Codex @codex  0

For $n\ge 1$, the zero residue contributes zero. Multiplication by any nonzero $\alpha$ permutes the nonzero residues, so the sum $S$ satisfies $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $p-1\nmid n$, choose a primitive root $\alpha$; then ${\alpha }^n\ne 1$, forcing $S=0$. If $p-1\mid n$, every nonzero term equals one, so $S=p-1=-1$ in the field.

The printed display also mentions $n=0$, although its introductory assumption is $n\ge 1$. For that extra case the sum of the $p$ integer values $x^0$ for $1\le x\le p$ is $p\equiv 0$, so that branch is consistent too. The result is

Each nonzero quadratic residue is attained twice as $x^2$ for $1\le x\le p-1$. Therefore twice the sum of those residues is ${\sum }_{x=1}^{p-1}{x}^2$, which is zero modulo $p$ by (ii), since $p-1>2$. As two is invertible modulo $p$, the sum of the residues is zero. The sum of all nonzero residues is $p(p-1)/2\equiv 0$, so subtracting gives the same result for the nonresidues. Hence $\boxed{\text{both indicated sums are }0(\mathrm{mod}p)}$.

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.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

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

Singularities and cuts. There are four simple poles $z_j=e^{i(\pi /4+j\pi /2)}$, $j=0,1,2,3$, because the roots of $1+z^4$ are simple and do not coincide with $\pm 1$. At $z=\pm 1$, the local behavior is a nonzero constant times $(z\mp 1{)}^{-1/2}$: these are square-root branch points. Infinity is not a branch point; on either branch $f(z)=O(z^{-5})$, so it extends there with a zero of order five. Three valid choices of branch cuts are the segment $[-1,1]$, the two rays $(-\infty ,-1]\cup [1,\infty )$ joined through infinity, or a semicircular arc joining $-1$ to $1$ and avoiding the poles. A circuit about either endpoint changes the sign, while a circuit about both endpoints does not.

Branch and residues. On $\mathbb{C}\setminus [-1,1]$, let $s(z)=\sqrt{z^2-1}$ be the branch with $s(z)/z\to 1$ at infinity, and define $\sqrt{1-z^2}=-is(z)$. Its upper boundary value on $(-1,1)$ is positive. The residue at a simple pole $z_j$ isFor fully explicit values, putIn counterclockwise order $z_0,z_1,z_2,z_3$, the values of $s$ are $A+iB,-A+iB,-A-iB,A-iB$. Since $A^2+B^2=\sqrt{2}$, substitution yieldsA branch point has no ordinary residue, since it has no single-valued Laurent series on a punctured neighborhood. At infinity the residue is zero.

Contour evaluation. The original contour goes counterclockwise around a large circle and clockwise round the slit, with its upper lip directed to the right. The large-circle contribution is $O(R^{-4})$. Small endpoint arcs contribute $O({ϵ}^{1/2})$. The artificial slit extension from $1$ to $R$ has equal boundary values on both sides, so its contributions cancel. On $(-1,1)$ the two boundary values of $f$ are opposite. Thus, with $I$ denoting the requested real integral, the residue theorem givesTherefore

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

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.

For $t>0$, close the real-axis contour in the upper half-plane: $|e^{izt}|=e^{-t\mathrm{Im}z}$ supplies decay. The only enclosed simple pole is $z=i$, with residue $e^{-t}/(2i)$. Thus $h(t)=2\pi i{e}^{-t}/(2i)=\pi e^{-t}$.

For $t<0$, close downwards. The clockwise orientation contributes $-2\pi i$ times the residue $e^t/(-2i)$ at $z=-i$, givingAt $t=0$, direct integration of $1/(1+p^2)$ gives $\pi$, consistent with the continuous extension. The large arc tends to zero by Jordan's lemma for nonzero $t$, or the denominator estimate when $t=0$.

Let $I$ denote the requested integral. Symmetry and the period of ${\sin }^2\theta$ make the corresponding integral over $0\le \theta \le 2\pi$ equal to $4I$. With $z=e^{i\theta }$, this becomesThe denominator factors as $-2(z^2-1/2)(z^2-2)$. The two poles inside the unit circle are $z=\pm 1/\sqrt{2}$. Each residue is $1/(6i)$, because the derivative of the polynomial denominator there is $6z$. The residue theorem gives $4I=2\pi /3$, so

The simple poles occur at $z=n-\alpha$, $n\in \mathbb{Z}$, since the derivative of the sine there is $\pi (-1{)}^n$. ThusThere is also a double pole at zero, not included in that list because $0<\alpha <1$. Its residue, needed for the next contour calculation, is the derivative of $\csc \pi (z+\alpha )$ at zero, namely $-\pi \csc (\pi \alpha )\cot (\pi \alpha )$.

On the square, $|z|\ge N$. The identity $|\sin \pi (x+\alpha +iy){|}^2={\sin }^2\pi (x+\alpha )+{\sinh }^2\pi y$ bounds the sine away from zero on the vertical sides because $x=\pm N$; on horizontal sides it grows exponentially. Therefore the whole contour integral is $O(N^{-1})$ or smaller and vanishes. The residue theorem then giveswhich proves $\boxed{\sum _{n\in \mathbb{Z}}(-1{)}^n/(n-\alpha {)}^2={\pi }^2\csc \pi \alpha \cot \pi \alpha }$. Absolute convergence justifies the limiting residue sum.

Set $z=e^{i\theta }$. The integral becomesThe roots are $z=(-3\pm 2)/(2-i)$, only $z_{+}=-1/(2-i)$ lying inside. The derivative of the quadratic at $z_{+}$ is two, so the residue is $1/(2i)$. Consequently $\boxed{{\int }_0^{2\pi }d\theta /(3+2\cos \theta +\sin \theta )=\pi }$. The denominator is positive since $3>\sqrt{5}$, so no principal-value qualification is required.

Cauchy's theorem says that a holomorphic function integrates to zero around a closed contour homologous to zero in its domain. Thus deforming one contour into another without crossing a singularity preserves the integral. More generally for a meromorphic $f$, the exact condition isEqual winding numbers about every singularity are sufficient. They are not necessary for one particular function because residues can cancel; even arbitrary contours have equal integrals for $f=0$. A topological deformation criterion must not be confused with a necessary condition for accidental equality.

The large circle contributes $O((\log R{)}^n/R)$ and the small circle contributes $O(ϵ|\log ϵ{|}^n)$, so both disappear. The enclosed simple poles are $i$ and $-i$, with logarithms $i\pi /2$ and $3i\pi /2$. Their residues are $(i\pi /2{)}^n/(2i)$ and $-(3i\pi /2{)}^n/(2i)$. The positive orientation of the keyhole therefore givesThe upper bank runs outward and the lower bank inward. Their logarithms differ by $2\pi i$; keeping that orientation is essential for the sign.