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Dan Romik works in number theory, combinatorics and probability theory.^[ref] His Complex Analysis Lecture Notes by Dan Romik also develops holomorphic functions, contour integrals and the geometry of the Riemann sphere, providing an expository connection between complex analysis and geometry.

Identity source: www.math.ucdavis.edu/people/general-profile?fac_id=romik

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.

An isolated singularity of a complex function at $a$ is a missing or nonholomorphic point surrounded by a punctured disk $0<|z-a|<r$ on which the function is holomorphic. The punctured-disk hypothesis is decisive: it gives a Laurent series centered at $a$ and the three-way classification of isolated singularities.

At zero, $\sin z/z$ has a removable singularity, $z^{-2}$ has a double pole, and $e^{1/z}$ has an essential singularity. Their principal parts contain respectively no negative powers, finitely many negative powers, and infinitely many negative powers. For the first example, assigning the limiting value one fills the puncture holomorphically. For the second, multiplying by $z^2$ leaves a nonzero regular value. The third has the convergent punctured-plane expansion ${\sum }_{n\ge 0}{z}^{-n}/n!$.

An accumulation of other singularities changes the question. In reciprocal cosine with accumulating poles, every disk about zero contains poles, so zero is a non-isolated singularity and the same Laurent classification does not apply there. At infinity use the reciprocal coordinate $w=1/z$, as in pole at infinity and removable singularity at infinity.^[ref]

The reciprocal chart makes infinity an ordinary coordinate origin. Its domain is the Riemann sphere minus the point $z=0$; $w(\infty )=0$ and $w(z)=1/z$ elsewhere. A neighbourhood $U_R$ of infinity becomes the open disc $|w|<1/R$. In sphere coordinates this is exactly the capFor $R=2$, the exterior $|z|>2$ together with infinity becomes $Z>3/5$ and $|w|<1/2$. The boundary itself is excluded from each open set.

The point $z=2+i$ lies in this neighbourhood because $|z|=\sqrt{5}>2$. Its reciprocal coordinate is $w=(2-i)/5$, with $|w|=1/\sqrt{5}<1/2$, and its sphere height is $Z=2/3>3/5$. The negative imaginary part of $w$ is the axis reversal required by the holomorphic stereographic atlas of the sphere.

For a complex-valued holomorphic function $f$ defined outside a large disc, having a removable singularity at infinity means that $g(w)=f(1/w)$ extends to a holomorphic function at $w=0$. This definition reduces questions at infinity to familiar questions at a finite point.^[ref] For example,The geometric series gives a Taylor series at the new origin, so $f(\infty )=1$ and $g^{\prime }(0)=-5$. The coefficient $-5$ is the derivative in the specified $w$ coordinate. In contrast, $f(z)=z^2$ becomes $g(w)=w^{-2}$, displaying a pole of order two at infinity when the target is the complex plane. As a holomorphic map to the Riemann sphere, it uses the target reciprocal coordinate $1/f(1/w)=w^2$, which is regular at zero.

The Riemann removable singularity theorem states that a holomorphic function on a punctured disk extends holomorphically over its center if and only if it is bounded sufficiently near that center. Necessity follows from continuity of the extension on a smaller closed disk.^[ref]

For sufficiency translate the puncture to zero and assume $|f(z)|\le M$. Define $g(z)=z^{2}f(z)$ for $z\ne 0$ and $g(0)=0$. Then $g$ is continuous at zero andIt is holomorphic at every other point too, hence holomorphic throughout the disk. Its Taylor series has zero constant and linear coefficients, so $g=z^{2}h$ for a holomorphic $h$. On the punctured disk $h=f$, furnishing the extension. The identity theorem makes that extension unique.

For example, $\sin z/z$ extends with value one. By contrast, $e^{1/z}$ is bounded along some approach directions but unbounded along others, so one-direction bounds cannot invoke the theorem. Boundedness must hold throughout a punctured neighborhood.

The Riemann sphere is $\mathbb{C}\cup \{\infty \}$ with two charts related by reciprocal coordinates.

The extra point completes the complex plane into a compact Riemann surface. It is one point shared by every direction of escape: $n$, $-n$ and $in$ all tend to $\infty$ as $n\to \infty$. A basis of neighbourhoods at $\infty$ isThis is the one-point compactification of the complex plane; convergence to $\infty$ means that the modulus tends to infinity.^[ref]

For a concrete picture, use the unit sphereand identify $z=x+iy$ with $(x,y,0)$ in its equatorial plane. The north pole $N=(0,0,1)$ represents $\infty$ and the south pole $S=(0,0,-1)$ represents $0$. The stereographic projection below gives the identification. This convention fixes the scale: the equator corresponds to $|z|=1$.^[ref]

Why add this point? The Möbius transformation $z\mapsto 1/z$ then exchanges $0$ and $\infty$ and becomes a holomorphic map everywhere on the Riemann sphere. Its apparent blow-up at zero disappears when its output is expressed in the reciprocal chart at infinity: $w=1/(1/z)=z$. This is a concrete instance of a meromorphic function as a holomorphic map to the Riemann sphere. The additional point makes the complex plane compact, while its second manifold chart lets us do ordinary complex analysis near infinity.