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On a domain carrying a branch of the complex logarithm, the holomorphic function $-i\mathrm{Log}z$ has real part equal to the chosen angle and imaginary part $-\log |z|$. These are harmonic conjugates; their level sets are rays and circles, meeting orthogonally. For the principal real arctangent $\varphi =\arctan (y/x)$, its domain has the two components $x>0$ and $x<0$. The corresponding holomorphic functions are $-i\mathrm{Log}z$ and $-i\mathrm{Log}(-z)$, using the complex logarithm on the right half-plane. This formulation does not assert a globally single-valued angle on the punctured plane.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

For an entire function $f$, let $F(z)=f(z)z^{-\alpha }$ with noninteger real $\alpha$, and choose the branch of the complex logarithm with $\varphi <\arg z<\varphi +2\pi$. On the cut ray $z=r{e}^{i\varphi }$, the bank whose argument approaches $\varphi$ has value $F_{+}=f(r{e}^{i\varphi })r^{-\alpha }{e}^{-i\alpha \varphi }$, while the other bank has argument approaching $\varphi +2\pi$ and value $F_{-}=e^{-2\pi i\alpha }{F}_{+}$. The signed jump is therefore $(1-e^{-2\pi i\alpha })F_{+}$ when defined as $F_{+}-F_{-}$. Stating the branch and order of subtraction removes the sign ambiguity. A keyhole contour uses this phase relation to convert its two bank integrals into a constant multiple of one real or radial integral.

For real $\beta >1$, choose a branch of the complex logarithm whose argument interval contains $[0,2\pi /\beta ]$ strictly, and integrate $\mathrm{Log}z/(1+z^{\beta })$ over an annular sector of angle $\alpha =2\pi /\beta$. The one enclosed simple pole is $e^{i\pi /\beta }$, with residue $-i\pi e^{i\pi /\beta }/{\beta }^2$. The circular arcs tend to zero, with bounds $O(ϵ|\log ϵ|)$ and $O(R^{1-\beta }\log R)$.

Writing $I$ for the title integral and $J={\int }_0^{\infty }(1+x^{\beta }{)}^{-1}dx$, the two radial segments and the residue theorem giveDivide by $e^{i\pi /\beta }$ and equate real and imaginary parts. This gives $J=\pi /[\beta \sin (\pi /\beta )]$ and the title formula. When $1<\beta <2$, the sector extends below the negative real axis; the principal complex logarithm therefore cannot be used throughout that sector. A cut along a ray just below the positive axis avoids this issue for every $\beta >1$.

For any value of the complex logarithm, $\log z=\ln |z|+i\arg z$ up to adding an integer multiple of $2\pi i$. Its real part therefore satisfiesThe locus is the circle centred at the origin with radius $e^2$. Different logarithm values change only the imaginary part. If a particular analytic branch of the complex logarithm with a cut is imposed, restrict the circle to that branch's domain; the modulus equation itself is branch independent.

The exponential series is absolutely norm-convergent, since ${\sum }_{n\ge 1}\Vert x^n\Vert /n!\le e^{\Vert x\Vert }-1$. Completeness defines the Banach algebra exponential $e^x\in A$. If $xy=yx$, absolute convergence permits rearrangement of the product series, and the binomial identity givesTaking $y=-x$ yields $(e^x{)}^{-1}=e^{-x}$.

For the spectral hypothesis, put $K={\sigma }_A(u)$. Its complement has an open, hence path-connected, unbounded component containing $0$. Join $0$ to an exterior point by a simple polygonal arc in that component, and continue it to infinity, avoiding $K$. Removing this polygonal slit gives a simply connected open neighborhood $\Omega$ of $K$ on which $z$ has a branch of the complex logarithm $L(z)$. The holomorphic functional calculus therefore defines $x=L(u)$ and its composition rule givesThe relevant slit is chosen through the resolvent, not assumed to be a fixed negative-real-axis cut. This is the logarithm from a spectral slit in a Banach algebra.

Let $H$ be the subgroup of finite products of exponentials. It is a subgroup because $e^a{e}^{-a}=1$ and reversing a product gives its inverse. Each product is joined to $1$ by $t\mapsto e^{t{a}_1}\cdots e^{t{a}_m}$, so $H\subseteq G_0$. Conversely, if $\Vert v-1\Vert <1$, the convergent logarithm seriessatisfies $e^{\log v}=v$, by the scalar analytic identity and functional calculus. Thus $H$ contains a neighborhood of $1$. Its translates make $H$ open, and all other cosets are open too, making $H$ closed. Connectedness forces $G_0\subseteq H$. Finally $g{e}^a{g}^{-1}=e^{ga{g}^{-1}}$, so $H$ is normal. Thereforean open-and-closed normal subgroup. This is the identity component of Banach-algebra invertibles; finite products are essential and are not asserted to be single exponentials.

The principal real arctangent is defined here on $x\ne 0$, so work on either of the two open half-planes. Put $r=\sqrt{x^2+y^2}$. Direct differentiation givesHence ${\varphi }_{xx}+{\varphi }_{yy}=0$: $\varphi$ is harmonic on its domain. The Cauchy-Riemann equations for $f=\varphi +i\psi$ requireIntegrating these partial derivatives gives the harmonic conjugateThe constants may be chosen independently on the two components. The harmonic angle and logarithmic radius representation supplies the holomorphic functionwhere $\mathrm{Log}$ is the branch of the complex logarithm on the right half-plane. On the left half-plane it is the angle of $-z$, rather than the principal angle of $z$, that equals $\arctan (y/x)$. This distinguishes the literal arctangent from a globally defined argument; no single-valued complex logarithm exists on the whole punctured plane.

With $C_0=0$, the level sets $\varphi =C$, for $-\pi /2<C<\pi /2$, are the two opposite rays of $y=x\tan C$ with the origin removed. On each half-plane only one ray remains. The level sets $\psi =K$ are circles $r=e^{-K}$, with their two points on $x=0$ excluded from the literal domain. The rays and circles meet at right angles, as expected for harmonic conjugates.

Put $a=1/n$, so $0<a<1$, and write $I(a)={\int }_0^{\infty }{x}^a/(1+x^2)dx$. Choose the branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, whose cut is the downward imaginary axis. Define $z^a=\exp (a\mathrm{Log}z)$ and integrate $z^a/(1+z^2)$ along the positively oriented upper semicircle, indented above zero. The straight pieces run from $-R$ to $-\epsilon$ and from $\epsilon$ to $R$; the small arc runs clockwise from argument $\pi$ to zero.

On the negative real axis $z^a=|z{|}^a{e}^{i\pi a}$, so the two straight pieces tend to $(1+e^{i\pi a})I(a)$. The outer arc is $O(R^{a-1})$ and the inner arc is $O({\epsilon }^{a+1})$, both tending to zero. The only enclosed pole is $i$, with residue $e^{i\pi a/2}/(2i)$. The residue theorem therefore givesUsing $1+e^{i\pi a}=2e^{i\pi a/2}\cos (\pi a/2)$ yieldsThe same bounds show convergence of the real integral at both zero and infinity.

Use the same indented upper-semicircle contour and branch of the complex logarithm, now for $F(z)=z^{1/2}\mathrm{Log}z/(1+z^2)$. Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. On the negative axis, $z^{1/2}=i|z{|}^{1/2}$ and $\mathrm{Log}z=\log |z|+i\pi$. Consequently the two straight contributions tend toThe outer arc is $O(R^{-1/2}\log R)$ and the small arc is $O({\epsilon }^{3/2}(|\log \epsilon |+1))$, so neither contributes in the limit. At the enclosed pole $i$,The residue theorem gives $(1+i)J-\pi I(1/2)=i{\pi }^2{e}^{i\pi /4}/2$. Part (a) gives $I(1/2)=\pi /\sqrt{2}$. Substituting and using $e^{i\pi /4}=(1+i)/\sqrt{2}$, the right side after moving the $I$ term is ${\pi }^2(1+i)/(2\sqrt{2})$. Hence

Take the branch of the complex logarithm with $-\pi <\arg s<\pi$ and write $s^{-1/2}=\exp [-\log s/2]$. Deform the Bromwich contour into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt{r}$, while on the lower lip it is $i/\sqrt{r}$. Keeping the lip orientations in the Bromwich inversion with a square-root branch cut givesWith $r=u^2$, the supplied Gaussian integral evaluates this as $2{\pi }^{-1}{\int }_0^{\infty }{e}^{-t{u}^2}du$. HenceThe inverse is singular but locally integrable at zero, which is compatible with existence of its Laplace transform.

Choose the principal branch of the complex logarithm on $H$, with $-\pi <\arg z<\pi$. ThenThus $-\pi <\mathrm{Re}w<\pi$ and the imaginary part ranges over all real numbers. Conversely, for $w=x+iy$ in the strip, $z=e^{iw}=e^{-y}{e}^{ix}$ is in $H$ and its principal logarithm is $iw$. Therefore $\boxed{w=-i\log z:H⟶S}$ is an analytic bijection, with analytic inverse $z=e^{iw}$. Every other continuous branch on the connected domain is $\log z+2\pi ik$ for some fixed integer $k$, so its image is the translated strip $\boxed{S+2\pi k}$.

For $G$, introduce the Möbius transformation $\zeta =(z-1)/(z+1)$, whose inverse is $z=(1+\zeta )/(1-\zeta )$. Writing $\zeta =a+ib$, its defining circle inequalities transform as follows:andHence $G$ maps bijectively onto the wedge $\pi /2<\arg \zeta <3\pi /4$. The principal branch of the complex logarithm is analytic throughout this wedge. Multiplying its logarithm by $-8i$ expands the interval of real parts to $(4\pi ,6\pi )$, and subtracting $5\pi$ centers it at zero. The required conformal map is thereforeIts real part is $8\arg \zeta -5\pi \in (-\pi ,\pi )$, and its imaginary part $-8\log |\zeta |$ takes every real value. More explicitly, its inverse on $S$ is obtained from $\zeta =\exp (i(W+5\pi )/8)$ and $z=(1+\zeta )/(1-\zeta )$, which verifies both surjectivity and injectivity.

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

Write $z=r{e}^{i\theta }$ with $r>0$. On a fixed branch of the complex logarithm on which $\log z=\log r+i\theta$ and $\log \bar{z}=\log r-i\theta$, the equation becomesIts real and imaginary parts give the same condition $\log r=\theta$, orThis is a logarithmic spiral. For the principal complex logarithm, it is the portion parametrized by $-\pi <\theta <\pi$ that avoids the branch cut.

For a positively oriented simple closed contour $C$, with no pole on $C$ and with $f$ holomorphic inside except at finitely many poles, the residue theorem statesFor a more general closed contour, each residue is multiplied by its winding number about the pole.

The powers must be defined using the same branch of the complex logarithm as the numerator. A branch suitable for both requested calculations isThe branch cut is a ray just below the positive real axis, and $z^{\beta }:=e^{\beta \mathrm{Log}z}$. Since $\beta >1$, the displayed choice of $\delta$ is possible. It includes the required upper half-plane and the sector used below. The origin remains an excluded branch point.

In $0\le \arg z<\pi$, all zeros of $1+z^{\beta }$ areEach is a simple pole, since $\beta {\zeta }_j^{\beta -1}=-\beta /{\zeta }_j\ne 0$. Its residue isThe strict inequality matters if a root lies on the negative real axis, which is outside the specified region.

To evaluate the two improper integrals, write them as $I$ for the logarithmic numerator and $J$ for numerator one, and set $\theta =\pi /\beta$, $\alpha =2\theta$. Use a counterclockwise annular sector with $ϵ<|z|<R$ and $0<\arg z<\alpha$. It encloses precisely ${\zeta }_0=e^{i\theta }$. The outward positive-axis segment contributes $I$ in the limit. On the inward segment $z=r{e}^{i\alpha }$, the logarithm is $\log r+i\alpha$ and $z^{\beta }=r^{\beta }$, so that segment contributes $-e^{i\alpha }(I+i\alpha J)$.

The small arc is $O(ϵ|\log ϵ|)$, and the large arc is $O(R^{1-\beta }\log R)$; both vanish. These estimates follow from the arc lengths and $|1+z^{\beta }|\ge ||z{|}^{\beta }-1|$. The real integrals also converge absolutely at both endpoints for $\beta >1$. The residue theorem now givesAfter division by $e^{i\theta }$, the real and imaginary parts areSince $0<\theta <\pi$, $\sin \theta >0$. Hence the logarithmic beta integral from a sector contour yieldsFor $1<\beta <2$ the sector angle exceeds $\pi$. This is why choosing the principal complex logarithm on the whole sector would be invalid; the displaced cut above handles that range as well as $\beta \ge 2$.

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.

The derivative of an analytic function must be independent of the direction of approach. Taking a real increment gives $f^{\prime }(z)=u_x+i{v}_x$; taking an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:A real harmonic function is a twice continuously differentiable function whose Laplacian vanishes. For the given logarithmic function and $x^2+y^2>0$,It is therefore harmonic everywhere on the punctured complex plane.

On any domain admitting a branch of the complex logarithm, chooseIts derivatives are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal complex logarithm gives such a harmonic conjugate on the plane cut along the negative real axis.

There is no single-valued harmonic conjugate on the whole punctured plane. Indeed its putative differential would satisfywhereas a globally defined function has zero integral of its differential around a closed curve. This is the log modulus has no global harmonic conjugate on the punctured plane obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.