Holomorphy and Calculus

Useful facts: \begin{align*} f' = {\frac{\partial f}{\partial z}\,} = {1\over i}{\frac{\partial f}{\partial y}\,} = {\frac{\partial f}{\partial x}\,} = {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} .\end{align*}

Complex Calculus

Various differentials: \begin{align*} dz &= dx + i~dy \\ d\overline{z} &= dx - i~dy \\ \\ f_z &= f_x = f_y / i .\end{align*}

Integral of a complex exponential: \begin{align*} \int_{0}^{2 \pi} e^{i \ell x} d x &=\left\{\begin{array}{ll} {2 \pi} & {\ell=0} \\ {0} & \text{else} \end{array}\right. .\end{align*}

\begin{align*} \int_{\gamma} f(z) \,dz= \int_0^{2\pi } f(Re^{i\theta}) \, iRe^{i\theta} \,d\theta .\end{align*}

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

You can show \(f(z) = \sqrt{z}\) is not holomorphic by showing its integral over \(S^1\) is nonzero. This is a direct computation: \begin{align*} \int_{S^1} z^{1/2} \,dz &= \int_0^{2\pi} (e^{i\theta})^{1/2} ie^{i\theta} \,d\theta\\ &= i \int_0^{2\pi} e^{i3\theta \over 2}\,d\theta\\ &= i \qty{2\over 3i} e^{i3\theta \over 2}\Big|_{0}^{2\pi} \\ &= {2\over 3}\qty{e^{3\pi i - 1}} \\ &= -{4\over 3} .\end{align*}

Note an issue: a different parameterization yields a different (still nonzero) number \begin{align*} \cdots &= \int_{-\pi}^{\pi} (e^{i\theta})^{1/2} ie^{i\theta} \,d\theta\\ &= {2\over 3}\qty{ e^{3\pi i \over 2} - e^{-3\pi i \over 2}} \\ &= -{4i\over 3} .\end{align*} This is these are paths that don’t lift to closed loops on the Riemann surface defined by \(z\mapsto z^2\).

The function \(e^z\) is entire, so “usual” calculus using it works. For example, \begin{align*} \int e^{3x}\cos(2x) \,dx &= \Re \int e^{3z}e^{2iz}\,dz\\ &= \Re \int e^{(3+2i)z} \,dz\\ &= \Re {e^{(3+2i)z} \over 3+2i} + C \\ &= e^{2x \over 13}\qty{\cos(2x) + i\sin(2x)} .\end{align*}


Let \(f:{\mathbf{C}}\to {\mathbf{C}}\) satisfy CR and consider it as a map \(f:{\mathbf{R}}^2\to{\mathbf{R}}^2\), writing \(f(x+iy) = u(x, y) + iv(x, y)\), the Jacobian is \begin{align*} J = u_xv_y -v_x u_y = u_x^2 + v_x^2 .\end{align*} OTOH, \begin{align*} {\left\lvert {f'(z)} \right\rvert}^2 = {\left\lvert {u_x + iv_y} \right\rvert}^2 = J .\end{align*}

A function \(f: {\mathbf{C}}\to {\mathbf{C}}\) is complex differentiable or holomorphic at \(z_0\) iff the following limit exists: \begin{align*} \lim_{h\to 0} { f(z_0 + h) - f(h) \over h } .\end{align*} A function that is holomorphic on \({\mathbf{C}}\) is said to be entire.

Equivalently, there exists an \(\alpha\in {\mathbf{C}}\) 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, \(\alpha = f'(z_0)\).

  • \(f(z) \coloneqq{\left\lvert {z} \right\rvert}\) 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 \({\mathbf{C}}\setminus\left\{{0}\right\}\) but not holomorphic on \({\mathbf{C}}\)
  • \(f(z) = \overline{z}\) is not holomorphic, but is real differentiable: \begin{align*} {f(z_0 + h) - f(z_0) \over h } = {\overline{z_0} + \overline{h} - \overline{z_0} \over h} = {\overline{h} \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.

A function \(F: {\mathbf{R}}^n\to {\mathbf{R}}^m\) 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*}

\(f\) is holomorphic at \(z_0\) iff there exists an \(a\in {\mathbf{C}}\) 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

Some properties:

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


Show that a real-valued holomorphic function is constant.


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

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


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.

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


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

#todo #complex/exercise/completed