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A complex number is $z=x+iy$, where $x,y$ are real numbers and the imaginary unit satisfies $i^2=-1$. The real part is $x$ and the imaginary part is $y$: for $3+4i$ they are $3$ and $4$, respectively. Equality requires both coordinates to agree. Thus the complex plane gives each complex number a concrete position, while ordinary addition and multiplication become geometric operations.

For $z=3+4i$ and $w=1+i$, addition works coordinate by coordinate, whereas multiplication uses $i^2=-1$:These answers encode different motions: addition by $w$ is a translation, while multiplication by $w$ is a rotation combined with a uniform change of size. Keeping these pictures distinct prevents a common beginner's confusion.

The complex numbers form a field: every nonzero $z$ has a multiplicative inverse, computable using its complex conjugate and modulus. They also supply the missing solutions of a quadratic equation such as $u^2+1=0$, namely $u=\pm i$. The surprising payoff is that extending arithmetic also makes Euclidean geometry easier to calculate. For a systematic introduction to these definitions, see Jeremy Orloff, MIT 18.04, topic 1, sections 1.3–1.5.

A branch point is a point around which a sufficiently small loop can change a germ of a holomorphic function under analytic continuation. This is a local condition: a large loop around a regular point might also enclose a distant branch point, so that large loop alone cannot classify the regular point. For the complex logarithm and square root, every nonzero point has a local holomorphic branch, but a loop once around $0$ changes the chosen branch. ^[ref]

For $z=r{e}^{it}$ with fixed $r>0$, start at $t=0$ with values $\sqrt{r}$ and $\log r$. Their continuations areOne counterclockwise turn gives $w(2\pi )=-\sqrt{r}$ and $L(2\pi )=\log r+2\pi i$. At $r=4$, the square root therefore changes from $2$ to $-2$, while the logarithm changes from $\log 4$ to $\log 4+2\pi i$. Two turns restore the square root to $2$; no nonzero number of turns restores the logarithm. This distinguishes a two-sheeted algebraic branch point from a logarithmic singularity.

A pole need not be a branch point: $1/z$ is unbounded near $0$, yet $1/(r{e}^{it})=r^{-1}{e}^{-it}$ returns to its starting value after one turn. Conversely, the square-root values tend to $0$ at their branch point. The issue is whether a single holomorphic choice survives going around the point, rather than whether the values become large. On the Riemann sphere, infinity is also a branch point of both $\sqrt{z}$ and $\log z$: in the local coordinate $\zeta =1/z$, these are ${\zeta }^{-1/2}$ and $-\log \zeta$, whose continuations around $\zeta =0$ change sign and subtract $2\pi i$, respectively. ^[ref]

A Bromwich contour is a vertical line in the complex frequency plane lying to the right of the singularities in the inverse Laplace integral.

A homotopy through closed paths avoiding the base point cannot change the integer winding number.

The winding number records how many net turns a closed piecewise smooth curve makes around a chosen point $w$. Counterclockwise turns contribute positively, clockwise turns negatively. It measures a traversal, including repeated visits; a drawing of the curve's outline alone can lose that information. For a closed curve $\gamma$ avoiding $w$,Equivalently, continuously follow an argument of $\gamma (t)-w$ along one journey and divide its total change by $2\pi$. The start and endpoint agree as complex numbers, so the change is an integer number of full turns. The principal argument at the endpoints cannot retain those turns; a continuous choice along the parametrized path can.^[ref]

For $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, the winding number about zero is one. Traversing the same unit circle twice makes it two; reversing its orientation of a contour makes it negative. The winding number about $w=2$ is zero: the same journey now stays to one side of the point being measured. Joining one counterclockwise loop and one clockwise loop around zero gives net winding number zero even though both loops encircle zero.

The winding number does not change during a homotopy of closed curves avoiding $w$. This is why it is useful: complicated curves can often be deformed to simple ones whose count is obvious. The argument principle transfers that robust geometric integer to a count of zeros of a function and poles. The example in boundary sampling can miss winding shows why finite samples of a curve need control between samples before they can certify the same winding number.

An elliptic function is a meromorphic function with two real-linearly independent periods ${\omega }_1,{\omega }_2$. A fundamental cell is a half-open parallelogramOpposite boundary integrals cancel. Consequently a nonconstant elliptic function has equally many zeros and poles in a cell, counted with multiplicity, and the sum of its pole residues in a cell is zero.

For a parameter $k$, one form of the incomplete elliptic integral of the first kind isIts value depends on the choices of square-root branches and, under analytic continuation around the branch points, on the integration path.

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the Gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the Gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

For real $a$ and nonzero real $\omega$,

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

If a continuous complex-valued function on a domain has zero integral around the boundary of every triangle contained in that domain, then it is holomorphic.

The principal complex inverse sine onis the branchwhose derivative equals one at zero.

It is the complex sine member of the inverse trigonometric functions.

A domain is simply connected when every closed path in it can be continuously contracted to a point while remaining in the domain.

Such a domain is a simply connected space.

A complex-valued function $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ when it has a complex derivative at every point of $U$. Requiring a derivative throughout a neighbourhood is much stronger than complex differentiability at a point. It forces locally convergent Taylor series and derivatives of every order, as follows from the Cauchy integral formula and Cauchy derivative formula.

To see the restriction in real coordinates, write $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. The real and imaginary increment quotients must agree, giving the Cauchy-Riemann equationsConversely, if $u,v$ have continuous partial derivatives on $U$ and satisfy the Cauchy-Riemann equations there, then $f$ is a holomorphic function. The equations at a lone point, without a differentiability hypothesis for the real coordinate map, are insufficient. Their role is to force the Jacobian matrix to have the special formSuch a matrix represents multiplication by a complex number; it has only two free real entries. ^[ref]

For $f(z)=z^2$, $u=x^2-y^2$ and $v=2xy$, so $u_x=v_y=2x$ and $u_y=-v_x=-2y$. Directly,At $z_0=1+i$, multiplication by $2+2i$ scales lengths by its modulus $2\sqrt{2}$ and rotates by its argument $\pi /4$. A horizontal increment $h=0.01$ gives the exact displacement $0.0201+0.02i$; a vertical increment $h=0.01i$ gives $-0.0201+0.02i$. The first-order predictions $0.02+0.02i$ and $-0.02+0.02i$ are perpendicular, and the errors have modulus $10^{-4}$.

At every nonzero $z_0$, this local rotation and scaling makes $z^2$ a local conformal map. At zero, the first-order map vanishes: $r{e}^{i\theta }$ becomes $r^2{e}^{2i\theta }$, so the argument doubles and the origin is a critical point. Globally $z$ and $-z$ have the same square. Local preservation of angles therefore coexists with global folding, a useful warning when using holomorphic functions as coordinate changes.

For $|z|<1$, the polylogarithm is ${\mathrm{Li}}_s(z)={\sum }_{n\ge 1}{z}^n/n^s$; contour formulas analytically continue it to a slit plane.

For susceptible, infective, and recovered populations $S,I,R$, mass-action infection coefficient $\beta$, recovery rate $\nu$, and immunity-loss rate $f$, the SIRS equations areThey conserve $S+I+R=N$ and have basic reproduction number ${\mathcal{R}}_0=\beta N/\nu$.

With susceptible, infective, and immune populations $X,Y,Z$, equal per-capita birth and death rate $\mu$, mass-action transmission coefficient $\beta$, and recovery rate $\nu$, the equations areThe conserved population is $N=X+Y+Z$, and the threshold population is $N_c=(\mu +\nu )/\beta$.

The basic reproduction number ${\mathcal{R}}_0$ is the expected number of secondary infections caused by one infective in an otherwise susceptible population.

The equationscombine mass-action infection with an exponential density-dependent birth model. Susceptibles reproduce at per-capita rate $b{e}^{-aS}$, while infectives have no reproductive term and die at total per-capita rate $d+\delta$.