Categories: Complex analysis, Mathematics.

Holomorphic function

In complex analysis, a complex function f(z)f(z) of a complex variable zz is called holomorphic or analytic if it is complex differentiable in the vicinity of every point of its domain. This is a very strong condition.

As a result, holomorphic functions are infinitely differentiable and equal their Taylor expansion at every point. In physicists’ terms, they are very “well-behaved” throughout their domain.

More formally, a given function f(z)f(z) is holomorphic in a certain region if the following limit exists for all zz in that region, and for all directions of Δz\Delta z:

f(z)=limΔz0f(z+Δz)f(z)Δz\begin{aligned} \boxed{ f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} } \end{aligned}

We decompose ff into the real functions uu and vv of real variables xx and yy:

f(z)=f(x+iy)=u(x,y)+iv(x,y)\begin{aligned} f(z) = f(x + i y) = u(x, y) + i v(x, y) \end{aligned}

Since we are free to choose the direction of Δz\Delta z, we choose Δx\Delta x and Δy\Delta y:

f(z)=limΔx0f(z+Δx)f(z)Δx=ux+ivx=limΔy0f(z+iΔy)f(z)iΔy=vyiuy\begin{aligned} f'(z) &= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x} = \pdv{u}{x} + i \pdv{v}{x} \\ &= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y} = \pdv{v}{y} - i \pdv{u}{y} \end{aligned}

For f(z)f(z) to be holomorphic, these two results must be equivalent. Because uu and vv are real by definition, we thus arrive at the Cauchy-Riemann equations:

ux=vyvx=uy\begin{aligned} \boxed{ \pdv{u}{x} = \pdv{v}{y} \qquad \pdv{v}{x} = - \pdv{u}{y} } \end{aligned}

Therefore, a given function f(z)f(z) is holomorphic if and only if its real and imaginary parts satisfy these equations. This gives an idea of how strict the criteria are to qualify as holomorphic.

Integration formulas

Holomorphic functions satisfy Cauchy’s integral theorem, which states that the integral of f(z)f(z) over any closed curve CC in the complex plane is zero, provided that f(z)f(z) is holomorphic for all zz in the area enclosed by CC:

Cf(z)dz=0\begin{aligned} \boxed{ \oint_C f(z) \dd{z} = 0 } \end{aligned}

Just like before, we decompose f(z)f(z) into its real and imaginary parts:

Cf(z)dz=C(u+iv)d(x+iy)=C(u+iv)(dx+idy)=Cudxvdy+iCvdx+udy\begin{aligned} \oint_C f(z) \dd{z} &= \oint_C (u + i v) \dd{(x + i y)} = \oint_C (u + i v) \:(\dd{x} + i \dd{y}) \\ &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y} \end{aligned}

Using Green’s theorem, we integrate over the area AA enclosed by CC:

Cf(z)dz=Avx+uydxdy+iAuxvydxdy\begin{aligned} \oint_C f(z) \dd{z} &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y} \end{aligned}

Since f(z)f(z) is holomorphic, uu and vv satisfy the Cauchy-Riemann equations, such that the integrands disappear and the final result is zero.

An interesting consequence is Cauchy’s integral formula, which states that the value of f(z)f(z) at an arbitrary point z0z_0 is determined by its values on an arbitrary contour CC around z0z_0:

f(z0)=12πiCf(z)zz0dz\begin{aligned} \boxed{ f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} } \end{aligned}

Thanks to the integral theorem, we know that the shape and size of CC are irrelevant. Therefore we choose it to be a circle with radius rr, such that the integration variable becomes z=z0+reiθz = z_0 + r e^{i \theta}. Then we integrate by substitution:

12πiCf(z)zz0dz=12πi02πf(z)ireiθreiθdθ=12π02πf(z0+reiθ)dθ\begin{aligned} \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} &= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta} = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} \end{aligned}

We may choose an arbitrarily small radius rr, such that the contour approaches z0z_0:

limr012π02πf(z0+reiθ)dθ=f(z0)2π02πdθ=f(z0)\begin{aligned} \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta} = f(z_0) \end{aligned}

Similarly, Cauchy’s differentiation formula, or Cauchy’s integral formula for derivatives gives all derivatives of a holomorphic function as follows, and also guarantees their existence:

f(n)(z0)=n!2πiCf(z)(zz0)n+1dz\begin{aligned} \boxed{ f^{(n)}(z_0) = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z} } \end{aligned}

By definition, the first derivative f(z)f'(z) of a holomorphic function exists and is:

f(z0)=limzz0f(z)f(z0)zz0\begin{aligned} f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \end{aligned}

We evaluate the numerator using Cauchy’s integral theorem as follows:

f(z0)=limzz01zz0(12πiCf(ζ)ζzdζ12πiCf(ζ)ζz0dζ)=12πilimzz01zz0Cf(ζ)ζzf(ζ)ζz0dζ=12πilimzz01zz0Cf(ζ)(zz0)(ζz)(ζz0)dζ\begin{aligned} f'(z_0) &= \lim_{z \to z_0} \frac{1}{z - z_0} \bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg) \\ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} \oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \\ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta} \end{aligned}

This contour integral converges uniformly, so we may apply the limit on the inside:

f(z0)=12πiC(limzz0f(ζ)(ζz)(ζz0))dζ=12πiCf(ζ)(ζz0)2dζ\begin{aligned} f'(z_0) &= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta} = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta} \end{aligned}

Since the second-order derivative f(z)f''(z) is simply the derivative of f(z)f'(z), this proof works inductively for all higher orders nn.