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Complex Differentiation

Two-dimensional complex functions have a two-dimensional complex ; the four-dimensional combination is difficult to visualize. The two-dimensional real approach of interpreting the derivative as the slope of a is no longer possible; instead, we must look separately at the two complex planes on which lie the domain and codomain of a given function. Consider a mapping (function) f(z) = f(x + iy) = u(x, y) + iv(x, y), and an infinitesimal vector dz = hdx, dyi on the xy-. The mapping sends the vector to the corresponding vector d f(z) = f 0(z) dz = f 0(z) hdx, dyi = hdu, dvi in the uv-plane. Hence the action of the complex derivative f 0(z) is an expansion and rotation of the vector dz into the vector d f(z). Because we are working in the two-dimensional , infinitely many vectors of a given magnitude can emanate from the single point (x, y), differing only in their direction. Generally speaking, the effect of the complex derivative f 0(z) may differ for each of these vectors, expanding and rotating each by a different amount. We call analytic the rare function for which the complex derivative expands and rotates each of these infinitesimal vectors by the same amount, that is, for which the complex derivative is isotropic and path-independent. [We call entire those functions analytic on the entire complex plane; holomorphic those functions analytic on their domain of definition; and meromorphic those functions analytic, save for a of isolated points, on their domain of definition.] The study of complex analysis deals primarily with analytic functions, for the requirement that the complex derivative f 0(z) expand and rotate each of these infinitesimal vectors by the same amount is a strong one, with far-reaching consequences. Furthermore, most of the functions with which we are familiar are analytic, and their the familiar ones from the of real variables!

Conformal Mapping

Conformal mappings are those that preserve the magnitude and sense of angles; that is, two vectors a and b forming an angle θ going from a to b are mapped to two vectors f(a) and f(b) forming an angle θ going from f(a) and f(b). [As example of a non-conformal mapping, consider f(z) = z, which preserves the magnitude but not the sense of angles.] Based on the definition of analyticity, we readily claim that all analytic functions are conformal except at points z where f 0(z) = 0, conformality not being well-defined for the zero vector. The converse is trickier: is a conformal function analytic? Analyticity requires that the action of the derivative be isotropic. We know that conformal mappings are isotropic with regard to angles, but are they also isotropic with regard to magnitudes? Consider an infinitesimal triangle centered around z, mapped conformally to another infinitesimal triangle centered around f(z). Because a conformal mapping preserves angles we know that the two triangles are similar and hence that their lengths are proportional, suggesting isotropy with regard to magnitudes. For isolated points we can imagine that magnitudes may not be preserved (draw it!), but for planar regions we see the need for such isotropy. The preservation of angles over a must imply a preservation of magnitudes: if a conformal mapping were anisotropic over a region, then its image is deformed and angles are no longer preserved, leading to a contradiction. Hence we claim that a conformal function having continuous partial derivatives (the requirement for conformality over a region instead of over isolated points) is analytic.

Cauchy-Riemann Equations

The Cauchy-Riemann equations are an algebraic representation of the path-independence of the derivative of analytic functions. In particular, we consider horizontal and vertical paths toward a point (x, y) for the function f(z) = f(x, y) = u(x, y) + iv(x, y):

0 0 f(x + ∆x, y) − f(x, y) f (z) = f (x, y) = lim = ux + ivx ∆x→0 ∆x

0 0 f(x, y + ∆y) − f(x, y) f (z) = f (x, y) = lim = vy − iuy ∆y→0 i∆y Hence equating real and imaginary parts gives us the Cauchy-Riemann equations,

ux = vy uy = −vx Complex Analysis a.cyclohexane.molecule

The Cauchy-Riemann equations provide another means of identifying an given its component functions u and v; alternatively, given one component u or v, the other can be determined up to an additive constant by finding its partial derivatives. Just as we claimed that conformal functions were analytic, so too must functions satisfying the Cauchy-Riemann equations be analytic (as long as their partial derivatives are continuous), conformality and the Cauchy-Riemann equations being two different representations of the same notion of isotropy. Further, note that the equivalence of partial derivatives on two paths is enough to guarantee analyticity and hence the equivalence of partial derivatives on all paths. We further make the observation that the component functions u and v of an analytic function are harmonic; that is, they satisfy the Laplace equation fxx + fyy = 0, as can be confirmed by suitable application of the Cauchy- Riemann equations. For a function f(z) = u(x, y) + iv(x, y), we call v the harmonic conjugate of u (though not the other way around) if u and v satisfy the Cauchy-Riemann equations. Harmonic functions, making use of conformal mapping, find applications in simplifying and solving boundary-value problems.

Complex Integration

Whereas on the real axis there is only one option for the path of integration between two points, on the complex plane we have infinitely many paths of integration, and we must work not with regular but with contour integrals, path integrals in the complex plane over piecewise-smooth . Just as with multivariable functions, we may evaluate these path integrals by parametrization or by the fundamental theorem of line integrals, if the integrands are analytic (and thus have antiderivatives) over the path. [While non-analytic functions can still be path-integrated provided their partial derivatives are piecewise continuous, and while some such integrals are even significant, they are ultimately of less concern. An interesting example is the

I I Z 2π Z 2π 2 Z 2π Z r −iθ iθ −iθ iθ r z¯dz = re d(re ) = re (ire ) dθ = 2i dθ = 2i r dr dθ = 2iAenc C C 0 0 2 0 0

which allows one to find Aenc, the area enclosed by the closed contour C.] Most complex functions share the same derivatives and antiderivatives as their real counterparts, but subtle issues arise when dealing with fractional exponents and , such functions being multi-valued and hence requir- ing the imposition of further restrictions before their antiderivatives can be well defined. Consider the complex log z = log[r exp(iθ)] = ln r + iθ, which has infinitely many possible values for θ. [If an angle θ satisfies the equation, so too does any integral multiple of θ + 2nπ.] In order for the necessarily single-valued path integral to make sense, we must make a branch cut, specifying an angular domain that bounds θ to an interval spanning at most 2π radians. Consider the closed contour C, a circle of radius r centered at the origin and once traversed in a positive (counterclockwise) sense, and a corresponding integral I dz r exp i(φ+2π) = [log(z)]r exp iφ = ln r + i(φ + 2π) − ln r − iφ = 2πi C z where we have implicitly defined a branch cut for which θ ∈ (φ, φ + 2π), for otherwise log z would not have been analytic over the path and we would not have been able to use the fundamental theorem for path integrals. Note that the value of the integral is independent of r, and also that because neither of the angular bounds of the path integral are within the branch cut, we are actually dealing with a doubly improper integral. Nevertheless, because φ and φ + 2π are respectively the right-handed and left-handed limits of the domain defined by the branch cut, the integral may still be evaluated—and, perhaps surprisingly, the resulting integral is non-zero. [We may evaluate a similar integral in the same manner, where C0 now represents a circle of radius r centered around ρ: I dz ρ+r exp i(φ+2π) = [log(z − ρ)]ρ+r exp iφ = ln r + i(φ + 2π) − ln r − iφ = 2πi C0 z − ρ obtaining the same result as before. Both these integrals will be of great importance in upcoming sections.] Complex Analysis a.cyclohexane.molecule

Cauchy-Goursat Theorem

Green’s theorem for line integrals applies equally well to contour integrals. In particular, I I I I f(z) dz = u(x, y) + iv(x, y) (dx + idy) = u(x, y) dx − v(x, y) dy + i v(x, y) dx + u(x, y) dy C C C C ZZ ZZ = ux − vy dx dy + i vx + uy dx dy S S = 0 where the second equality follows from application of the Cauchy-Riemann equations. In applying the Cauchy- Riemann equations we have made the implicit assumption that the function is analytic within and on the simply connected region—one in which every simple closed contour within it encloses only points within the region— bounded by C and also that the derivative of f is continuous; the latter assumption is actually unnecessary. In extending the theorem to multiply connected regions, we need only split up each multiply connected region with line segments to form multiple simply connected regions; the theorem then applies to each of these regions. Of particular importance is an annular region centered around a singular point, a point at which a given function is not analytic, but infinitesimally close to a point or points at which the function is analytic. By the extension to the Cauchy-Goursat theorem, then, the path integral around the outer and inner contours are equivalent, as is any contour enclosing said singular point. This is the principle of deformation of paths. To get a better understanding of what the Cauchy-Goursat theorem does and does not convey, let us here consider the path integrals in integral powers of z about the closed contour C going once clockwise about the origin.  I  0 for n ≥ 0 zn dz = 2πi for n = −1 C  0 for n ≤ −2 For n ≥ 0, the antiderivative is an entire in z, and hence analytic over any choice of C. By the Cauchy- Goursat theorem, the path integral over C must be 0. For n = −1, the logarithmic antiderivative is analytic save for a singularity at 0, and, as calculated prior, the path integral over C is 2πi. Because of the singularity at 0, the Cauchy-Goursat theorem does not apply. For n ≤ −2, there is once again a singularity at 0, and the Cauchy- Goursat theorem does not apply. Integrating via the fundamental theorem results in a result of 0: the application of the Cauchy-Goursat theorem is sufficient but not necessary for the integral to have a value of 0. These results are also interesting independent of the Cauchy-Goursat theorem. Particularly, why should n = −1 be the only integral value of n for which the path integral about the origin does not vanish identically? The answer readily reveals itself upon visual consideration of the antiderivative. Recall that the action of the complex derivative was an expansion (contraction) and a rotation of a given vector. The action of the complex antiderivative is the inverse of that of the complex derivative, and is thus a contraction (expansion) and a rotation of said vector. Integrating over the unit circle, the antiderivative vectors are rotated due to the curvature of the path as well as the rotating effect of the antiderivative; only for n = −1 do these effects cancel out, each infinitesimal vector on the path pointing straight up. Since we integrate over the unit circle, the magnitude of each vector is unchanged by inversion; the final result, then, is a vector pointing in the positive y-direction with magnitude equal to the circumference of the circle, or 2πi. [The choice of path is arbitrary, for by the principle of deformation of paths, the path integral must be the same for all paths. Integrating over the unit circle only serves to make the resulting antiderivative easier to visualize.]

Cauchy Integral Formula

Consider an bounded on the outside by the contour C and on the inside by the contour Γ, both enclosing the point z0. Then by the principle of deformation of paths, I f(z) I f(z) I f(z) − f(z ) I f(z ) dz = dz = 0 dz + 0 dz C z − z0 Γ z − z0 Γ z − z0 Γ z − z0

As we make Γ infinitesimal, the first term on the right becomes vanishingly small. [z − z0 is proportional to the radius, as is the path around which we integrate; the two factors cancel each other out. Meanwhile, f(z) − f(z0) Complex Analysis a.cyclohexane.molecule

also tends to zero, leaving only the second term.] Hence we obtain the Cauchy integral formula, I I I f(z) f(z0) dz dz = dz = f(z0) = 2πif(z0) C z − z0 Γ z − z0 Γ z − z0 The rigidity of analytic functions, given the need for the isotropy of their derivatives, allows us to determine path integrals around a point knowing only the value of the function at that point!

The Cauchy integral formula may be further extended to other powers of z − z0 by formally differentiating under the integral with respect to z0, hence obtaining I I I d f(z) ∂ f(z) f(z) 0 dz = dz = 2 dz = 2πif (z0) dz0 C z − z0 C ∂z0 z − z0 C (z − z0) d(n) I f(z) I ∂(n) f(z) I f(z) 2πi dz = dz = dz = f (n)(z ) (n) (n) (n+1) 0 z − z0 z − z0 (z − z ) n! dz0 C C ∂z0 C 0

As a corollary to this result, note that all analytic functions must be infinitely differentiable, for the equation immediately prior allow us to construct formulae for each derivative of a given analytic function.

Maxima and Minima

First, a more rigorous justification of the limiting process applied in establishing the Cauchy integral formula. We consider the contour Γ as a small circle of radius ρ centered around z0. An upper bound for the vanishing term in the Cauchy integral formula, is given by I f(z) − f(z0)  dz ≤ |M||L| = (2πρ) = 2π Γ z − z0 ρ where M is the maximum value of the integrand along the path, and L the length of the path. [ bars assure that we work with the magnitude of each of these quantities, and are unconcerned regarding sign.] In this case we have neatly made use of the δ −  limits for the derivative. If we bound |z − z0| = ρ < δ, then we must have |f(z) − f(z0)| < , giving us the bound |M| < /δ < /ρ, and L is simply the circumference of the circle of radius ρ. Since  can be made arbitrarily small, the value of the integral must be identically zero. The limiting arguments made here are widely applicable, and play a significant role in further applications of . The requirement of analyticity leads to many remarkable theorems regarding maxima and minima, as can be established with the help of the Cauchy integral formula. Consider a circular contour C with radius ρ oriented in the positive sense and centered about a point z0. Then, by parametric evaluation, we obtain I Z 2π iθ Z 2π 1 f(z) 1 f(z0 + ρe ) iθ 1 iθ f(z0) = dz = iθ (iρe ) dθ = f(z0 + ρe ) dθ 2πi C z − z0 2πi 0 ρe 2π 0

We have proven Gauss’ mean value theorem, which states that the value of f(z0) is the average value of f about any circle centered around z0.

Now consider a circular region centered around z0 over which the modulus of the analytic function f is bounded by f(z0); that is, |f(z)| ≤ |f(z0)|. By application of Gauss’ mean value theorem we note that the average value of f along any circular path centered at z0 in the region is equal to f(z0) and simultaneously bounded by f(z0); hence, |f(z)| must be identically |f(z0)|. It turns out that this implies that f(z) = f(z0) throughout the domain of f; hence, if there exists a maximum modulus for f within the domain, then f is constant. This is the maximum modulus principle. Taking its contrapositive, if f is not a , then its maximum modulus does not lie within the domain, and must hence lie on the boundary of the domain. [The same argument with the function g := 1/f shows that the minimum modulus of f must likewise lie on the boundary of the domain.]

Calculus of Residues

Just as we may represent (complex) functions analytic on a disk by their Taylor , so too may we represent functions analytic on an annulus by their , a in powers of n from −∞ to ∞. We will Complex Analysis a.cyclohexane.molecule

not be concerned here with proving the existence of Laurent series for the general case, but will simply show their existence by constructing them on a case-by-case basis by manipulation of familiar . Consider the function f(z) = z2 sin(1/z), which is singular at z = 0, sin(1/z) not being well-defined there. Then we claim the existence of a Laurent series centered around 0 in the domain 0 < |z| < ∞, and construct it as follows:

z3 z5 1 1 1 1 1 1 1 1 1 sin(z) = z − + · · · ⇐⇒ sin(1/z) = − + · · · ⇐⇒ z2 sin(1/z) = z − + ··· 3! 5! z z3 3! z5 5! z 3! z3 5! To evaluate the path integral for f(z) about a path C that encloses the center of the annulus, we may then substitute our Laurent series for f(z) and write

I I 1 1 1 1 1 I dz 1 πi f(z) dz = z − + 3 ··· dz = − = − (2πi) = − C C z 3! z 5! 3! C z 6 3 where we have appealed to an earlier result stating that the path integral about powers of z not equal to −1 is necessarily zero. The evaluation of this path integral suggests the amazingly simple general case I I ∞ ∞ I I X n X n dz f(z) dz = cnz dz = cnz dz = c−1 = 2πi c−1 = 2πi Res f(z) z z=0 C C n=−∞ n=−∞ C C

where the switching of the order of integration and summation is justified by the integrand being well-defined (finite) along the path, and we have re-defined the coefficient c−1 as the residue of the Laurent series centered around z = 0, notated as shown. Let us next consider the paths C : |z − 2| = 1 and Γ: |z| = 1 and the path integrals

I dz I dz 4 and 4 C z(z − 2) Γ z(z − 2) In the first case, our path is centered around the fourth-order singular point at z = 2; in the second, our path is centered around the singular point at z = 0. Our Laurent series in the two cases, centered respectively at z = 2 and z = 0, are thus 1 1 1 1 = = z(z − 2)4 (2 + z − 2)(z − 2)4 2 (1 + (z − 2)/2) (z − 2)4 ! 1 1 z − 2 z − 22 z − 23 = 1 − + − + ··· 2 (z − 2)4 2 2 2 1 1 1  z −4 1 = (−2 + z)−4 = 1 − = (1 + 2z − · · · ) z(z − 2)4 z 16z 2 16z

In the evaluation of the first series we have used the well-known result that 1/(1 − z) = 1 + z + z2 + ···; in the evaluation of the second series we have used the generalized binomial theorem. Note how different the two series are, just by centering z around a different position! From these series representations, we can easily read off the residues −1/16 and 1/16 and hence the values of the path integrals −πi/8 and πi/8 at z = 2 and z = 0. Alternatively, we could have obtained the residues by means of Cauchy’s integral theorem, obtaining respectively

I  3  I −4 1/z 1 d 1 πi (z − 2) −4 πi 4 dz = 2πi 3 = − and dz = 2πi[(z − 2) ]z=0 = C (z − 2) 3! dz z z=2 8 Γ z 8

Next, consider the value of the path integral over a path γ that encloses both singular points. By the principle of deformation of paths we may deform this path into the contours C and Γ, and hence I dz I dz I dz πi πi 4 = 4 + 4 = − + = 0 γ z(z − 2) C z(z − 2) Γ z(z − 2) 8 8

When the integral encloses many singularities, it may be more efficient to evaluate the integral by means of the residue at infinity, defined by the path integral along a clockwise contour −C that encloses all the singularities of f. [The is clockwise in order that the point at infinity be kept to the left of the curve; the curve thus Complex Analysis a.cyclohexane.molecule

“encloses” the point at infinity. Note that, in our previous examples, the curves have all been implicitly defined in the positive, counter-clockwise sense.] In evaluating our desired integral, we make the substitution z → 1/z, an inversion about the origin mapping ∞ to 0 and the clockwise path −C to the counter-clockwise path C0. Thus we have I I I 1 1 I 1 1  1 1 f(z) dz = − f(z) dz = − f d = 2 f dz = 2πi Res 2 f C −C C0 z z C0 z z z=0 z z the path integral along C being what we are ultimately interested in. Contour integration even finds use in evaluating real integrals: by treating the real integral we wish to evaluate as part of a contour in the complex plane and evaluating the path integral by means of residues, we may hence obtain our desired integral.

References

[1] Complex Variables and Applications, Brown and Churchill, 8th ed. [2] Visual Complex Analysis, Needham.