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COMPLEX · 1

Holomorphic functions

9 concepts · 19 questions

Complex differentiability, the Cauchy–Riemann equations, harmonic functions, and power series as holomorphic functions.

Overview 1The plan of this section

Complex differentiability is far stronger than real differentiability; this section sets up the definitions and first consequences.

  1. Complex differentiability and holomorphic functions (Def. 1.1), with the usual rules (Prop. 1.2).
  2. The Cauchy–Riemann equations are necessary (Thm 1.3) and, with continuous partials, sufficient (Thm 1.4).
  3. Real and imaginary parts are harmonic (Cor. 1.5); harmonic conjugates (Example 1.6).
  4. Power series are holomorphic inside their disc of convergence (Thm 1.7).

f′(a)=lim⁡f(a+h)−f(a)hf'(a) = \lim\frac{f(a + h) - f(a)}h with complex hh; Cauchy–Riemann; harmonic parts; power series are holomorphic.

Definition 1.1Complex differentiability and holomorphic functions

Let U⊆CU\subseteq\mathbb{C} be open. f:U→Cf: U\to\mathbb{C} is complex differentiable at a∈Ua\in U if

f′(a)=lim⁡h→0, h∈Cf(a+h)−f(a)hf'(a) = \lim_{h\to0,\ h\in\mathbb{C}}\frac{f(a + h) - f(a)}h

exists. ff is holomorphic on UU if it is complex differentiable at every point of UU; entire if holomorphic on C\mathbb{C}.

Holomorphic: complex derivative exists at every point of an open set; the limit is over h→0h\to0 in every direction.

Proposition 1.2Rules for complex derivatives

Complex differentiable functions are continuous. Sums, products, quotients (where the denominator is nonzero) and compositions of holomorphic functions are holomorphic, with the usual formulas. Polynomials are entire and rational functions are holomorphic away from the zeros of the denominator.

All the familiar rules hold; polynomials entire, rational functions holomorphic off their poles.

Proof

Theorem 1.3The Cauchy–Riemann equations are necessary

If f=u+ivf = u + iv is complex differentiable at a=x0+iy0a = x_0 + iy_0, then the partial derivatives of u,vu, v exist at (x0,y0)(x_0, y_0) and

ux=vy,uy=−vx,u_x = v_y,\qquad u_y = -v_x,

with f′(a)=ux+ivx=vy−iuyf'(a) = u_x + iv_x = v_y - iu_y.

ux=vyu_x = v_y, uy=−vxu_y = -v_x; f′=ux+ivxf' = u_x + iv_x.

Proof

Theorem 1.4Cauchy–Riemann with continuous partials implies holomorphic

If u,vu, v have continuous first partial derivatives on UU satisfying the Cauchy–Riemann equations, then f=u+ivf = u + iv is holomorphic on UU.

Complex differentiable ⇔ real differentiable with derivative = multiplication by a complex number ⇔ CR (given continuous partials).

Proof

Corollary 1.5Real and imaginary parts are harmonic

If f=u+ivf = u + iv is holomorphic and u,vu, v are C2C^2 (true automatically, see Section 4), then ∇2u=uxx+uyy=0\nabla^2u = u_{xx} + u_{yy} = 0 and ∇2v=0\nabla^2v = 0. A pair u,vu, v satisfying Cauchy–Riemann are harmonic conjugates.

Holomorphic ⇒ uu and vv harmonic; vv is a harmonic conjugate of uu.

Proof

Example 1.6Finding a harmonic conjugate

u=x3−3xy2u = x^3 - 3xy^2 is harmonic. From vy=ux=3x2−3y2v_y = u_x = 3x^2 - 3y^2, v=3x2y−y3+g(x)v = 3x^2y - y^3 + g(x); then vx=6xy+g′(x)=−uy=6xyv_x = 6xy + g'(x) = -u_y = 6xy, so gg is constant. Hence f=u+iv=z3+icf = u + iv = z^3 + ic.

Method: integrate vy=uxv_y = u_x in yy, fix the function of xx with vx=−uyv_x = -u_y.

Theorem 1.7Power series are holomorphic

If ∑an(z−c)n\sum a_n(z - c)^n has radius of convergence R>0R > 0 (given by 1R=lim sup⁡∣an∣1/n\frac1R = \limsup|a_n|^{1/n}), its sum ff is holomorphic on D(c,R)D(c, R) with f′(z)=∑nan(z−c)n−1f'(z) = \sum na_n(z - c)^{n-1}, which has the same radius. Hence ff is infinitely complex differentiable there and an=f(n)(c)n!a_n = \frac{f^{(n)}(c)}{n!}.

Inside the disc of convergence, power series are holomorphic and differentiate termwise.

Proof

Check 1.8Check: is this function holomorphic?

  • Built from zz by algebra, exp⁡\exp, power series: holomorphic where defined.
  • Involves zˉ\bar z, ∣z∣|z|, Re⁡z\operatorname{Re}z explicitly: test Cauchy–Riemann; usually fails.
  • Differentiable at isolated points or along a curve is not holomorphic (needs an open set).
  • Check ∇2u=0\nabla^2u = 0 before looking for a conjugate.

Algebra in zz ⇒ holomorphic; zˉ\bar z ⇒ test CR; points/curves ≠ open set.