Start with a fast massive projectile following an approximately straight path. Its time-dependent electric field perturbs bound absorber electrons. At large impact parameter, treating the perturbation to first order and summing excitation and ionization transition probabilities gives an energy transfer weighted by the absorber's oscillator strengths. The oscillator-strength sum rule reduces the leading coefficient to the number of target electrons, while their logarithmically weighted excitation energies define $I$.

The classical impulse picture explains the logarithm: a transverse impulse scales as $1/(bv)$, so transferred energy scales as $1/(b^2{v}^2)$; multiplying by the number of encountered electrons in $2\pi bdbdx$ gives $db/b$. Quantum scattering replaces ad hoc short- and long-distance cutoffs by recoil kinematics and bound-state excitation. Relativistic projectile kinematics supplies $T_{\max }$ and the $-{\beta }^2$ term. Summation over many independent encounters gives the mean stopping power, rather than the energy-loss straggling of individual tracks.

The leading approximation assumes $M\gg m_e$, a sufficiently fast projectile for perturbation theory, a largely fixed projectile charge, and weak correlations between collisions. At low velocity, electron binding, charge exchange and nuclear stopping become important; at high velocity, polarization screens distant interactions and gives the density correction. The projectile's bremsstrahlung is suppressed by its large mass and is excluded from the collision stopping power.

Define the positive mass collision stopping power as $S_{\mathrm{col}}/\rho =-(1/\rho )dT/dx$. For a projectile of charge $ze$, mass $M\gg m_e$, speed $v=\beta c$ and Lorentz factor $\Gamma =(1-{\beta }^2{)}^{-1/2}$, the Bethe stopping formula can be writtenHere $Z$ is the absorber atomic number, $A$ its molar mass, $I$ its mean excitation energy, $r_e=e^2/(4\pi {ϵ}_0{m}_e{c}^2)$, and $N_A$ the Avogadro constant. AlsoThe optional $\delta$ and $C/Z$ represent density and shell corrections. For a very heavy projectile $T_{\max }\simeq 2m_e{c}^2{\beta }^2{\Gamma }^2$, and the bracket reduces to $\ln (2m_e{c}^2{\beta }^2{\Gamma }^2/I)-{\beta }^2$ when those corrections are neglected. If $T$ denotes remaining projectile kinetic energy, the printed $dT/(\rho dx)$ is negative; the conventional positive stopping power has the minus sign above.^[ref]

After thermalization, approximate the electron and Positron as initially at rest in their centre-of-momentum frame. Their total energy is $2m_e{c}^2$ and total momentum is zero. Two photons must therefore have equal energies and opposite momenta, since $|{\mathbf{p}}_{\gamma }|=E_{\gamma }/c$. Conservation of energy givesResidual pair momentum and electron binding cause small departures from exact collinearity and exact photon energy; the standard PET calculation neglects them.

Single-photon emission computed tomography detects individual gamma rays from a tracer. Mechanically collimated gamma cameras acquire projections from multiple angles for tomography. The collimator discards most emitted photons, creating a strong sensitivity–spatial resolution trade-off.

Positron emission tomography uses a Positron-emitting tracer. The emitted Positron travels a short distance before electron-positron annihilation produces an approximately back-to-back photon pair. Coincident detection in a detector ring defines a line of response, giving electronic collimation and greater detection efficiency; time-of-flight PET further constrains position along that line. The positron range, non-collinearity, detector size, scatter and accidental coincidences limit spatial resolution and quantification. In both hybrids, computed tomography supplies anatomy and an attenuation correction map, with attention to registration and respiratory motion.^[ref]

A diagnostic example is technetium-99m methylene diphosphonate, which accumulates at sites of bone turnover for bone scintigraphy. A therapeutic example is sodium iodide iodine-131, taken up by thyroid tissue and used for selected thyroid disease including differentiated thyroid cancer.

For diagnosis, an escaping gamma ray or a Positron followed by annihilation photons allows external detection, with an appropriate short half-life and limited deposited energy. For molecular radiotherapy, short-range charged emissions such as beta particles, alpha particles or Auger electrons deposit substantial local energy in the targeted tissue. Iodine-131 emits therapeutic beta particles as well as imageable gamma rays, so these categories are functional rather than mutually exclusive isotope classes.

After injection, the radiopharmaceutical circulates and accumulates according to its biological targeting and clearance. Technetium-99m undergoes an isomeric nuclear transition, producing approximately $140\mathrm{keV}$ gamma rays; its roughly six-hour half-life permits useful counting with limited persistence. Some photons undergo absorption or Compton scattering in the patient. A lead collimator in the gamma camera accepts a narrow range of directions so the accepted event retains positional information.

An accepted gamma ray deposits energy, commonly in a NaI(Tl) scintillator, by photoelectric effect or Compton scattering. The resulting visible particle scintillation light is collected by photomultipliers. Their relative pulse sizes determine an event position and their summed pulse estimates its energy. A pulse-height analyser accepts an energy window around the photopeak to reject many scattered events. Position and uniformity corrections are applied; accepted events increment a two-dimensional pixel histogram. After acquisition the computer displays this count map as a planar projection of tracer activity. Finite counting statistics, attenuation, scattering and the collimator limit the image's quantitative accuracy and spatial resolution.^[ref]

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

The square root produces branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$. One branch cut can join them along $[-1,1]$. Another choice uses the two rays $(-\infty ,-1]\cup [1,\infty )$. Each choice prevents a loop from enclosing just one branch point within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a pole of order three, not a branch point, since

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

Cauchy's integral theorem states that a holomorphic function on a simply connected open region has zero contour integral around every closed piecewise smooth contour in that region. More generally the same holds for a contour homologous to zero in its domain of holomorphy. Having no singularity on the curve alone is insufficient: $1/z$ has a nonzero contour integral around the unit circle.

If $f(z)=\sum a_j{z}^j$, then $f(-z)=\sum (-1{)}^j{a}_j{z}^j$ and its residue at zero is $-a_{-1}$. The circle $|z-1|=2$ encloses zero and no other finite singularity. Thus the residue theorem givesfor positive orientation.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.