A map \(f = u + iv\) that is real differentiable at \(z_0\), regarded as a map \(\RR^2\to\RR^2\), is complex differentiable at \(z_0\) if and only if \(u_x = v_y\) and \(u_y = -v_x\) at \(z_0\).
Analytic function (Tag D-V6UQJ)
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 \(\Omega\).
Holomorphic and entire functions (Tag D-E7A5W)
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].
Real differentiability of maps $\RR^n\to\RR^m$ (Tag D-KLTBZ)
Let \(U\subseteq\RR^n\) be open, let \(F\colon U\to\RR^m\), and let \(\vector p\in U\). The map \(F\) is real-differentiable at \(\vector p\) if there exists a linear map \(A\colon\RR^n\to\RR^m\) such that \[ \lim_{\vector h\to0}\frac{\norm{F(\vector p+\vector h)-F(\vector p)-A(\vector h)}}{\norm{\vector h}}=0. \]
Equivalently, there is a function \(R\), defined for \(\vector h\) with \(\vector p+\vector h\in U\), such that \(\norm{R(\vector h)}\to0\) as \(\vector h\to0\) and \[ F(\vector p+\vector h)=F(\vector p)+A(\vector h)+\norm{\vector h}\,R(\vector h). \] In little-\(o\) notation, \(F(\vector p+\vector h)=F(\vector p)+A(\vector h)+o(\norm{\vector h})\) as \(\vector h\to0\).
Complex differentiable implies Cauchy--Riemann (Tag PR-37QA5)
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 \(\dd{v}{x} = -\dd{u}{y}\).
Polar form
Polar Cauchy--Riemann equations (Tag PR-OOHUL)
Let \(\Omega\subseteq\CC\sm\ts{0}\) be open and let \(f=u+iv\) be holomorphic on \(\Omega\), with \(u,v\) real-valued. Writing \(u\) and \(v\) as functions of \((r,\theta)\) through \(z=re^{i\theta}\), \(r>0\), \[ \frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial\theta}\qquad\text{and}\qquad\frac{1}{r}\frac{\partial u}{\partial\theta}=-\frac{\partial v}{\partial r}. \]
With \(x=r\cos\theta\) and \(y=r\sin\theta\), the chain rule gives \[ \begin{aligned} u_r&=u_x\cos\theta+u_y\sin\theta, & u_\theta&=r(-u_x\sin\theta+u_y\cos\theta),\\ v_r&=v_x\cos\theta+v_y\sin\theta, & v_\theta&=r(-v_x\sin\theta+v_y\cos\theta). \end{aligned} \] Substituting the Cauchy–Riemann equations \(v_x=-u_y\) and \(v_y=u_x\) gives \[ v_\theta=r(u_y\sin\theta+u_x\cos\theta)=ru_r,\qquad v_r=-u_y\cos\theta+u_x\sin\theta=-\frac1r u_\theta. \]
Setting \[ z = re^{i\theta} = r(\cos\theta + i\sin\theta ) = x+iy \] gives \(x=r\cos\theta\) and \(y=r\sin\theta\), so \[ \begin{aligned} x_r = \cos(\theta)&, x_\theta = -r\sin(\theta) \\ y_r = \sin(\theta)&, y_\theta = r\cos(\theta) \end{aligned}.\]
By the chain rule, \[ \begin{aligned} u_r &= u_x x_r + u_y y_r \\ &= v_y x_r -v_x y_r && \text{CR}\\ &= v_y \cos(\theta) - v_x \sin(\theta) \\ &= {1\over r}\qty{ v_y r\cos(\theta) - v_x r\sin(\theta) } \\ &= {1\over r}\qty { v_y y_\theta + v_x x_\theta} \\ &= {1\over r} v_\theta \end{aligned},\] and similarly \[ \begin{aligned} v_r &= v_x x_r + v_y y_r \\ &= v_x \cos(\theta) + v_y\sin(\theta) \\ &= -u_y\cos(\theta) + u_x\sin(\theta) && \text{CR} \\ &= -{1\over r} u_\theta \end{aligned}.\] Thus \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} .\]
The polar form applies directly to functions given in terms of \(r\) and \(\theta\).
Wirtinger derivatives
In terms of the Wirtinger derivatives \(\del\) and \(\delbar\), the Cauchy–Riemann equations are the single equation \(\delbar f = 0\).
Wirtinger operators $\partial$ and $\bar\partial$ (Tag D-OAFF5)
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 (Tag PR-TVPCM)
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 \(z_0\).
(Computing with Wirtinger derivatives)
\(\overline{\delbar f(z)} = \del \overline f(z)\), and the basic differentials are \(d(cz) = c\dz\) and \(d(c\bar z) = c\dzbar\). For example, \[ \del \abs z^2 = \del (z\bar z) = \bar z, \qquad \delbar \abs z^2 = z, \qquad d\qty{\abs z^2} = \bar z \dz + z\dzbar ,\] and \[ \del \exp\qty{-\abs z^2} = \del \exp\qty{-z\bar z} = e^{-\abs z^2}\cdot\del(z\bar z) = \bar z e^{-\abs z^2} .\]
Derivatives in coordinates
For a holomorphic \(f\), \[ f' = \dd{f}{z} = {1\over i}\dd{f}{y} = \dd{f}{x} = \dd u x + i \dd v x .\]
(Differentials)
\[ \begin{aligned} dz &= dx + i~dy \\ d\bar z &= dx - i~dy \\ f_z &= f_x = f_y / i \end{aligned}.\]
(The Jacobian is $\abs{f'}^2$)
Regarding \(f\) as a map \(\RR^2\to\RR^2\) with \(f(x+iy) = u+iv\), the Jacobian determinant is \[ J = u_xv_y -v_x u_y = u_x^2 + v_x^2 \] after applying CR, and on the other hand \(\abs{f'(z)}^2 = \abs{u_x + iv_x}^2 = J\). So a holomorphic map scales area infinitesimally by \(\abs{f'}^2\) and is orientation preserving wherever \(f'\neq 0\).
Linear approximation criterion for complex differentiability, and continuity (Tag PR-Y36BS)
Let \(f\) be defined on a neighborhood of \(z_0\in\CC\). Then \(f\) is holomorphic at \(z_0\) if and only if there exist \(a\in\CC\) and a function \(\psi\) defined for small \(h\neq0\) such that \[ f(z_0+h)-f(z_0)-ah=h\,\psi(h),\qquad \psi(h)\to0\text{ as }h\to0. \] In this case \(a=f'(z_0)\), and \(f\) is continuous at \(z_0\).
If \(f'(z_0)\) exists, put \(a\coloneqq f'(z_0)\) and \(\psi(h)\coloneqq\frac{f(z_0+h)-f(z_0)}{h}-a\), which tends to \(0\). Conversely, dividing the displayed identity by \(h\) gives \(\frac{f(z_0+h)-f(z_0)}{h}=a+\psi(h)\to a\), so \(f'(z_0)=a\). Finally \(f(z_0+h)-f(z_0)=h\big(a+\psi(h)\big)\to0\) as \(h\to0\).
(Holomorphic iff linear approximation)
If \(f'(z_0)\) exists, set \(a\coloneqq f'(z_0)\) and \[ \psi(h) \coloneqq \begin{cases} \dfrac{f(z_0+h)-f(z_0)}{h}-a & h\neq 0,\\ 0 & h=0. \end{cases} \] Then \(\psi(h)\to 0\) as \(h\to 0\) by definition of \(f'(z_0)\), and \(f(z_0+h)-f(z_0)-ah = h\psi(h)\).
Conversely, if such \(a\) and \(\psi\) exist then for \(h\neq 0\) \[ \frac{f(z_0+h)-f(z_0)}{h} = a + \psi(h) \to a ,\] so \(f'(z_0)=a\). Continuity at \(z_0\) is the same identity: \(f(z_0+h)=f(z_0)+h\bigl(a+\psi(h)\bigr)\to f(z_0)\).
Solving real integrals with complex calculus
Since \(e^{(3+2i)z}\) has antiderivative \(e^{(3+2i)z}/(3+2i)\), restricting to real \(z=x\) and taking real parts evaluates a real integral: \[ \begin{aligned} \int e^{3x}\cos(2x) \dx &= \Re \int e^{3x}e^{2ix}\dx \\ &= \Re \int e^{(3+2i)x} \dx \\ &= \Re {e^{(3+2i)x} \over 3+2i} + C \end{aligned}.\]
Exercises
A $C^1$ function satisfies the Cauchy--Riemann equations if and only if it is holomorphic (Tag E-BGKED)
Show that \(f = u+iv\) with \(u, v\in C^1(\RR)\) satisfying the Cauchy-Riemann equations on \(\Omega\), then \(f\) is holomorphic on \(\Omega\) with \[ f'(z) = \dd{f}{x} = {1\over i} \dd{f}{y} = {1\over 2}\qty{u_x + iv_x} .\] Conversely, show that if \(f\) is holomorphic, then \(f\) satisfies the Cauchy-Riemann equations.
Solution
Direction 1: Holomorphic \(\implies\) CR.
Suppose \(f'(z_0)\) exists for some \(z_0 = x_0 + iy_0 \in \Omega\). Since the limit \[f'(z_0) = \lim_{h\to 0,\, h\in \CC} \frac{f(z_0 + h) - f(z_0)}{h}\] exists, we may approach along the real and imaginary axes.
Along \(h = t \in \RR\): \[f'(z_0) = \lim_{t\to 0} \frac{f(x_0 + t, y_0) - f(x_0, y_0)}{t} = f_x(x_0, y_0).\] Along \(h = it\), \(t \in \RR\): \[f'(z_0) = \lim_{t\to 0} \frac{f(x_0, y_0 + t) - f(x_0, y_0)}{it} = \frac{1}{i} f_y(x_0, y_0).\] Equating the two expressions for \(f'(z_0)\) and writing \(f = u + iv\): \[f_x = \frac{1}{i} f_y \implies i(u_x + iv_x) = u_y + iv_y \implies -v_x + iu_x = u_y + iv_y.\] Comparing real and imaginary parts: \(u_x = v_y\) and \(u_y = -v_x\).
Direction 2: CR \(\implies\) holomorphic.
Suppose \(u, v \in C^1(\Omega)\) satisfy \(u_x = v_y\) and \(u_y = -v_x\) on \(\Omega\). Write \(h = \Delta x + i\Delta y\). By the multivariable Taylor theorem (since \(u, v\) are \(C^1\)): \[u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0) = u_x \Delta x + u_y \Delta y + o(|h|),\] \[v(x_0 + \Delta x, y_0 + \Delta y) - v(x_0, y_0) = v_x \Delta x + v_y \Delta y + o(|h|),\] where the partial derivatives are evaluated at \((x_0, y_0)\). Using the CR equations \(u_y = -v_x\) and \(v_y = u_x\): \[f(z_0 + h) - f(z_0) = (u_x \Delta x - v_x \Delta y) + i(v_x \Delta x + u_x \Delta y) + o(|h|).\] The first two terms factor as \((u_x + iv_x)(\Delta x + i\Delta y) = (u_x + iv_x) h\), so: \[\frac{f(z_0 + h) - f(z_0)}{h} = (u_x + iv_x) + \frac{o(|h|)}{h}.\] Since \(|o(|h|)/h| \to 0\) as \(h \to 0\), the limit exists and \(f'(z_0) = u_x + iv_x = u_x - iu_y\).
The last expression in the statement has a spurious factor: \(f'(z)=\dd{f}{x}=u_x+iv_x\). The factor \({1\over2}\) belongs to the Wirtinger derivative \({\partial f\over\partial z}={1\over2}\qty{\dd{f}{x}-i\dd{f}{y}}\), which equals \(f'(z)\) for holomorphic \(f\).
Polar CR equations (Tag E-PAQPF)
Show that in polar coordinates, the CR equations take the following form: \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{1}{r} \frac{\partial u}{\partial \theta}=-\frac{\partial v}{\partial r} .\]
Solution
Setting \[ z = re^{i\theta} = r(\cos(\theta) + i\sin(\theta) ) = x+iy \] yields \(x=r\cos(\theta), y=r\sin(\theta)\), one can identify \[ x_r = \cos(\theta)&, x_\theta = -r\sin(\theta) \\ y_r = \sin(\theta)&, y_\theta = r\cos(\theta) .\]
Now apply the chain rule: \[ u_r &= u_x x_r + u_y y_r \\ &= v_y x_r -v_x y_r && \text{CR}\\ &= v_y \cos(\theta) - v_x \sin(\theta) \\ &= {1\over r}\qty{ v_y r\cos(\theta) - v_x r\sin(\theta) } \\ &= {1\over r}\qty { v_y y_\theta + v_x x_\theta} \\ &= {1\over r} v_\theta .\] Similarly, \[ v_r &= v_x x_r + v_y y_r \\ &= v_x \cos(\theta) + v_y\sin(\theta) \\ &= -u_y\cos(\theta) + u_x\sin(\theta) && \text{CR} \\ &= {1\over r}\qty{ -u_y r\cos(\theta) + u_x r\sin(\theta)} \\ &= {1\over r}\qty{ -u_y y_\theta - u_x x_0 } \\ &= -{1\over r} u_\theta .\]
Thus \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} \\ .\]
Real-valued holomorphic functions are constant (Tag E-6F2HU)
Show that a real-valued holomorphic function is constant.
Solution
Let \(f\) be holomorphic and real-valued on a domain \(\Omega\). If \(f\) were nonconstant, the open mapping theorem would make \(f(\Omega)\) a nonempty open subset of \(\CC\) contained in \(\RR\), but \(\RR\) contains no nonempty open subset of \(\CC\). So \(f\) is constant.
Holomorphic functions on a domain: $fg\equiv 0$ implies $f\equiv 0$ or $g\equiv 0$ (Tag E-FXYTL)
Show that if \(fg \equiv 0\) on a domain \(\Omega\) then either \(f\equiv 0\) or \(g\equiv 0\).
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.
Entire functions agreeing with $e^x$ on $\RR$ (Tag E-WKY7C)
Find all entire functions \(f\) such that \(f(x) = e^x\) on \(\RR\).
Solution
The function \(g(z) \da f(z) - e^z\) is entire and identically zero on \(\RR\), which contains a limit point. So \(g(z) \equiv 0\) on \(\CC\), meaning \(f(z) = e^z\) is the only such function.
Injectivity relates to derivatives (Tag E-UVNVV)
Show that if \(z_0\) is a zero of \(f'\) of order \(n\), then \(f\) is \((n+1)\)-to-one in a neighborhood of \(z_0\).
Solution
Goal: Show that if \(z_0\) is a zero of \(f'\) of order \(n\), then \(f\) is \((n+1)\)-to-one in a neighborhood of \(z_0\) (for \(z\) near \(z_0\), the equation \(f(z) = w\) has exactly \(n+1\) solutions in a punctured neighborhood of \(z_0\), for \(w\) near \(f(z_0)\), \(w \neq f(z_0)\)).
<1>1. Setup: expand \(f\) about \(z_0\).
Since \(z_0\) is a zero of \(f'\) of order \(n\), we have \(f'(z) = (z - z_0)^n h(z)\) with \(h\) analytic near \(z_0\), \(h(z_0) \neq 0\).
Integrating, \(f(z) - f(z_0) = (z - z_0)^{n+1} \phi(z)\) where \(\phi\) is analytic near \(z_0\) and \(\phi(z_0) = h(z_0)/(n+1) \neq 0\).
<1>2. \(\phi\) is nonzero on a neighborhood of \(z_0\), so \(\phi\) has an \((n+1)\)-st root \(\psi\) with \(\psi^{n+1} = \phi\), \(\psi\) analytic, \(\psi(z_0) \neq 0\).
\(\phi(z_0) \neq 0\) by <1>1; by continuity \(\phi \neq 0\) in a disk around \(z_0\), where a holomorphic branch of the \((n+1)\)-st root exists (the disk is simply connected and \(\phi\) avoids \(0\)).
<1>3. Define \(\zeta(z) = (z - z_0)\psi(z)\); then \(\zeta\) is a biholomorphism from a neighborhood of \(z_0\) onto a neighborhood of \(0\), with \(\zeta(z_0) = 0\) and \(\zeta'(z_0) = \psi(z_0) \neq 0\).
\(\zeta'(z_0) = \psi(z_0) \neq 0\), so the inverse function theorem gives the local biholomorphism.
<1>4. In the \(\zeta\) coordinate, \(f(z) = f(z_0) + \zeta(z)^{n+1}\).
<1>1 and <1>2: \(f(z) - f(z_0) = (z - z_0)^{n+1}\phi(z) = ((z - z_0)\psi(z))^{n+1} = \zeta(z)^{n+1}\).
<1>5. The map \(w \mapsto f(z_0) + w^{n+1}\) is \((n+1)\)-to-one on a punctured neighborhood of \(0\).
For \(w' \neq 0\) near \(0\), the equation \(w^{n+1} = w'\) has exactly \(n+1\) distinct solutions (the \(n+1\) roots of \(w'\)), all lying in a small disk around \(0\).
<1>6. Q.E.D.
<1>3 shows the change of variables \(z \leftrightarrow \zeta\) is one-to-one, and <1>5 shows \(f\) in these coordinates is \((n+1)\)-to-one; composing, \(f\) is \((n+1)\)-to-one near \(z_0\).