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The improper integral converges: its integrand is bounded near zero and at most $x^{-n}$ for $x\ge 1$, which is integrable because $n\ge 2$.

Set $\theta =\pi /n$, $\omega =e^{2i\theta }$ and $\zeta =e^{i\theta }$. Apply the residue theorem to $F(z)=1/(1+z^n)$ on the positively oriented sector contour consisting of the segment $0\to R$, the arc $R{e}^{it}$ for $0\le t\le 2\theta$, and the segment $R\omega \to 0$, with $R>1$. No pole lies on the boundary. Exactly one root of $z^n=-1$ is inside this sector, namely $\zeta$; it is a simple pole. Part (a) givesOn the returning ray, put $z=\omega x$ with $x$ decreasing from $R$ to zero. Since ${\omega }^n=1$, the two ray contributions areOn the arc, $|1+z^n|\ge R^n-1$, so the estimation lemma bounds its contour integral byThe residue theorem and the convergence already established now giveUsing $1-e^{2i\theta }=-2i{e}^{i\theta }\sin \theta$, we concludeThe sector also works for $n=2$, when it is the upper semicircle. Its angle is chosen so that $z^n$ has the same value on the two rays.

Factor the denominator as $(z^2+1)(z^2+4)$ and integrateover $[-R,R]$ followed by the upper semicircle, for $R>2$. The enclosed simple poles are $i$ and $2i$. The local quotient formula from part (a) givesOn this arc, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+1|\ge R^2-1$, $|z^2+4|\ge R^2-4$. Thus the estimation lemma givesOn the real line, $|F(x)|\le 1/[(x^2+1)(x^2+4)]$, so the improper integral is absolutely convergent. Hence the residue theorem yieldsTaking real parts is justified by absolute convergence. ThereforeThe imaginary part vanishes as a further check, since the sine integrand is odd.

Start from the uniform-flow complex potential of a two-dimensional potential flow $Uz$. Milne-Thomson's circle theorem adds the image doublet $U{a}^2/z$, yielding constant stream function on $|z|=a$. Add a line vortex with positive counterclockwise circulation $\Gamma$:Indeed, on $z=a{e}^{i\theta }$ the uniform-flow and doublet terms sum to the real number $2Ua\cos \theta$, while the vortex has imaginary part $-\Gamma \log a/(2\pi )$, also constant. Thus the cylinder is a streamline and satisfies the impermeability condition. The vortex velocity is $\Gamma /(2\pi r)$ in the counterclockwise direction. Although the potential is multivalued when $\Gamma \ne 0$, its velocity is single-valued.

Differentiate and locate the coefficient of $z^{-1}$ in the square:For the counterclockwise contour in the printed Blasius force theorem, the residue theorem gives ${\oint }_C(w^{\prime }{)}^{2}dz=2U\Gamma$. ConsequentlyThese are force components per unit cylinder length; with the stated circulation convention the lift is downward for positive $U\Gamma$.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.

Milne-Thomson's circle theorem inserts an impermeable circular boundary into a two-dimensional irrotational incompressible flow. If $f(z)$ is analytic in a neighborhood of the closed disc and its flow singularities lie outside, the new complex potential outside $|z|=a$ isReflection and coefficient conjugation make the second term analytic outside the circle, with its image singularities inside. On $|z|=a$, $a^2/\bar{z}=z$, so the two terms sum to $2\mathrm{Re}f(z)$: the streamfunction is constant there. Thus no velocity crosses the circle. At infinity the image term tends to the constant $\overline{f(0)}$, which has no effect on velocity, and the far-field flow is retained. This proves the construction, with an additive potential constant immaterial.

Put $A=\mu e^{i\alpha }$, $B=\mu e^{-i\alpha }$. For the dipole the reflected potential is $B/(a^2/z-b)=Bz/(a^2-bz)$, givingIts image pole is $c=a^2/b$, inside the cylinder. The complex velocity isNear $c$, the second term is $C(z-c{)}^{-2}$ with $C=B{a}^2/b^2$, while $v_0(z)=-A/(z-b{)}^2$ is analytic there. The square of the pole term has no residue, nor does $v_0^2$. The cross term has residue $2C{v}_0^{\prime }(c)$, soTake the contour counterclockwise around the cylinder, as in the supplied Blasius force theorem. The residue theorem gives $\frac{1}{2}i\rho \oint (w^{\prime }{)}^{2}dz=-\pi \rho \mathrm{Res}(w^{\prime }{)}^2$. ThereforeThe force points toward the dipole. Its independence of dipole orientation follows from the product $AB={\mu }^2$ in the sole residue contribution.

Orient the body boundary counterclockwise. The outward normal from the body satisfies $nds=(dy,-dx)$, so pressure givesFor steady irrotational flow, Bernoulli's equation makes $p+\rho |u{|}^2/2$ constant in the connected fluid region. Its constant contribution integrates to zero. Write $q=w^{\prime }=u-iv$. Impermeability says the velocity is tangent to the boundary, so $qdz=\bar{q}d\bar{z}$ there. Consequentlywhich proves Blasius' theorem with this contour convention.

The background stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. Since its coefficients are real, the circle theorem givesIts derivative isIn $q^2$ the coefficient of $z^{-1}$ comes from the exponent pairs $(2,-3)$ and $(1,-2)$, including both orders. It is $36{\alpha }^{2}c{a}^4+36{\alpha }^2{c}^3{a}^2$. The residue theorem then givesThe vanishing transverse force also agrees with reflection symmetry about the real axis.

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.

The Milne-Thomson circle theorem says that if a complex potential $f(z)$ has no singularities on or inside $|z|=a$, then a circular impermeable boundary can be introduced by takingThe new singularities are the reflected ones inside the circle. On $|z|=a$, $a^2/z=\overline{z}$, so $W(z)=f(z)+\overline{f(z)}$ is real; the stream function is constant and there is no normal velocity. At infinity the added term tends to $f^{\ast }(0)$, a harmless potential constant, so it preserves the far-field flow. This proves the theorem with zero added net circulation about the cylinder. A logarithmic term may then specify a different cylinder circulation.

For $f(z)=-i\Gamma \log (z-b)/(2\pi )$, put $c=a^2/b$. The reflected term iswith locally consistent logarithm branches. The circle-theorem solution has zero cylinder circulation: its image vortex at $c$ and its central vortex cancel. To impose circulation $-\Gamma$, add $i\Gamma \log z/(2\pi )$. The result isIt consists of the physical vortex and an opposite image vortex inside the obstacle. On the circle, $|z-b|=(b/a)|z-c|$, so its stream function is indeed constant.

The complex velocity isInside the positively oriented obstacle contour, $q^2$ has only the double pole at $c$. Its residue isThe residue theorem and the supplied Blasius force theorem therefore giveConsequentlyThe force points toward the held vortex, independent of the sign of its circulation.

The residue calculation is the force on a circular cylinder from a held exterior vortex.

In the analytic continuation inside the circle, only the image location $c=a^2/b$ is singular; $b$ lies outside. Squaring the velocity givesThe last term has no residue, and the middle term has residuebecause $AB=m^2{a}^2/b^2$ and $c-b=-(b^2-a^2)/b$. The residue theorem and part (a) therefore yieldThus $F_y=0$ and $F_x>0$. The force points towards the planar potential doublet and is independent of its orientation, since the factors $e^{i\alpha }$ and $e^{-i\alpha }$ cancel in $AB$.

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.

Conformality implies $f^{\prime }(0)\ne 0$, so $a_1\ne 0$. Apply the stated Green's theorem with $F=z^n\bar{z}$:Here $\bar{\zeta }={\zeta }^{-1}$ on the unit circle. The lowest power in $f^n{f}^{\prime }$ is ${\zeta }^n$, while the most negative power in the conjugate polynomial is ${\zeta }^{-N}$. For $n\ge N$ their product has no ${\zeta }^{-1}$ term. The residue theorem gives $M_n=0$ for all $n\ge N$.

When $N=2$, the residue for $n=0$ is $|a_1{|}^2+2|a_2{|}^2$. For $n=1$, expandOnly $a_1^2\zeta {\bar{a}}_2{\zeta }^{-2}$ contributes. ThereforeThe first is the area; the second records its complex first moment, and need not be real.

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.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

If $f(z)=\sum a_j{z}^j$, then $f(-z)=\sum (-1{)}^j{a}_j{z}^j$ and its residue at zero is $-a_{-1}$. The circle $|z-1|=2$ encloses zero and no other finite singularity. Thus the residue theorem givesfor positive orientation.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

Assume positive, counterclockwise orientation of a contour. Factor the polynomial denominator asOnly $z=\pm (\sqrt{3}-\sqrt{2})$ lie inside the unit circle. These are simple poles. Writing the denominator as $P(z)$, their residues areThe residue theorem therefore givesReversing orientation of a contour changes the sign.

Let $f$ be holomorphic on a neighbourhood of a positively oriented, piecewise smooth simple closed curve and its interior except at finitely many isolated singularities $a_j$ strictly inside. The residue theorem statesThe residue is the coefficient of $(z-a_j{)}^{-1}$ in the local Laurent series. No singularity may lie on the contour. For more general closed contours, each residue is multiplied by the corresponding winding number.

The poles are the four roots of $z^4=-1$. The upper half-plane contains $a=e^{i\pi /4}$ and $b=e^{3i\pi /4}$, both enclosed when $R>1$. At a simple pole $q$,The sum of the two residues isThe residue theorem, with the real segment traversed from $-R$ to $R$ and the upper arc from $R$ to $-R$, yields