| the higher-order-pole-terms control the modulus of a contour integration through a 1/z-pole of an otherwise analytic aka-holomorphic complex function so-much-so that the pole-itself may cleave or leave its limit-point-position-width/closure |
FORENOTE:
This discussion also relates to essential singularities, Picard's theorems {Little, Great}, Casorati-Weierstrass theorem [aka Weierstraß, sometimes Weierstrauss], But this result seems unknown...
BACKGROUND:
The Cauchy-Riemann contour integral modulus (2πi × residue) value initially comes from a convenient circle integral of fixed radius = r around a 1/z-pole: ∮(1/z) dz ≡ ∫(1/r eiθ) r eiθ i dθ [0,2π] = 2πi , wherefor contours not-enclosing the 1/z-pole, and as well functions with only higher-order 1/zn-poles (n>1), are deemed contributing naught to the modulus—but—nevertheless otherwise necessary for a proper understanding of poles and especially when the contour splits-through a pole (a pole on the contour curvilinear line), which split-modulus is usually-not its half-value though the partial-∮(1/z) dz is symmetric proportionally around the contour origin angle:
N.B. The pole-itself is usually referring to its definitively undefined-division-by-zero point-position, i.e. undefined 1/0 is exactly-one point where the denominator = 0 , But we discover this-one-point may be moved a split-infinitesimal off its point position, to open-space adjacent its previously-(sort-of)-definitive-point. Also note a pole has been only-at-best associated with its poling-process, and deemed to contribute nothing of itself... unlike δ-functions...
Background #2. A complex function of a complex parameter, f(z) , is analytic where its derivative, df(z)/dz , is fully complex-consistent, well-defined, continuous...
(which can also help integrate related real, functions, infinite, improper, integrals, by integration of the same, as complex, functions, over an inclusive contour, and extending to ∞ .)
The complex function 1/z ≡ 1/x+iy ≡ x-iy/x2+y2 ≡ x-iy/r2 ≡ (1/r) e-iθ where z ≡ x+iy ≡ r eiθ ≡ r cos(θ) + i r sin(θ) ; r2 ≡ x2+y2; θ ≡ arctan(y,x) ; is analytic except for a 1/0 "pole" at the origin z=0 ...
Background #3. Complex functions are uniquely equivalent to their Laurent-Taylor Expansion [*] power-series = ∑ fi zi (i∈ℤ) (all integer exponents), e.g. exp(-1/z) ≡ e-1/z = 1 + ∑(-z)-i/i! (i∈ℕ) ...
A PARADOXIC POLE:
The single-pole complex function e-1/z→0 exponentially fast as z→0+ on every radiant through the origin, squeezing against the y-axis, so although the Cauchy-Riemann contour integral definition takes only the first-order 1/z-term, this example function is not-so-symmetrically/antisymmetrically spread and in fact has no steep-poling-process contribution on the positive-x-half-plane, near its residue pole at z=0, nor even on the y-axis, where |e-1/iy| = 1 and its integral ∫ [-r,r]→0 parabolically fast (*) as r→0 , as sin(1/y) is antisymmetric fully-self-canceling and cos(1/y) tends mostly-self-canceling bounded positively-and-negatively, except = 0 if taken instead as the limit from the positive-x-side of the origin complex-zero, but so in either case contributes ∫→0 , And so the close-in contour integral, briefly along the y-axis including the origin and around the connecting positive-x-half-circle, →0 enclosing no 1/z-pole at all—nor even a wedge, of a pole,—and likewise at greater distance the larger positive-x-half-circle integral ∫e-1/z dz→∫1-(1/z) dz = ∫(1-(1/r eiθ)) r eiθ i dθ [-π/2,π/2] = r2i-πi , which also = ∫e-1/iy d(iy) = ∫cos(1/y) i dy [-r,r]→r2i-πi (half-pole **), so the poling-process but-not-the-pole must have passed-through the y-axis and the origin—so,
* (Note, still checking, it should be significantly faster than parabolically fast...)
**(quick calculus, ∫cos(1/y) dy [-r,r] ≡ (ξ≡1/y) ∫cos(ξ)(-1/ξ2) dξ ≡ cos(ξ)/ξ+∫sin(ξ)/ξ dξ [-1/r,1/r]-[-∞,∞]→(r→∞)→2r+0-0-π (∫sinc())...)
The contour-integral modulus on the negative-x-half-circle and through the origin, of e-1/z, includes the whole 2πi ... But is it really possible to split a line widthwise, (N.B. the e-1/0-pole if taken as the limit from the negative, side, ∫{∞}[0-,0]→∞ or positive-indeterminate having no negative-pole to compensate), Or, Did we just-now move infinity infinitesimally-off its origin-position-point...
TO WIT:
e-1/z ≡ e-1/x+iy ≡ e-x+iy/x2+y2 ≡ e-x+iy/r2 ≡ e-x/r2 eiy/r2 ≡ e-x/r2 [ cos(y/r2) + i sin(y/r2) ] which convergences →0 exponentially rapidly as r→0 along every y=mx+0 radiant from the origin, on the (0+) positive-x-right-half-plane, and squeezing against the y-axis finite-value as m→±∞ for uniform convergence to zero—so that there is, no, effectual 1/z-poling process anywhere on the positive-x-right-half-plane, and naught but the ei/0-pole (complex zero) at the origin on the y-axis as cos(y/r2) + i sin(y/r2) is alternating-and-tending-to-self-canceling symmetric (positive-symmetric) cos() and fully canceling antisymmetric (negative-symmetric) sin()... its steep-poling-process is all on the open negative-x-left-half-plane, exponentially-high near the origin 'left-side' point, and yet-paradoxically capable of reaching around the pole to the far-positive-x-right-half-plane, tailing out its process...
CLEFT, LEFT, OR DISAPPEARED...NO—
Not-disappeared, for the coefficients do-not-cancel: the exact 1/0-pole-point, not the limit, of e-1/z where z ≡ 0 , 1/0 ≡ 1/0n (∀n≫0) , as-on-the-positive-x-side, has undefined value→1 + ∑(-1/0)i/i! (i∈ℕ) = 1 + (e-1-1)(1/0) ≈ -0.632/0 (negative pole), (which is at-least-as-strange as some analytically-extended Riemann sums), unless exponentiation increases the value of the compounded-pole to larger-than the ℵ1-infinity of Reals... meanwhile as-on-the-negative-x-side it has undefined value→1 + ∑(1/0)i/i! (i∈ℕ) = 1 + (e+1-1)(1/0) ≈ +1.718/0 (positive pole), with the same concern for exponentiation increasing beyond ℵ1-infinity...
THE COUNTEREXAMPLE: (to whomever/whatever it belongs)
So, the pole is still there, but just-not-on-the-origin nor-y-axis, though all the pole-terms, are, on the origin-zero and do not cancel; So, the steep-poling-process has shifted its 1/0-pole-point-position, infinitesimally, either, inside, the y-axis closure-line and origin-point, (which might yet be counted as On-the-line-and-point), or, Off-the-line and-point entirely onto the immediately-adjacent, infinitesimally-nearest point of open,-space that had been previously designated 100% analytic—and whence a counterexample, in reality... (or else we've split the line width, which is equally paradoxic and counterexamplish but may be not for analyticity... TBD).
N.B. Green's Function, which was used to prove the Cauchy-Riemann theorem, appears to be safe from this boundary-point-anomaly because it equates area-integration with contour-integration and both appear to pick up the 'pole shift'— so the resolution appears to be in the definition of the pole for Cauchy-Riemann contour-integration and its 'pole shift' effect—just where the pole is, in the case of higher-order poles and-or essential singularities....
POLE MOMENTS:
The 1/z-pole supplies the contour integral modulus 'mean value', the 1/z2-pole is its 'tilt moment' ∫(1/r2 ei2θ) [r eiθ] r eiθ i dθ [0,2π] = 2πi , the 1/z3-pole its 'spreading/splitting moment', etc., and all 1/zn-poles must be computed (summed) to determine the polar inclusion-portion in contour integration through a pole...
Example #1. At fixed radius = r , halfway-around a 1/z2-pole, ∫(1/r2 ei2θ) r eiθ i dθ ≡ ∫(1/r eiθ) i dθ ≡ -e-iθ/r which on [-π/2,π/2] = 2i/r , but on [0,π] = 2/r , which is the effect of 2πi tilt-moment in a specific direction and inversely proportional to leverage-distance, (infinitely strong 'tilt' at the pole)...
Example #2. f(z) ≡ (1/z)+(1/z2) : ∫((1/r eiθ)+(1/r2 ei2θ)) r eiθ i dθ ≡ ∫(1/r eiθ) i dθ [-π/2,π/2] = i2/r ...[under construction] the positive-x-half-circle
Cordially, sincerely, yours,
'The man who moved infinity'
(coined humor intended no brag)
NOTES & COMMENTS:
Note. The e-1/0-pole is better-known as an "essential singularity" but this analyticity paradox seems otherwise unknown.
Note, cos(1/0) + i sin(1/0) appears to be exactly (1+i0) as sin() is antisymmetric (negative-symmetric), sin(-1/y) = -sin(1/y) always, and ever exactly midway between is sin() = 0 , ergo cos() = 1 (not -1 as 1/y is also antisymmetric so ever exactly midway between 1/y and 1/-y is 0 not π , and so cos() = 1), albeit immediately-infinitesimally-aside is infinite-slope ergo discontinuous but an 'island discontinuity' because itself is finite and definite...
Notes for deeper study: 1. Surface integration of the 1/z2-pole, (avoid overloading on the 1/z-pole), 2. complexor pole analysis, (quaternion-/octonion-space), 3. does this creation/tilt/existence of a pole-point in the presumed-fully-analytic adjacent-space induce a 3rd-dimension as infinitesimally-wound around the 2nd, (or 2.b. can 3D map to 2D, whereas 2D could-not map to 1D but by citified-metric 'neighborhooding')...
* [Note: Beware of historian obfuscation: The name of the Expansion is oft-listed as Laurent-Taylor, but historians in their efforts to sort-and-span history regularly confuse mathematics by assuming the Laurent Expansion extended the insufficient Taylor Expansion to include negative exponents, aka poles, in full power series, But mathematician methods are otherwise, e.g. in the closely related Maclaurin Series, historians have assumed Maclaurin's zero-centering is a special case of Taylor's any-constant-z0-centering, whereas mathematicians translate without special attribution, 0→z0 by z' = z-z0 , and in fact typical examples for Maclaurin Expansions are exactly that, such as log(1+x) about x=0 translates to log(x) about x=1 , and some mathematicians have thus attributed the power series Expansion to Maclaurin because his method of undetermined coefficients was distinct-from-and-never-a-special-case-of Taylor Expansion which followed Newton's methods...]
A premise discovery under the title,