Holomorphy and Calculus

fact:

Useful facts: f=fz=1ify=fx=ux+ivx.

Complex Calculus

fact:

Various differentials: dz=dx+i dyd¯z=dxi dyfz=fx=fy/i.

Integral of a complex exponential: 2π0eixdx={2π=00else.

γf(z)dz=2π0f(Reiθ)iReiθdθ.

warnings:

f(z)=sin(z),cos(z) are unbounded on C! An easy way to see this: they are nonconstant and entire, thus unbounded by Liouville.

example (Square root is not holomorphic):

You can show f(z)=z is not holomorphic by showing its integral over S1 is nonzero. This is a direct computation: S1z1/2dz=2π0(eiθ)1/2ieiθdθ=i2π0ei3θ2dθ=i(23i)ei3θ2|2π0=23(e3πi1)=43.

Note an issue: a different parameterization yields a different (still nonzero) number =ππ(eiθ)1/2ieiθdθ=23(e3πi2e3πi2)=4i3. This is these are paths that don’t lift to closed loops on the Riemann surface defined by zz2.

remark (Solving real integrals with complex calculus):

The function ez is entire, so “usual” calculus using it works. For example, e3xcos(2x)dx=e3ze2izdz=e(3+2i)zdz=e(3+2i)z3+2i+C=e2x13(cos(2x)+isin(2x)).

Holomorphy

remark:

Let f:CC satisfy CR and consider it as a map f:R2R2, writing f(x+iy)=u(x,y)+iv(x,y), the Jacobian is J=uxvyvxuy=u2x+v2x. OTOH, |f(z)|2=|ux+ivy|2=J.

definition (Complex differentiable / holomorphic /entire):

A function f:CC is complex differentiable or holomorphic at z0 iff the following limit exists: lim A function that is holomorphic on {\mathbb{C}} is said to be entire.

Equivalently, there exists an \alpha\in {\mathbb{C}} such that \begin{align*} f(z_0+h) - f(z_0) = \alpha h + R(h) && R(h) \overset{h\to 0}\longrightarrow 0 .\end{align*} In this case, \alpha = f'(z_0).

example (Holomorphic vs non-holomorphic):

    
  • f(z) \coloneqq{\left\lvert {z} \right\rvert} is not holomorphic.
  • f(z) \coloneqq\arg{z} is not holomorphic.
  • f(z) \coloneqq\Re{z} is not holomorphic.
  • f(z) \coloneqq\Im{z} is not holomorphic.
  • f(z) = {1\over z} is holomorphic on {\mathbb{C}}\setminus\left\{{0}\right\} but not holomorphic on {\mathbb{C}}
  • f(z) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu is not holomorphic, but is real differentiable: \begin{align*} {f(z_0 + h) - f(z_0) \over h } = {\mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu + \mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu \over h} = {\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu \over h} = {re^{-i\theta} \over re^{i\theta}} = e^{-2i\theta} \overset{h\to 0}\longrightarrow e^{-2i\theta} ,\end{align*} which is a complex number that depends on \theta and is thus not a single value.
definition (Real (multivariate) differentiable):

A function F: {\mathbb{R}}^n\to {\mathbb{R}}^m is real-differentiable at \mathbf{p} iff there exists a linear transformation A such that \begin{align*} { {\left\lVert { F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p}) - A(\mathbf{h}) } \right\rVert} \over {\left\lVert { \mathbf{h} } \right\rVert} } \overset{{\left\lVert {\mathbf{h}} \right\rVert}\to 0}\longrightarrow 0 .\end{align*} Rewriting, \begin{align*} {\left\lVert { F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p}) - A(\mathbf{h}) } \right\rVert} = {\left\lVert { \mathbf{h} } \right\rVert} {\left\lVert { R(\mathbf{h}) } \right\rVert} && {\left\lVert {R(\mathbf{h}) } \right\rVert}\overset{{\left\lVert {\mathbf{h} } \right\rVert} \to 0}\longrightarrow 0 .\end{align*}

Equivalently, \begin{align*} F(\mathbf{p} + \mathbf{h}) - F(\mathbf{p}) = A(\mathbf{h}) + {\left\lVert {\mathbf{h}} \right\rVert} R(\mathbf{h}) && {\left\lVert {R(\mathbf{h}) } \right\rVert}\overset{{\left\lVert {\mathbf{h} } \right\rVert} \to 0}\longrightarrow 0 .\end{align*}

Or in a slightly more useful form, \begin{align*} F(\mathbf{p} + \mathbf{h}) = F(\mathbf{p}) + A(\mathbf{h}) + R(\mathbf{h}) && R\in o( {\left\lVert {\mathbf{h}} \right\rVert}), \text{ i.e. } { {\left\lVert { R(\mathbf{h}) } \right\rVert} \over {\left\lVert {\mathbf{h}} \right\rVert}} \overset{\mathbf{h}\to 0}\longrightarrow 0 .\end{align*}

proposition (Holomorphic implies continuous.):

f is holomorphic at z_0 iff there exists an a\in {\mathbb{C}} such that \begin{align*} f(z_0 + h) - f(z_0) - ah = h \psi(h), \quad \psi(h) \overset{h\to 0}\to 0 .\end{align*} In this case, a = f'(z_0).

#todo proof

Wirtinger Calculus

remark:

Some properties:

  • \mkern 1.5mu\overline{\mkern-1.5mu{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f(z)\mkern-1.5mu}\mkern 1.5mu = {\partial}\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu(z), or {\partial}f^*(z) = ({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f(z))^*
    • E.g. d(cz) = c,dz+ 0,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu and d(c\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) = 0,dz+ c,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu
    • E.g. {\partial}{\left\lvert {z} \right\rvert}^2 = {\partial}z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu and { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{\left\lvert {z} \right\rvert}^2 = z.
      • So d({\left\lvert {z} \right\rvert}^2) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu,dz+ z,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu
    • E.g. {\partial}\exp\qty{ - {\left\lvert {z} \right\rvert}^2 } = {\partial}\exp\qty{-z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = e^{-{\left\lvert {z} \right\rvert}^2}\cdot {\partial}(z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu e^{-{\left\lvert {z} \right\rvert}^2}.

Exercises

exercise (?):

Show that a real-valued holomorphic function is constant.

#complex/exercise/completed

solution:

Use that {\mathbb{R}} is not open in {\mathbb{C}} and apply the open mapping theorem to conclude: f({\mathbb{C}}) must be open in {\mathbb{C}} if f is holomorphic and nonconstant.

exercise (?):

Show that if fg \equiv 0 on a domain \Omega then either f\equiv 0 or g\equiv 0.

#complex/exercise/completed

solution:

If f\not\equiv 0, there is a point z_0 where f(z_0)\neq 0, and thus a neighborhood U\ni z_0 where f is nonvanishing. This forces g\equiv 0 on U, however U is a set with a limit point, so g\equiv 0 on \Omega by the identity principle.

exercise (?):

Find all entire functions f such that f(x) = e^x on {\mathbb{R}}.

#complex/exercise/completed

solution:

The function g(z) \coloneqq f(z) - e^z is entire and identically zero on {\mathbb{R}}, which contains a limit point. So g(z) \equiv 0 on {\mathbb{C}}, meaning f(z) = e^z is the only such function.

#todo #complex/exercise/completed