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    Holomorphy and analyticity

    A function \(f=u+iv\) on an open set \(\Omega\subseteq\CC\) is holomorphic if \(f'(z)=\lim_{h\to0}\frac{f(z+h)-f(z)}{h}\) exists at every \(z\in\Omega\). Taking the limit along real and along imaginary \(h\) gives the Cauchy–Riemann equations \(u_x=v_y\), \(u_y=-v_x\), equivalently \(\partial f/\partial\bar z=0\); conversely, if \(u\) and \(v\) are continuously differentiable and satisfy these equations, then \(f\) is holomorphic. A function is analytic on \(\Omega\) if it is given by a convergent power series on a neighbourhood of each point. A power series is holomorphic on its disc of convergence, with derivative the termwise derivative, so analytic functions are holomorphic; the converse follows from the Cauchy integral formula (Cauchy theory). If \(u\) and \(v\) are twice continuously differentiable, the Cauchy–Riemann equations give \(\Delta u=\Delta v=0\).

    Holomorphic and entire functions

    Let \(\Omega\subseteq\CC\) be open, let \(f\colon\Omega\to\CC\), and let \(z_0\in\Omega\). The function \(f\) is complex differentiable, or holomorphic, at \(z_0\) if the limit \[ f'(z_0)\coloneqq\lim_{\substack{h\to0\\ h\in\CC\setminus\{0\}}}\frac{f(z_0+h)-f(z_0)}{h} \] exists. The function \(f\) is holomorphic on \(\Omega\) if it is holomorphic at every point of \(\Omega\), and a function \(f\colon\CC\to\CC\) that is holomorphic on \(\CC\) is entire.

    Let \(\Omega\subseteq\CC\) be open, \(f\colon\Omega\to\CC\), and \(z_0\in\Omega\). Then \(f\) is holomorphic at \(z_0\) if and only if there exist \(\alpha\in\CC\) and a function \(\psi\), defined for \(h\in\CC\) with \(z_0+h\in\Omega\), such that \(\psi(h)\to0\) as \(h\to0\) and \[ f(z_0+h)-f(z_0)=\alpha h+h\,\psi(h). \] In this case \(\alpha=f'(z_0)\).

    If \(f'(z_0)\) exists, put \(\alpha\coloneqq f'(z_0)\), \(\psi(0)\coloneqq0\), and \(\psi(h)\coloneqq\frac{f(z_0+h)-f(z_0)}{h}-\alpha\) for \(h\neq0\); then \(\psi(h)\to0\) and the identity holds. Conversely, dividing the identity by \(h\neq0\) gives \(\frac{f(z_0+h)-f(z_0)}{h}=\alpha+\psi(h)\to\alpha\), so \(f'(z_0)\) exists and equals \(\alpha\).

    See [@SS03].

    Complex differentiable implies Cauchy--Riemann

    A complex derivative has the same directional limit along the real and imaginary axes, forcing the Cauchy--Riemann equations.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be open, let \(f=u+iv\colon\Omega\to\CC\) with \(u,v\) real-valued, and regard \(f\), \(u\), \(v\) as functions of \((x,y)\) where \(z=x+iy\). If \(f\) is complex differentiable at \(z_0=x_0+iy_0\in\Omega\), then the partial derivatives of \(u\) and \(v\) exist at \((x_0,y_0)\), \[ f'(z_0)=\dd{f}{x}(x_0,y_0)={1\over i}\dd{f}{y}(x_0,y_0), \] and at \((x_0,y_0)\) \[ \dd{u}{x} = \dd{v}{y}, \qquad \dd{u}{y} = -\dd{v}{x}. \]

    The limit \(f'(z_0)=\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}{h}\) exists, so it may be computed along any direction. Along real \(h=h_1\in\RR\), \[ f'(z_0) = \lim_{h_1\to 0} { f(x_0+h_1, y_0) - f(x_0, y_0) \over h_1} = \dd{f}{x}(x_0, y_0). \] Along purely imaginary \(h=ih_2\) with \(h_2\in\RR\), \[ f'(z_0) = \lim_{h_2\to 0} { f(x_0, y_0+h_2) - f(x_0, y_0) \over ih_2 } = {1\over i} \dd{f}{y}(x_0, y_0). \] Taking real and imaginary parts of these limits shows that the partial derivatives of \(u\) and \(v\) exist. Writing \(1/i=-i\), \[ \begin{aligned} \dd{f}{x} &= \dd{u}{x} + i \dd{v}{x}, \\ {1\over i} \dd{f}{y} &= -i \qty{ \dd{u}{y} + i \dd{v}{y}} = \dd{v}{y} - i\dd{u}{y}. \end{aligned} \] Equating real and imaginary parts of the two expressions for \(f'(z_0)\) gives \(\dd{u}{x} = \dd{v}{y}\) and \(\dd{v}{x} = -\dd{u}{y}\).

    Cauchy--Riemann implies holomorphic

    $C^1$ plus Cauchy--Riemann turns real differentiability into complex differentiability.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be open and let \(f = u+iv\colon\Omega\to\CC\), where \(u,v\colon\Omega\to\RR\) are continuously differentiable as functions of \((x,y)\), \(z=x+iy\). If \(u_x=v_y\) and \(u_y=-v_x\) on \(\Omega\), then \(f\) is holomorphic on \(\Omega\) and \[ f'(z) = \del f(z) = u_x(z) + iv_x(z). \]

    Fix \(z\in\Omega\) and write \(h=h_1+ih_2\). Since \(u\) and \(v\) are continuously differentiable, they are real differentiable at \(z\), so \[ f(z+h)-f(z)=u_xh_1+u_yh_2+i\qty{v_xh_1+v_yh_2}+o(\abs{h}), \] with the partial derivatives evaluated at \(z\). Substituting \(u_y=-v_x\) and \(v_y=u_x\) gives \(u_xh_1-v_xh_2+i\qty{v_xh_1+u_xh_2}=(u_x+iv_x)(h_1+ih_2)\), so \(\frac{f(z+h)-f(z)}{h}\to u_x+iv_x\) as \(h\to0\). Finally, \(\del f=\frac12\qty{f_x-if_y}=\frac12\qty{u_x+v_y+i(v_x-u_y)}=u_x+iv_x\) by the same equations.

    Wirtinger operators $\partial$ and $\bar\partial$

    Let \(U\subseteq\CC\) be open, identify \(\CC\) with \(\RR^2\) by \(z=x+iy\), and let \(F\colon U\to\CC\) be \(C^1\) as a function of \((x,y)\). The Wirtinger derivatives of \(F\) are \[ \frac{\partial F}{\partial z}\coloneqq\frac12\qty{\frac{\partial F}{\partial x}-i\frac{\partial F}{\partial y}},\qquad \frac{\partial F}{\partial\bar z}\coloneqq\frac12\qty{\frac{\partial F}{\partial x}+i\frac{\partial F}{\partial y}}, \] and the operators \(\del\coloneqq\del_z\coloneqq\frac12(\del_x-i\del_y)\) and \(\delbar\coloneqq\del_{\bar z}\coloneqq\frac12(\del_x+i\del_y)\) are the del and del-bar operators.

    Since \(\frac1i=-i\), the same formulas read \(\frac{\partial F}{\partial z}=\frac12\qty{\frac{\partial F}{\partial x}+\frac1i\frac{\partial F}{\partial y}}\) and \(\frac{\partial F}{\partial\bar z}=\frac12\qty{\frac{\partial F}{\partial x}-\frac1i\frac{\partial F}{\partial y}}\).

    With \(F\) as in the definition, \(dz=dx+i\,dy\), and \(d\bar z=dx-i\,dy\), the differential of \(F\) is \[ dF=\frac{\partial F}{\partial z}\dz+\frac{\partial F}{\partial\bar z}\dzbar. \]

    Substituting \(dz\) and \(d\bar z\), the right side equals \[ \frac12\qty{F_x-iF_y}(dx+i\,dy)+\frac12\qty{F_x+iF_y}(dx-i\,dy)=F_x\dx+F_y\dy=dF. \]

    Holomorphic at a point if and only if $\bar\partial f$ vanishes there

    For a real-differentiable map, complex differentiability is exactly the equation $\bar\partial f=0$.Suggest a slogan

    Let \(f=u+iv\) be defined on a neighborhood of \(z_0\in\CC\), with \(u,v\) real-valued and real-differentiable at \(z_0\). Then \(f\) is holomorphic at \(z_0\) if and only if \[ \delbar f(z_0)=\frac12(\del_x+i\del_y)f(z_0)=0. \]

    At \(z_0\), \[ \begin{aligned} 2\delbar f&=(\del_x+i\del_y)(u+iv)\\ &=u_x+iv_x+iu_y-v_y\\ &=(u_x-v_y)+i(u_y+v_x). \end{aligned} \] So \(\delbar f(z_0)=0\) exactly when \(u_x=v_y\) and \(u_y=-v_x\) at \(z_0\), the Cauchy–Riemann equations. For \(f\) real-differentiable at \(z_0\), these hold if and only if the real derivative of \(f\) at \(z_0\) is multiplication by a complex number, that is, if and only if \(f\) is complex differentiable at \(z_0\).

    Laplacian and harmonic functions

    Let \(\Omega\subseteq\RR^2\) be open and let \(u\colon\Omega\to\RR\) be a \(C^2\) function of \((x,y)\). The Laplacian of \(u\) is \[ \Delta u\coloneqq\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}, \] and \(u\) is harmonic on \(\Omega\) if \(\Delta u=0\) on \(\Omega\).

    Holomorphic functions have harmonic components

    Holomorphicity couples real and imaginary parts by Cauchy--Riemann, forcing both to be harmonic.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be open and let \(f(z) = u(x, y) + iv(x, y)\) be holomorphic on \(\Omega\), with \(u,v\) real-valued and \(z=x+iy\). Then \(u\) and \(v\) are harmonic on \(\Omega\).

    The derivative of a holomorphic function is holomorphic, so \(f\) is infinitely differentiable and \(u,v\) have continuous partial derivatives of all orders. By the Cauchy–Riemann equations \(u_x=v_y\) and \(u_y=-v_x\), so \(u_{xx}=v_{yx}=v_{xy}=-u_{yy}\) and \(v_{xx}=-u_{yx}=-u_{xy}=-v_{yy}\), using equality of mixed partials. Hence \(\Delta u=\Delta v=0\).

    Analytic function

    Let \(\Omega\subseteq\CC\) be open, let \(f\colon\Omega\to\CC\), and let \(z_0\in\Omega\). The function \(f\) is analytic at \(z_0\) if there exist a power series \(\sum_{n\ge0}a_n(z-z_0)^n\) with radius of convergence \(R>0\) and an open neighborhood \(U\subseteq\Omega\) of \(z_0\) with \(\abs{z-z_0}<R\) for all \(z\in U\), such that \[ f(z)=\sum_{n\ge0}a_n(z-z_0)^n\qquad\text{for all } z\in U. \] The function \(f\) is analytic on \(\Omega\) if it is analytic at every point of \(\Omega\).

    Power series are smooth on their disc of convergence

    On its open disc of convergence, a power series is holomorphic, differentiates term by term, and has coefficients $c_k=f^{(k)}(z_0)/k!$.Suggest a slogan

    Let \(f(z)=\sum_{k\ge0}c_k(z-z_0)^k\) be a power series with radius of convergence \(R>0\). Then \(f\) is holomorphic and infinitely differentiable on \(D_R(z_0)\), its derivative is obtained by term-by-term differentiation, \[ f'(z)=\sum_{k\ge1}kc_k(z-z_0)^{k-1},\qquad z\in D_R(z_0), \] where this series also has radius of convergence \(R\), and the coefficients are \[ c_k=\frac{f^{(k)}(z_0)}{k!},\qquad k\ge0. \]