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A germ of a holomorphic function at $z\in D$ is an equivalence class $[f{]}_z$ of pairs consisting of a neighbourhood $U$ of $z$ and a holomorphic function $f:U\to \mathbb{C}$. Two pairs represent the same germ if their functions agree on some neighbourhood of $z$.

The space of germs of holomorphic functions isFor every holomorphic $f$ on an open $U\subseteq D$, the setis declared open; these sets form a basis. The forgetful mapmaps each ${\tilde{U}}_f$ homeomorphically onto $U$. The inverses of these restrictions give the charts defining the complex structure, so $\pi$ is a local biholomorphism.

The evaluation map isIt is well-defined by the germ equivalence relation. On the chart ${\tilde{U}}_f$, its coordinate expression iswhich is holomorphic. Hence $\mathcal{E}$ is analytic, as stated in evaluation map on a space of germs.

Now putThis gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch $q(z)$ of $\sqrt{z^8-1}$. Take two copies, labelled by $q$ and $-q$, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in $D$.

On this surface defineFor each $(z,w)\in R$, the holomorphic function $z^8-1$ is nonzero near $z$, so it has a unique local square root $q$ with $q(z)=w$. DefineThis is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is $[q{]}_z\mapsto (z,q(z))$. Thus $\Phi$ is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently

Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map $\bar{\pi }:\overline{R}\to {\mathbb{C}}_{\infty }$. Its only ramification consists of the eight simple finite branch points, soThereforeThis is the compactification of y squared equals x to the eighth minus one.

PutThis slit complex plane is a simply connected domain. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

Fora finite point $z_0$ is an ordinary point exactly when $p$ and $q$ are holomorphic there. It is a regular singular point exactly whenare holomorphic at $z_0$.

Put $\zeta =1/z$ and $W(\zeta )=w(1/\zeta )$. Direct differentiation givesso the transformed equation isConsequently $z=\infty$ is ordinary precisely whenare holomorphic at $\zeta =0$. It is a regular singular point at infinity precisely when $zp(z)$ and $z^{2}q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the Laurent series of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hencefor constants $a,b$. The equation is a Cauchy-Euler differential equation. Its indicial equation isFor distinct roots ${\lambda }_1,{\lambda }_2$, the general solution on a domain with a chosen logarithm branch isFor a repeated root $\lambda$, the general solution of an Euler-Cauchy equation is

Finally require infinity to be ordinary. In the transformed equation its coefficients becomeBoth are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is

Suppose a $C^1$ classical solution existed and set $w=u+iv$. The system is preciselywhich are the Cauchy-Riemann equations for $w$ as a function of the complex number $z=x+it$. Hence $w$ is holomorphic near $z=0$. Every holomorphic function is real analytic, so its restrictionto the real axis is real analytic near zero. Its real part and imaginary part show that both $f$ and $g$ must be real analytic, contradicting the hypothesis. Therefore no such $C^1$ solution exists.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.

Write $M(x)={\sum }_{n\le x}\mu (n)$ for the Mertens function. Suppose, to the contrary, that for some $\epsilon >0$ the quotient $|M(x)|/x^{1/2-\epsilon }$ were bounded. Partial summation would then makeconverge and define a holomorphic function throughout $\Re s>1/2-\epsilon$. In $\Re s>1$ the Euler product identifies this function with $1/\zeta (s)$, so analytic continuation would make $1/\zeta (s)$ holomorphic in that larger half-plane.

By assumption, $\zeta$ has a nontrivial zero. The functional equation of the Riemann zeta function reflects one of that zero and its partner into $\Re s\ge 1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\epsilon }$ is unbounded, which gives an $x\ge 1$ exceeding any prescribed constant $C$.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Compute the complex derivative along horizontal and vertical increments. Along the real direction, $f^{\prime }=u_x+i{v}_x$; along the imaginary direction, $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equationsThe real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, proving that both are harmonic functions.

For the first real part, $u_x=2x$ and $u_y=-2y$. The Cauchy-Riemann equations require $v_y=2x$ and $v_x=2y$, so its harmonic conjugate is $v=2xy+C$. ThereforeFor the second real part, $u=x/(x^2+y^2)$ is the real part of $1/z$. Its harmonic conjugate is $-y/(x^2+y^2)+C$, givingThese are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.

Compute the complex derivative along real increments to get $f^{\prime }=u_x+i{v}_x$, and along imaginary increments to get $f^{\prime }=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equations, $\boxed{u_x=v_y,u_y=-v_x}$.

For the specified real part, observe that $iz{e}^{-z}=e^{-x}[(x\sin y-y\cos y)+i(x\cos y+y\sin y)]$. Thus the most general holomorphic function with this real part on the connected plane isThe difference of two such holomorphic functions has zero real part, so the Cauchy-Riemann equations make its imaginary part constant.

Next, $\cosh z=\cosh x\cos y+i\sinh x\sin y$. Its real and imaginary parts satisfy $u_x=\sinh x\cos y=v_y$ and $u_y=-\cosh x\sin y=-v_x$. Their partial derivatives are continuous everywhere, proving that the complex hyperbolic cosine is entire.

For the Laurent series, set $h=z-1$. Since $z^2-1=h(2+h)$,The regular part at a simple pole therefore gives

A holomorphic function is complex differentiable throughout an open region; equivalently it is locally represented by a convergent complex power series. For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$, $u_y=-v_x$. With continuous first partial derivatives, these equations imply holomorphicity and $f^{\prime }=u_x+i{v}_x$.

For $e^z=e^x\cos y+i{e}^x\sin y$, the equations hold and $u_x+i{v}_x=e^x(\cos y+i\sin y)=e^z$. For $\sin z=\sin x\cosh y+i\cos x\sinh y$, they also hold, and $u_x+i{v}_x=\cos x\cosh y-i\sin x\sinh y=\cos z$. Therefore

At a point of complex differentiability, the complex derivative is independent of the direction of approach. Taking increments along the real axis gives $f^{\prime }=u_x+i{v}_x$; taking imaginary increments gives $f^{\prime }=v_y-i{u}_y$. Equality of their real parts and imaginary parts proves the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

For the specified real part, observe that $z^2{e}^z$ has exactly that real part. Equivalently, integrating $v_y=u_x$ and checking $v_x=-u_y$ givesThusThe freedom is only a constant imaginary part: the difference of two such holomorphic functions has zero real part, so its Cauchy-Riemann equations make it constant on each connected domain.

Cauchy's integral theorem states that a holomorphic function on a simply connected domain has zero contour integral around every closed piecewise smooth curve in that domain. One proof begins with the Cauchy theorem for a triangle. Subdivide a triangle into four equal triangles. Their oriented boundary integrals sum to the original integral because internal edges cancel, so choose a nested subtriangle carrying at least one quarter of the preceding integral's modulus. At stage $n$, its perimeter and diameter are $P/2^n,d/2^n$, and the nested triangles converge to a point $z_0$. Complex differentiability gives $f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$, with $ϵ(z)\to 0$. The affine terms have primitives and integrate to zero. Therefore $|I_n|\le \sup |ϵ|Pd/4^n$, while $|I_0|\le 4^n|I_n|$. Letting $n\to \infty$ proves $I_0=0$. Triangulation cancels internal edges for polygonal contours; local primitives and deformation through the domain extend the result to the stated piecewise smooth curves. Simple connectedness ensures the necessary contractions do not pass through a missing point.

Apply the theorem to $e^{-a{z}^2}$ on the rectangle between heights zero and $\rho$, with vertical sides at $\mathrm{Re}z=\pm R$. On either vertical side,The modulus of each side integral is at most $|\rho |e^{-a{R}^2+a{\rho }^2}\to 0$. Both horizontal integrals converge absolutely. The two horizontal orientations are opposite, so the limit gives $I(\rho ,a)=I(0,a)$, also for negative $\rho$. This is Gaussian contour translation.

Now complete the square:The substitution $z=x-i\nu /(2a)$ moves the path to height $-\nu /(2a)$. Its Gaussian integral equals the real-axis integral by the preceding result. HenceThis also gives the Fourier transform of a Gaussian in the convention with no prefactor in the forward transform.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives, which holds because holomorphic functions are smooth:Thus both parts are harmonic functions.

Evaluate the derivative along real and imaginary increments. For a holomorphic function, both give the same value:Equating real and imaginary parts yields the Cauchy-Riemann equations

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

The derivative of a holomorphic function can be evaluated using an increment along the real axis:Identifying the components of the gradient $\mathbf{V}=\nabla u$ gives

For $\mathrm{Re}z>0$, the defining Gamma integral converges at zero and at infinity and defines a holomorphic function. The Gamma function recurrence then supplies its meromorphic continuation to the rest of the complex plane, as used below.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives:Thus the real part is a harmonic function. The same calculation gives $\Delta v=0$. Every holomorphic function has the required smoothness.