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Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.

For the stochastic differential equation $d{X}_s=b(s,X_s)ds+\Sigma (s,X_s)d{B}_s$, the diffusion generator on smooth functions isOne form of the Feynman-Kac formula is the following. Suppose the stochastic differential equation is well posed and nonexplosive, $c\ge 0$ is bounded, the terminal data $g$ and source $h$ are bounded, and $F$ is a classical solution bounded on finite time intervals ofWith the regularity needed for the Itô formula, the parabolic Feynman-Kac formula with a source statesHere ${\mathbb{E}}_{t,x}$ means expectation for the diffusion process started at $X_t=x$. Under hypotheses ensuring existence of a classical solution, this representation also identifies that solution uniquely in the stated bounded class. In the remaining parts, $c=h=0$, so $F(t,x)={\mathbb{E}}_{t,x}g(X_T)$ solves the terminal-value Kolmogorov backward equation $F_t+{\mathcal{L}}_{t}F=0$. Complex terminal data are allowed by applying the real formula to their real part and imaginary part separately.

For an unstable Taylor–Goldstein equation mode in a finite channel, take $a=1$ in the power-transformed Taylor–Goldstein energy identity. Its imaginary part givesThus the real phase velocity lies strictly between the extremes of a nonconstant smooth shear profile. Equality would force the regular eigenfunction to vanish on an interval, hence everywhere by uniqueness for its ordinary differential equation.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

Let $a>0$, $t>0$ and $z=t+ia/t$. The real and imaginary coordinates obey $XY=a$, a hyperbola. Put $s=t^2-a^2/t^2$, which is strictly increasing. Then $z^2=s+2ia$ traces a horizontal line, and $z^4=s^2-4a^2+4ias$ traces a parabola. Eliminating $s$ gives the displayed quadratic relation between real part and imaginary part. The value of $t$ fixes an orientation and determines which segment is traversed; the full real parabola is obtained as $t$ ranges over $(0,\infty )$.

The principal argument selects a single complex argument of a nonzero complex number, conventionally in $(-\pi ,\pi ]$. It is the imaginary part of the principal logarithm. The alternative interval $[-\pi ,\pi )$ changes the convention on the negative real axis. Neither convention makes the selected complex argument continuous across that axis.

The real axis consists of the points of the complex plane whose imaginary part is zero. These points represent the real numbers.

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.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.

The upper half-plane is the open setIt contains precisely the complex numbers with positive imaginary part: $1+i$ is inside, $1$ lies on its boundary, and $1-i$ belongs to the lower half-plane. In the complex plane its boundary is the real line; in the Riemann sphere the boundary also includes infinity. It is a convex set, hence a simply connected domain, and the Cayley transform between the half-plane and disk makes it biholomorphic to the unit disc.

With the hyperbolic metric $ds=|dz|/\mathrm{Im}z$, the same domain becomes the Poincaré half-plane model. Its hyperbolic geodesics are vertical lines and upper semicircles whose centres lie on the real line. Under the Cayley transform, these become arcs of circles orthogonal to the unit circle, with diameters as the straight-line cases. ^[ref]

Matching colours identify each hyperbolic geodesic and its image. The vertical line through zero maps to a diameter; the vertical line through one maps to an arc ending at $-i$ and $1$. A semicircle from $-1$ to $1$ also maps to a diameter, whereas the semicircle from $0$ to $2$ maps to a curved arc. Their endpoints lie on the boundary, outside the domains: the endpoints are infinitely far from interior points in hyperbolic distance.

The upper half-plane is useful because its straight boundary and explicit upper half-plane automorphisms make conformal maps and hyperbolic geometry calculable. The picture explains why shortest hyperbolic geodesic routes can look curved in Euclidean coordinates; their lengths are measured with the position-dependent hyperbolic metric.

For historical context, in his letter to Klein of 22 September 1882, Poincaré constructs a quadrilateral bounded by four circles orthogonal to a fundamental circle, and studies transformations generating repeated copies. In modern language, the sides inside that circle are hyperbolic geodesics in a disk model. The letter directly exhibits the connection between transformation groups, geometry and complex functions. ^[ref]