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No. The nonvanishing holomorphic functions $f_n(z)=e^{nz}$ satisfy $f_n(0)=1$, butAvoiding zero alone permits an unbounded holomorphic logarithm's real part; the omitted ray in part (c) was what constrained the image after taking a square root.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

Put $w=e^x-1=x+x^2/2+x^3/6+O(x^4)$. The local logarithmic series $\ln (1-w)=-w-w^2/2-w^3/3+O(w^4)$ givesFor the infinite Taylor series, it is necessary to distinguish the real domain of the logarithm, $x<\ln 2$, from its convergence interval about zero. Write $a=\ln 2$. The complex zeros of $2-e^z$ are $a+2\pi ik$ for integers $k$, so the nearest logarithmic singularity to zero is at $a$. The radius of convergence is therefore $a$, giving convergence for $|x|<a$ and divergence for $|x|>a$.

The two endpoints need separate checking. Factor off the nearest singularity:The quotient has a nonzero removable value $2a$ at $z=a$; its other zeros are at $a+2\pi ik$ with $k\ne 0$. Consequently it has a holomorphic logarithm on a zero-free disk larger than $|z|\le a$, which is a simply connected domain, and the Taylor series of $H$ converges absolutely at both real endpoints. At $x=-a$, the remaining series $-{\sum }_{n\ge 1}(-1{)}^n/n$ converges by the alternating series test. At $x=a$, it is $-{\sum }_{n\ge 1}1/n$ and diverges. Thus the full real convergence set isConvergence is absolute in the open interval and conditional at the left endpoint, where the sum equals $\ln (3/2)$. This endpoint distinction is an instance of isolating a logarithmic singularity to test a power-series endpoint.

For the upper Apollonius circle, let its constant ratio be $P>1$. Matching the center and radius from part (i) givesSubtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. AlsoThe relevant inverse hyperbolic cosine is positive. Now $\log |(z+ib)/(z-ib)|$ is the real part of a local holomorphic logarithm, hence a harmonic function away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.

On $y=0$ the numerator and denominator have equal modulus, so the logarithm is zero. On the circular boundary its value is $\log P=\mathrm{arcosh}(y_0/R)$. Therefore the harmonic potential between a grounded line and a circular boundary isThis verifies Laplace's equation and both prescribed Dirichlet boundary conditions. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.

On the slit domain $D$, choose the complex argument $\theta ={\mathrm{Arg}}_{(0,2\pi )}z$. The resulting holomorphic logarithm is $L(z)=\log |z|+i\theta$, with $L^{\prime }(z)=1/z$. DefineIt is holomorphic as a composition of holomorphic functions, and $f(z{)}^2=e^{L(z)}=z$. At $z=-1$ the chosen argument is $\pi$, so $f(-1)=i$.

Suppose a holomorphic square root $g$ existed on the whole punctured complex plane. It would be nonzero, and differentiating $g^2=z$ would give $2g{g}^{\prime }=1$. There are several ways to express the resulting obstruction; a direct one uses the unit circle. SetThe chain rule and $g^{\prime }(z)=g(z)/(2z)$ imply $q^{\prime }(t)=0$, so $q$ is constant. ButThese cannot agree because $g(1{)}^2=1$. Thus no such global holomorphic square root exists. Going once around zero changes the sign of the locally chosen root.

Use the same principal square root $S(z)=\sqrt{(z-a)/(z-b)}$, which is holomorphic on $A$ because its argument avoids the nonpositive real axis. Thenis a holomorphic function satisfyingThus it is a holomorphic branch of the requested square root. This construction does not require a global holomorphic logarithm of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to 1$ at infinity, so $g(z)/z\to 1$. The negative of this function is the other holomorphic branch; any two branches have a continuous quotient taking values in $\{1,-1\}$, which is constant because $A$ is connected.

The two boundary circles meet at $z=\pm 1$: subtracting their equations gives $\mathrm{Im}z=0$, and either equation then gives $(\mathrm{Re}z{)}^2=1$. Send these points to zero and infinity by the Möbius transformationThe point $-1$ is on the boundary, so $T$ is holomorphic throughout $B$. Write $w=u+iv$. Substituting the inverse gives the exact equivalencesThusThere is no missing point from the inverse here: $w=1$ lies outside this sector. The image is therefore exactly the stated sector, rather than merely a subset of it.

Rotate by $e^{-i\pi /4}$ to obtain the first quadrant, and take the power $2/3$ using its holomorphic logarithm with $0<\arg \zeta <\pi /2$. The required map isHere the principal logarithm works because the expression inside it lies in the first quadrant. It multiplies its argument by $2/3$, giving $0<\arg \Phi <\pi /3$, and maps its modulus $r>0$ to $r^{2/3}$, which ranges over all positive real numbers.

For explicit surjectivity and injectivity, given $\xi$ with $0<\arg \xi <\pi /3$, definewhere ${\xi }^{3/2}$ uses $0<\arg \xi <\pi /3$. Then $\pi /4<\arg w<3\pi /4$, so $w\ne 1$ and $z\in B$, and these are inverse operations. Each stage is holomorphic with nonzero derivative, proving that $\Phi$ is the requested conformal bijection.

Let $F=e^g$. It is holomorphic and nowhere zero on a punctured neighbourhood of $a$, and the chain rule givesIf $F$ had a pole of order $n\ge 1$, part (c) would imply ${\mathrm{Res}}_a(g^{\prime })=-n$. But the derivative of a holomorphic function on a punctured disc has residue zero. Indeed, write its Laurent seriesTermwise differentiation is valid on every smaller closed annulus. The only possible $(z-a{)}^{-1}$ term comes from $k=0$ and has coefficient zero. Thusa contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves an exponential has no isolated pole, including when $g$ has an essential singularity; no choice of a holomorphic logarithm on the entire punctured disc was assumed.

Every closed discrete subset of a plane domain is the zero set of a holomorphic function, with assigned finite positive multiplicities if desired. Exhaust the domain by relatively filled compact sets. For a prescribed point outside an earlier compact, choose a rational factor with that zero and its possible pole outside the domain in the same complementary component. A slit joining zero and pole allows a holomorphic logarithm near the earlier compact. Approximate that logarithm by Runge theorem and multiply the rational factor by the negative exponential of the approximant. The resulting factors tend to one with summable errors on each compact. Their infinite product has the prescribed zeros and no additional zeros.

// Destination: analysis.bigb

For real $s>1$, the logarithm defined by the Euler product of a Dirichlet L-function differs from $F_{\chi }(s)$ by a quantity of absolute value at most ${\sum }_{n\ge 2}1/(n(n-1))=1$. If $L(\chi ,1)\ne 0$, a local holomorphic logarithm differs from this continuous logarithm by one fixed multiple of $2\pi i$ on a short real interval, so $F_{\chi }(s)$ stays bounded as $s\downarrow 1$. For the principal Dirichlet character, the pole of the Riemann zeta function instead gives $F_{{\chi }_0}(s)=\log (1/(s-1))+O(1)$. Orthogonality of Dirichlet characters then makes the sum over any reduced residue class diverge, proving infinitude of its primes.

For real boundary data $u$, the displayed holomorphic integral has real part equal to the disk Poisson integral of $u$ and imaginary part its normalized harmonic conjugate. If $u\ge 0$ is not identically zero, the real part is strictly positive inside the disk. A holomorphic logarithm then has imaginary part between $-\pi /2$ and $\pi /2$, while its real part records the logarithm of the magnitude. This separates uniform imaginary-part control from large real parts on small boundary sets.

For a zero-free holomorphic function $f:\mathbb{D}\to \mathbb{D}$, put $c=f(0)$ and $\alpha =-\log |c|$. A holomorphic logarithm makes $-\log f$ have positive real part. The Carathéodory derivative bound for the right half-plane yields the displayed sharp bound. Equality functions are $f(z)=c\exp [-2\alpha \omega z/(1-\omega z)]$, $|\omega |=1$. The Harnack inequality on the unit disk also yields the sharp pointwise bounds $|c{|}^{(1+r)/(1-r)}\le |f(w)|\le |c{|}^{(1-r)/(1+r)}$ for $r=|w|$. Aligning $\omega w=\pm r$ attains the two endpoints.