThe complex exponential is injective on $S$, and its derivative $i{e}^{iz}$ never vanishes.

On $D_{+}$, use the Joukowski map with a sign change,Sinceit maps into the upper half-plane. To prove injectivity, $J(w_1)=J(w_2)$ givesBoth moduli are below one, so $w_1{w}_2\ne 1$ and $w_1=w_2$.

For surjectivity, take any $\zeta$ in the upper half-plane and solve $w^2+\zeta w+1=0$. The two roots have product one. Neither can lie on the unit circle, since there $-(w+w^{-1})$ is real. Exactly one root therefore has modulus below one. For this root, the displayed imaginary-part identity forces $\mathrm{Im}w>0$, so it belongs to $D_{+}$. Finally $J^{\prime }(w)=w^{-2}-1$ has no zero in $D_{+}$. This proves the Joukowski map from the upper half-disc to the upper half-plane is a conformal bijection.

Composing gives the required conformal mapIts derivative $2\sin z$ is nonzero on $S$, since the zeros of the complex sine are the real integer multiples of $\pi$.

Parameterize the unit circle by $\zeta =e^{i\theta }$. Under the Joukowski map,Consequently its image is the ellipsewith horizontal and vertical semi-axesBoth are positive because $0<a<1$. The horizontal axis is the major axis. The derivative $dz/d\zeta =1-a^2/{\zeta }^2$ is nonzero for $|\zeta |\ge 1$. Also, equality of two images implies $({\zeta }_1-{\zeta }_2)(1-a^2/({\zeta }_1{\zeta }_2))=0$; two distinct exterior points cannot satisfy the second factor since $|{\zeta }_1{\zeta }_2|>1>a^2$. Thus the exterior map is one-to-one, mapping the exterior of the unit circle conformally onto the exterior of the ellipse.

The Joukowski map is a holomorphic function for $z\ne 0$, withA holomorphic function is a local conformal map wherever its derivative is nonzero. Hence the two finite nonzero critical points areThey lie on the boundary of the fluid region. At $z=0$ the displayed formula has a pole and is undefined as a map to the finite complex plane; zero is outside the fluid region. Thus $g$ is locally conformal everywhere in the fluid.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover dates this paper to 11 June 2021 despite the attachment filename. Solutions identify a false printed biharmonic equation identity, retain the stipulated drop curvature approximation explicitly, and state the conditions needed for the final Joukowski map.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.

An explicit disk map of a slit half disk makes the two boundary approaches visible without assuming an extension theorem for the boundary. We construct the map as a succession of bijections and then determine its limits.

First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the square root whose argument lies in $(0,\pi )$. The removed segment on the imaginary axis becomes $[-1,0)$, so the image of $D$ is preciselyThe Möbius transformation $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the unit disc, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an automorphism of the unit disk after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued square root $u=\sqrt{v}$ and maps that slit disc bijectively ontoAll these maps have nonzero derivatives on their respective domains, since their potentially problematic zero or branch point is on the removed slit.

Next the Joukowski map $J(u)=u+1/u$ is a conformal bijection from $U$ onto the lower half-plane. For $u\in U$,If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the lower half-plane and consider $u^2-qu+1=0$. Neither root lies on the unit circle, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\mathrm{Im}u>0$. Thus it lies in $U$. Also $J^{\prime }(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the lower half-plane bijectively onto the unit disc: the inequality $|q+i|<|q-i|$ is exactly $\mathrm{Im}q<0$, and its inverse is $q=i(\zeta +1)/(\zeta -1)$. Consequentlyis a conformal bijection of $D$ onto the disc.

At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero real part. Since $\mathrm{Im}(z^2)=2\mathrm{Re}z\mathrm{Im}z$ and the real-coefficient Möbius map $w\mapsto v$ preserves the sign of the imaginary part, right-side approach gives $\arg v\to 0$, while left-side approach gives $\arg v\to 2\pi$. HenceThe two limits of $F$ are thereforeThey have modulus one and are distinct, since $L>0$ and the Möbius map $C$ is injective on the extended real line. The two-sided cluster set at an interior slit point is a geometric consequence of opening the slit into two boundary intervals.

For the given map $f$, the composition $B=f\circ F^{-1}$ is an automorphism of the unit disk. To see its precise boundary behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta )=(\zeta -a)/(1-\overline{a}\zeta )$. This gives a disc automorphism fixing zero. Applying Schwarz lemma to it and its inverse shows it is a rotation. ThusIts denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous bijection of the unit circle. Put $\alpha =B({\gamma }_{+})$ and $\beta =B({\gamma }_{-})$. These are distinct points of that circle, independent of the approaching sequence.

For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore exactly $\{\alpha \}$, $\{\beta \}$ or $\{\alpha ,\beta \}$, according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm {\epsilon }_n$ with positive ${\epsilon }_n\downarrow 0$.

For the Joukowski map $g(z)=z+z^{-1}$,It maps the exterior upper half-disc $D_0=\{z\in \mathbb{H}:|z|>1\}$ into $\mathbb{H}$. It is injective there: $g(z)=g(w)$ gives $(z-w)(1-1/(zw))=0$, and the second factor cannot vanish when both moduli exceed one. It is onto as well. For $v\in \mathbb{H}$, the two roots of $z^2-vz+1=0$ have product one. Neither lies on the unit circle, since $z+z^{-1}$ is real there. Exactly one root has modulus greater than one, and the displayed imaginary-part formula makes its imaginary part positive. That root belongs to $D_0$ and maps to $v$.

Since $g_{K_0}$ takes values in $\mathbb{H}$, its domain must be contained in $D_0$. The established injectivity and surjectivity of $g:D_0\to \mathbb{H}$ mean that no proper subdomain of $D_0$ can map onto all of $\mathbb{H}$ using this same function. ThereforeThe half-plane capacity is the coefficient of $1/z$ in the normalized expansion. If hulls are regarded as closed subsets of $\overline{\mathbb{H}}$, take the closure of this half-disc, including its diameter; its relative half-plane hull and capacity are unchanged. This is the half-plane capacity of a half-disc.

For the Joukowski map $F(z)=z+b^2/z$, the derivative is $F^{\prime }(z)=1-b^2/z^2$. Thus it is a conformal map at every finite $z$ except its pole at zero and its critical pointsThe pole maps to $\infty$; as a map of Riemann spheres it is a simple pole and is not a branch point. Neither is the point at infinity, where the map behaves like $z$.

An appropriate domain is the upper exterior half-discFor $z=x+iy$, $\mathrm{Im}F(z)=y(1-b^2/|z{|}^2)>0$ on $D$. To prove surjectivity and injectivity, solve $F(z)=\zeta$ as $z^2-\zeta z+b^2=0$. Its two roots have product $b^2$. For $\mathrm{Im}\zeta >0$, neither root lies on the real axis or on $|z|=b$, since either boundary maps to the real axis. Exactly one root therefore has modulus greater than $b$. The imaginary-part identity forces that root to have positive imaginary part. It is the unique root in $D$, so $F$ maps $D$ conformally and bijectively onto the upper half-plane.

There is a small closure issue in the printed set: the unit disc is open, so its half-disc is not relatively closed in $\mathbb{H}$. As written, the set is not a compact H-hull. We compute the intended half-plane capacity after taking its relative closure in $\mathbb{H}$.

First remove the closed unit half-disc $K$. Its mapping-out function of a compact H-hull is the Joukowski mapFor $z=iy$ on the remaining portion of the vertical slit, $1<y\le 2$,Thus the image of the remaining slit is $(0,3i/2]$. By the half-plane capacity of a vertical slit, its capacity is $(3/2{)}^2/2=9/8$. The half-plane-capacity composition rule givesAs a direct check, composing with the slit map gives $g_A(z)=\sqrt{(z+z^{-1}{)}^2+9/4}$, with the branch asymptotic to $z$. Its $z^{-1}$ coefficient is $17/8$.

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.

On the circle put $\zeta =a{e}^{i\theta }$. The Joukowski map gives $z=2a\cos \theta$, so its image is the slit $S=[-2a,2a]$. Away from that circle, the inverse equation is ${\zeta }^2-z\zeta +a^2=0$. For $z\notin S$, neither root has modulus $a$, and their product is $a^2$, so exactly one root has modulus greater than $a$. Selecting that root defines the exterior inverse.

More explicitly, define $s(z)=\sqrt{z^2-4a^2}$ to be the branch analytic on $\mathbb{C}\setminus S$ normalized by $s(z)/z\to 1$ at infinity. It is positive for real $z>2a$ and negative for real $z<-2a$. ThenExistence and analyticity of this branch also follow from the exterior inverse: $J^{\prime }(\zeta )=1-a^2/{\zeta }^2$ is nonzero for $|\zeta |>a$, so the selected simple root varies analytically; $s=2g-z$ has the stated normalization. The exterior map is injective because $J({\zeta }_1)=J({\zeta }_2)$ with distinct roots would imply ${\zeta }_1{\zeta }_2=a^2$, impossible for two exterior points. Thus $g$ is a conformal bijection throughout $\mathbb{C}\setminus S$.

The proposed potential is a uniform stream plus its cylinder doublet and the circulation term. On $\zeta =a{e}^{i\theta }$, the first two terms sum to the real quantity $2Ua\cos (\theta -\alpha )$, while the imaginary part of the logarithmic term is the constant $-\Gamma \log a/(2\pi )$. Thus the cylinder is a streamline. At infinity the complex velocity tends to $U{e}^{-i\alpha }$ and the circulation is $\Gamma$, verifying both far-field conditions. Logarithms may be chosen locally; their derivative gives a single-valued velocity.

Composition therefore gives the plate potentialAt the edge $z=2a$, corresponding to $\zeta =a$, the physical complex velocity is $W^{\prime }(\zeta )/J^{\prime }(\zeta )$. Since the denominator has a simple zero, finiteness requireswhich is exactlyWith this cancellation, the limiting ratio is $W^{\prime \prime }(a)/J^{\prime \prime }(a)$. The Kutta value of the circulation gives $W^{\prime \prime }(a)=2U\cos \alpha /a$ and $J^{\prime \prime }(a)=2/a$. Hence the finite edge velocity is

The Joukowski map $f(z)=z+z^{-1}$ is holomorphic on $\mathbb{C}\setminus \{0\}$. A holomorphic function is locally conformal where its derivative is nonzero. Here $f^{\prime }(z)=1-z^{-2}$ vanishes at $z=\pm 1$. Thereforewhere the exclusion of zero is because the function is undefined there. Holomorphy alone does not supply conformality at the two critical points.

For $0<a<1$, the Joukowski map $z=\zeta +a^2/\zeta$ maps the exterior unit circle to an ellipse of horizontal and vertical semiaxes $1+a^2$ and $1-a^2$. The Milne-Thomson circle theorem gives $W(\zeta )=U(e^{-i\alpha }\zeta +e^{i\alpha }/\zeta )$ for a uniform flow inclined by $\alpha$. The chain rule produces the displayed complex velocity in the physical plane. Since $z\sim \zeta$, the far-field flow retains its speed and direction. Its two stagnation points areobtained from $\zeta =\pm e^{i\alpha }$. The map derivative is nonzero on and outside the unit circle, so these zeros are preserved without introducing a boundary singularity.