Note: These are notes live-tex’d from a graduate course on Complex Analysis taught by Jingzhi Tie at the University of Georgia in Spring 2020. As such, any errors or inaccuracies are almost certainly my own.
Last updated: 2020-10-25
Recall that \({\mathbb{C}}\) is a field, where \[\begin{align*}z = x + iy \implies \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = x - iy\end{align*}\] and if \(z\neq 0\) then \[\begin{align*}z^{-1}= \frac{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} {{\left\lvert {z} \right\rvert}^2}\end{align*}\]
Note that \(z_n \to z\) and \(z_n = x_n + iy_n\), and \[\begin{align*}{\left\lvert {z_n - z} \right\rvert} = \sqrt{(x_n - x)^2 - (y_n - y)^2} < \varepsilon \implies {\left\lvert {x - x_n} \right\rvert}, {\left\lvert {y - y_n} \right\rvert} < \varepsilon.\end{align*}\]
Since \({\mathbb{R}}\) is complete iff every Cauchy sequence converges iff every bounded monotone sequence has a limit.
Note: This is useful precisely when you don’t know the limiting term.
Note that \(\sum_k z_k\) thus converges if \({\left\lvert {\sum_{k=m}^n z_k} \right\rvert} < \varepsilon\) for \(m, n\) large enough, so sums converges iff they have small tails.
Note that the partial sums \(\sum^N {\left\lvert {z_k} \right\rvert}\) are monotone, so \(\tilde S_N\) converges iff the partial sums are bounded above.
Examples:
\[\begin{align*} \sum x^k &= \frac 1 {1-x} \\ \sum (-x^2)^k &= \frac 1 {1+x^2} .\end{align*}\]
Note that both of these have a radius of convergence equal to 1, since the first has a pole at \(x=1\) and the second as a pole at \(x = i\).
Recall that \(\sum z_k\) converges iff \(s_n = \sum_{k=1}^n z_k\) converges.
The most interesting series: \(f(z) = \sum a_k z^k\), i.e. power series.
Proposition: If \(\sum a_k z_k\) converges at some point \(z_0\), then it converges for all \({\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert}_0\).
Note that this inequality is necessarily strict. For example, \(\sum \frac{z^{n-1}}{n}\) converges at \(z=-1\) (alternating harmonic series) but not at \(z=1\) (harmonic series).
Suppose \(\sum a_k z_1^k\) converges. The terms are uniformly bounded, so \({\left\lvert {a_k z_1^k} \right\rvert} \leq C\) for all \(k\). Then we have \[\begin{align*}{\left\lvert {a_k} \right\rvert} \leq C/{\left\lvert {z_1} \right\rvert}^k\end{align*}\], so if \({\left\lvert {z} \right\rvert} < {\left\lvert {z_1} \right\rvert}\) we have \[\begin{align*}{\left\lvert {a_k z^k} \right\rvert} \leq {\left\lvert {z} \right\rvert}^k \frac{C}{{\left\lvert {z_1} \right\rvert}^k} = C ({\left\lvert {z} \right\rvert} / {\left\lvert {z_1} \right\rvert} )^k.\end{align*}\] So if \({\left\lvert {z} \right\rvert} < {\left\lvert {z_1} \right\rvert}\), the parenthesized quantity is less than 1, and the original series is bounded by a geometric series. Letting \(r = {\left\lvert {z} \right\rvert} / {\left\lvert {z_1} \right\rvert}\), we have
\[\begin{align*} \sum {\left\lvert {a_k z^k} \right\rvert} \leq \sum c r^k = \frac{c}{1-r} ,\end{align*}\]
and so we have absolute convergence.
The radius of convergence of a series is the real number \(R\) such that \(f(z) = \sum a_k z^k\) converges precisely for \({\left\lvert {z} \right\rvert} < R\) and diverges for \({\left\lvert {z} \right\rvert} > R\).
We denote a disc of radius \(R\) centered at zero by \(D_R\). If \(R=\infty\), then \(f\) is said to be entire.
Recall that \(S_n(z) \to S(z)\) uniformly on \(\Omega\) iff \(\forall \varepsilon > 0\), there exists a \(M\in {\mathbb{N}}\) such that \[\begin{align*}n> M \implies {\left\lvert {S_n(z) - S(z)} \right\rvert} < \varepsilon\end{align*}\] for all \(z\in \Omega\)
Note that arbitrary limits of continuous functions may not be continuous. Counterexample: \(f_n(x) = x^n\) on \([0, 1]\); then \(f_n \to \delta(1)\). This uniformly converges on \([0, 1-\varepsilon]\) for any \(\varepsilon > 0\).
Show that the uniform limit of continuous functions is continuous.
Hint: Use the triangle inequality.
Write \(f(z) = \sum_{k=0}^N a_k z^k + \sum_{N+1}^\infty a_k z^k \mathrel{\vcenter{:}}= S_N(z) + R_N(z)\). Note that if \({\left\lvert {z} \right\rvert} < R\), then there exists a \(T\) such that \({\left\lvert {z} \right\rvert} < T < R\) where \(f(z)\) converges uniformly on \(D_T\).
Check!
We need to show that \({\left\lvert {R_N(z)} \right\rvert}\) is uniformly small for \({\left\lvert {z} \right\rvert} < s < T\). Note that \(\sum a_k z^k\) converges on \(D_T\), so we can find a \(C\) such that \({\left\lvert {a_k z^k} \right\rvert} \leq C\) for all \(k\). Then \({\left\lvert {a_k} \right\rvert} \leq C/T^k\) for all \(k\), and so
\[\begin{align*} {\left\lvert {\sum_{k=N+1}^\infty a_k z^k} \right\rvert} &\leq \sum_{k=N+1}^\infty {\left\lvert {a_k} \right\rvert} {\left\lvert {z} \right\rvert}^k \\ &\leq \sum_{k=N+1}^\infty (c/T^k) s^k \\ &= c\sum {\left\lvert {s/T} \right\rvert}^k \\ &= c \frac{r^{N+!}}{1-r} &= C \varepsilon_n \to 0 ,\end{align*}\]
which follows because \(0 < r = s/T < 1\).
So \(S_N(z) \to f(z)\) uniformly on \({\left\lvert {z} \right\rvert} < s\) and \(S_N(z)\) are all continuous, so \(f(z)\) is continuous.
There are two ways to compute the radius of convergence:
As long as these series converge, we can compute derivatives and integrals term-by-term, and they have the same radius of convergence.
See references: Taylor’s Complex Analysis, Stein, Barry Simon (5 volume set), Hormander (technically a PDEs book, but mostly analysis)
Good Paper: Hormander 1955
We’ll mostly be working from Simon Vol. 2A, most problems from from Stein’s Complex.
To do analysis, we’ll need the following notions:
Continuity of a complex-valued function \(f: \Omega \to \Omega\)
Complex-differentiability: For \(\Omega \subset {\mathbb{C}}\) open and \(z_0 \in \Omega\), there exists \(\varepsilon > 0\) such that \(D_\varepsilon = \left\{{z {~\mathrel{\Big|}~}{\left\lvert {z - z_0} \right\rvert} < \varepsilon}\right\} \subset \Omega\), and \(f\) is holomorphic (complex-differentiable) at \(z_0\) iff \[\begin{align*}\lim_{h\to 0} \frac 1 h (f(z_0 + h) - f(z_0))\end{align*}\] exists; if so we denote it by \(f'(z_0)\).
Given \(f(z) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\), we have \(f(z+h)-f(z) = \mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu\), so the ratio is \(\frac{\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu}{h}\) and the limit doesn’t exist.
Note that if \(h\in {\mathbb{R}}\), then \(\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu = h\) and the ratio is identically 1, while if \(h\) is purely imaginary, then \(\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu = -h\) and the limit is identically \(-1\).
We say \(f\) is holomorphic on an open set \(\Omega\) iff it is holomorphic at every point, and is holomorphic on a closed set \(C\) iff there exists an open \(\Omega \supset C\) such that \(f\) is holomorphic on \(\Omega\).
\[\begin{align*} \lim_{h_1 \to 0} \frac{f(x_0 + iy_0 + h_1) - f(x_0 + iy_0)}{h_1} = {\frac{\partial f}{\partial x}\,}(x_0, y_0) \\ \lim_{h_2 \to 0} \frac{f(x_0 + iy_0 + ih_2) - f(x_0 + iy_0)}{ih_2} = \frac 1 i {\frac{\partial f}{\partial y}\,}(x_0, y_0) \\ \implies {\frac{\partial f}{\partial x}\,} = \frac 1 i {\frac{\partial f}{\partial y}\,} .\end{align*}\]
So if we write \(f(z) = u(x, y) + i v(x, y)\), we have
\[\begin{align*} {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} \mathrel{\Big|}_{(x_0, y_0)} = \frac 1 i \qty{ {\frac{\partial u}{\partial y}\,} + i {\frac{\partial v}{\partial y}\,} } \mathrel{\Big|}_{(x_0, y_0)} ,\end{align*}\]
and equating real and imaginary parts yields the Cauchy-Riemann equations:
\[\begin{align*} {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} = -i {\frac{\partial u}{\partial y}\,} + {\frac{\partial v}{\partial y}\,} \\ \iff {\frac{\partial u}{\partial x}\,} = {\frac{\partial v}{\partial y}\,} \quad\text{and}\quad {\frac{\partial u}{\partial y}\,} = - {\frac{\partial v}{\partial x}\,} .\end{align*}\]
The usual rules of derivatives apply:
Use the fact that if \(f'(g(z)) = a\), then \[\begin{align*}f(z+h) - f(z) = a h + r(z, h),\quad {\left\lvert {r(z, h)} \right\rvert} = o({\left\lvert {h} \right\rvert}) \to 0.\end{align*}\]
Write \(b = g'(z)\), then \[\begin{align*}f(g(z + h)) = f(g(z) + b h + r_1 ) = f(g(z)) + f'(g(z))bh + r_2\end{align*}\] by considering error terms, and so \[\begin{align*}\frac 1 h (f(g(z+h)) - f(g(z))) \to f'(g(z)) g'(z)\end{align*}\].
Reference: See Lang’s Complex Analysis, there are plenty of solution manuals. Note: look for 13 statements equivalent to holomorphic: Springer GTM Lipman.
Let \(f; \Omega \to {\mathbb{C}}\) be a complex-valued function. Recall that \(f\) is complex differentiable iff the usual ratio/limit exists. Note that \(h = x+iy\) and \(h\to 0 \iff x,y\to 0\).
We can write \[\begin{align*}f'(z) = {\frac{\partial f}{\partial x}\,} = \frac 1 i {\frac{\partial f}{\partial y}\,}.\end{align*}\] This follows from Cauchy-Riemann since \(u_x = v_y\) and \(u_y = -v_x\).
We want to define \({\partial}, \mkern 1.5mu\overline{\mkern-1.5mu\partial\mkern-1.5mu}\mkern 1.5mu\) operators. We have the identities
\[\begin{align*} x = \frac{z + \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}{z} \quad y = \frac{z - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}{iz} .\end{align*}\]
We can then write
\[\begin{align*} dz &= dx + idy \\ d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu &= dx - i dy .\end{align*}\]
We define the dual operators by \({\left\langle {{\frac{\partial }{\partial z}\,}},~{dz} \right\rangle} = 1\) and similarly \({\left\langle { {\frac{\partial }{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} },~{d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} \right\rangle} = 1\).
By the chain rule, we can write
\[\begin{align*} f_z &= {\frac{\partial f}{\partial x}\,} {\frac{\partial x}{\partial z}\,} + {\frac{\partial f}{\partial y}\,} {\frac{\partial y}{\partial z}\,} \\ &= \frac 1 2 {\frac{\partial f}{\partial x}\,} + {\frac{\partial f}{\partial y}\,} \frac{1}{2i} \\ &= \frac 1 2 \qty{{\frac{\partial }{\partial x}\,} + i {\frac{\partial }{\partial y}\,} }f ,\end{align*}\]
and similarly \[\begin{align*} f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} &= {\frac{\partial f}{\partial x}\,} {\frac{\partial x}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} + {\frac{\partial f}{\partial y}\,} {\frac{\partial z}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} \\ &= \frac 1 2 \qty{ {\frac{\partial }{\partial x}\,} - \frac{1}{2i} {\frac{\partial }{\partial y}\,} }f .\end{align*}\]
We thus find \({\partial}_x = {\partial}_z + {\partial}_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\) and \({\partial}_y = i\qty{ {\partial}_z - {\partial}_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} }\), so define
\[\begin{align*} {\partial}f &\coloneqq{\frac{\partial f}{\partial z}\,} dz \\ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu f &\coloneqq{\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \\ \implies df &= {\frac{\partial f}{\partial z}\,} dz + {\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu .\end{align*}\]
This means that \(f\) depends on \(z\) alone and not \(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\).
Application to PDEs: we can write \[\begin{align*}u_{xx} = v_{xy} \quad u_{yy} = v_{yx}\end{align*}\] and so \[\begin{align*}u_{xx} + u_{yy} = 0 = v_{xx} + v_{yy}.\end{align*}\]
Thus \(\Delta f = 0\), sp \(f\) satisfies Laplace’s equation and is said to be harmonic.
Let \(w = f(z)\) and \(g(w) = g(f(z))\). Then
\[\begin{align*} h_z &= g_w f_z + g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_z \\ h_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} &= g_w f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} + g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} .\end{align*}\]
If \(f, g\) are holomorphic, \(f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} = 0\), so \(h_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = 0\) and \(h\) is holomorphic and \[\begin{align*}h_z = g_w f_z.\end{align*}\]
Given a power series \(f=\sum a_n (z- z_0)^n\). Then
Recall the ratio test: \[\begin{align*}\sum {\left\lvert {a_k} \right\rvert} <\infty \iff \limsup {\left\lvert {a_{k+1} / a_k} \right\rvert} < 1\end{align*}\]
Given \({\left\lvert {z} \right\rvert} < R\), fix \(r>0\) such that \({\left\lvert {z} \right\rvert} < r < R\). Suppose that \({\left\lvert {w-z} \right\rvert} < r - {\left\lvert {z} \right\rvert}\), so \({\left\lvert {w} \right\rvert} < r\).
We want to show \[\begin{align*} {\left\lvert {S} \right\rvert} = {\left\lvert {\frac{f(w) - f(z)}{w - z} - \sum_{n=1} a_n n z^{n-1}} \right\rvert} \to 0 \quad \text{as } w\to z .\end{align*}\]
Idea: write everything in terms of power series. Use the fact that \(a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots)\), and so \({\left\lvert {(w^k-z^k)/(w-z)} \right\rvert} \leq k r^{k-1}\).
\[\begin{align*} S &= \sum_{n=1} a_n \qty{ \frac{w^n - z^n}{w-z} - n z^{n-1} } \\ &= \sum a_n \qty{ w^{n-1} + w^{n-2}z + \cdots + z^{n-1} + nz^{n-1} } \\ &= \sum a_n \qty{ (w^{n-1} - z^{n-1}) + (w^{n-2} - z^{n-2})z + \cdots + (w-z) z^{n-2} } &= \sum a_n (w-z) \qty{ \cdots + z^{n-2} }\\ &\leq \sum_{n=2} {\left\lvert {a_n} \right\rvert} \frac 1 2 n(n-1) r^{n-2} {\left\lvert {z-w} \right\rvert} .\end{align*}\]
Show \(\lim_n n^{\frac 1 n} = 1\).
Also tricky: show \(\lim \sin(n)\) doesn’t exist, and \(\sin(n)\) is dense in \([-1, 1]\).
Note that an analytic function is holomorphic in its domain of convergence, so analytic implies holomorphic. The converse requires Cauchy’s integral formula.
Next time: trying to prove holomorphic functions are analytic.
Note: multiple complex variables, see Hormander or Steven Krantz
Recall from last time that if \[\begin{align*}f(z) = \sum_{n=0}^\infty a_n z^n\end{align*}\] with \(z_0 \neq 0\) has radius of convergence \[\begin{align*}R = (\limsup {\left\lvert {a_n} \right\rvert}^{1/n})^{-1}> 0\end{align*}\] then \(f'\) exists and is obtained by differentiating term-by-term.
We know that \(f\) analytic \(\implies\) \(f\) holomorphic (and smooth), and we want to show the converse. For this, we need integration.
Note that such a curve may fail to have tangent lines at \(a_i\).
Note that \(s' > 0\) preserves orientation and \(s'<0\) reverses orientation.
The integral of \(f\) over \(\gamma\) is defined as
\[\begin{align*} \int_\gamma f ~dz = \int_a^b f(z(t)) z'(t)~dt .\end{align*}\]
Note: this doesn’t depend on parameterization, since if \(t = t(s)\), then a change of variables yields \[\begin{align*} \int_\gamma f ~dz - \int_c^d f(z(t(s)))~z'(t(s))~t'(s) ~ds = \int_c^d f(\tilde z(s)) ~\tilde z'(s) ~ds .\end{align*}\]
We can extend this definition to piecewise smooth curves by \[\begin{align*} \int_\gamma f~dz = \sum \int_{a_k}^{a_{k+1}} f ~dz \end{align*}\]
This integral is linear and \(\int_\gamma f = -\int_{\gamma^-} f\).
We have an inequality
\[\begin{align*} {\left\lvert {\int_\gamma f} \right\rvert} \leq \max_{a\leq t \leq b} {\left\lvert {f(z(t))} \right\rvert} {\left\lvert {\gamma} \right\rvert} .\end{align*}\]
Recall that in \({\mathbb{R}}\), we have \[\begin{align*}F(x) =\int_a^x f(t)~dt\end{align*}\] as an antiderivative with \(F'(x) = f(x)\), and \(\int f = F(b) - F(a)\).
Use definitions, write \(z(t)\) where \(z(a) = w_1, z(b) = w_2\). Then
\[\begin{align*} \int_\gamma f &= \int_a^b f(z(t)) z'(t) ~ dt \\ &= \int_a^b F'(z(t)) z'(t) ~dt \\ &= \int_a^b F_t ~dt \\ &= F(z(b)) - F(z(a)) \quad\text{by FTC}\\ &= F(w_1) - F(w_2) .\end{align*}\]
Note that if \(\gamma\) is piecewise smooth, the sum of the integrals telescopes to yield the same conclusion.
Pick \(w_0 \in \Omega\); we want to fix \(w_0 \in \Omega\) and show \(f(w) = f(w_0)\) for all \(w\in \Omega\).
Take any path \(\gamma: w_0 \to w\), then
\[\begin{align*} 0 = \int_\gamma f' = f(w) - f(w_0) .\end{align*}\]
Let \(f(z) = e^{-z^2}\), this is holomorphic. Write \[\begin{align*} f(z) = \sum \frac{(-1)^nz^{2n}}{n!} ,\end{align*}\] so \[\begin{align*} \int f = \sum \frac{ (-1)^n z^{2n+1} }{ n! (2n+1) } .\end{align*}\]
Since \(f\) is entire, \(\int f\) is entire, and \((\int f)' = f\) so this function has a primitive. Thus \(\int_\gamma f(z) = 0\) for any closed curve. So take \(\gamma\) a rectangle with vertices \(\pm a , \pm a + ib\).
So
\[\begin{align*} \int_\gamma f = \int_{-a}^a e^{-x^2} ~dx + \int e^{-(a+iy)^2} i ~dy - \int_{-a}^a e^{-(x+ib)^2} ~dx - \int_0^b e^{-(a+iy)^2} i dy = 0 .\end{align*}\]
We can do some estimates,
\[\begin{align*} e^{-(a+iy)^2} &= e^{-(a^2 + 2iay - y^2)} \\ &= e^{-a^2 + y^2} e^{2iay} \\ &\leq e^{-a^2 + y^2} \\ &\leq e^{-a^2 + b^2}, \\ \\ {\left\lvert {\int_0^b e^{-(a+ib)^2} i ~dy} \right\rvert} &\leq e^{-a^2 + b^2} \cdot b \\ \\ \int_{-a}^a e^{-(x^2 + 2ib x)-b^2} &= e^{b^2} \int_{-a}^a e^{-x^2} ( \cos(2bx) - i \sin(2bx) ) \\ \\ &\stackrel{\mathclap{\scriptscriptstyle{odd fn}}}{=} e^{b^2} \int_{-a}^a e^{-x^2} \cos(2bx) ~dx .\end{align*}\]
Now take \(a\to \infty\) to obtain
\[\begin{align*} \int_{\mathbb{R}}e^{-x^2} ~dx = e^{b^2} \int_{\mathbb{R}}e^{-x^2} \cos(2bx) ~dx .\end{align*}\]
We can compute
\[\begin{align*} \int_{\mathbb{R}}e^{-x^2} = \left[ \qty{\int_{\mathbb{R}}e^{-x^2}}^2 \right]^{1/2} = \qty{ \int_0^{2\pi} \int_0^\infty e^{r^2} r~dr~d\theta} = \sqrt{\pi} .\end{align*}\]
and then conclude
\[\begin{align*} \int_{\mathbb{R}}e^{-x^2} \cos(2bx) = \sqrt{\pi} e^{-b^2} .\end{align*}\]
Make a change of variables \(2b = 2\pi \xi\), so \(b = \pi \xi\), then
\[\begin{align*} \int_{\mathbb{R}}e^{-x^2} \cos(2\pi \xi x) ~dx = \sqrt{\pi} e^{-\pi^2 \xi^2} .\end{align*}\]
Thus \({\mathcal{F}}(e^{-x^2}) = \sqrt{\pi} e^{-\pi^2 \xi^2}\), allowing computation of the Fourier transform. Note that this can be used to prove the Fourier inversion formula.
Take \(\Omega \in {\mathbb{R}}^2\) bounded with \({{\partial}}\Omega\) piecewise smooth. If \(f, g\in C^1{\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu}\), then
\[\begin{align*} \int_{{{\partial}}\Omega} f ~dx + g ~dy = \iint_{\Omega} \qty{g_x - f_y} ~dA .\end{align*}\]
Write \(\gamma = {{\partial}}\Gamma\), and noting that \(f_z = f_x = \frac 1 i f_y\) implies that \({\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,}\), so
\[\begin{align*} \int_\gamma f ~dz &= \int_\gamma f(z) ~(dx + i dy) \\ &= \int f(z) ~dx + i f(z) ~dy \\ &= \iint_\Gamma \qty{if_x - f_y} ~dA \\ &= i \iint_\Gamma \qty{f_x - \frac 1 i f_y} ~dA \\ &= i \iint 0 ~dA \\ &= 0 .\end{align*}\]
Next up, we’ll prove that this integral over any triangle is zero by a limiting process.
Open question: does a PDE involving analytic functions always have solutions? Or does this hold with analytic replaced by smooth?
Fix a connected domain \(\Omega\) which is bounded with a piecewise \(C^1\) boundary.
Given \(f, g \in C^1 \mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu\), we can take a vector field \(F = \left\langle{f, g}\right\rangle\) and have
\[\begin{align*} \int_{{{\partial}}\Omega} f~dx + g~dy &= \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} - {\frac{\partial f}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} - f~dx + g~dy &= \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} + {\frac{\partial f}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} f~dy - g~dy &= \iint_{\Omega} \qty{ {\frac{\partial f}{\partial x}\,} + {\frac{\partial g}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} F \cdot \mathbf{n} ~ds &= \iint_{\Omega} \nabla \cdot F ~dA \\ \int_{{{\partial}}\Omega} \mathrm{curl}(F) ~ds &= \iint_{\Omega} \mathrm{div}(F) ~dA ,\end{align*}\]
where we take \(\mathbf{n}\) to be orthogonal to \({{\partial}}\Omega\). The quantities appearing on the RHS are referred to as the flux.
For \(f(z) \in C^1(\Omega)\) holomorphic, we can then write
\[\begin{align*} \int_{{{\partial}}\Omega} f ~dz &= \int_{{{\partial}}\Omega} f ~(dx + idy) \\ &= \int_{{{\partial}}\Omega} f ~dx + if~dy \\ &= \iint_\Omega \qty{if_x - f_y} ~dA \\ &= 0 ,\end{align*}\]
which follows since \(f\) holomorphic, we can write \[\begin{align*}f'(z) = f_x = \frac 1 i f_y,\end{align*}\] so \(i f_x = f_y\) and thus \({\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} = 0\).
See Taylor’s Introduction to Complex Analysis
If \(f\in C^1(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu)\) and \(f\) is holomorphic, then for any \(z\in \Omega\)
\[\begin{align*} f(z) = \frac{1}{2\pi i} \int_{{{\partial}}\Omega} \frac{d(\xi)}{\xi - z} ~d\xi .\end{align*}\]
Since \(z\in \Omega\) an open set, we can find some \(r> 0\) such that \(D_r(z) \subset \Omega\). Then \(\frac{f(\xi)}{\xi - z}\) is holomorphic on \(\Omega\setminus D_r(z)\). Let \(C_r = {{\partial}}D_r(z)\).
Claim: \[\begin{align*}\int_{{{\partial}}\Omega} \frac{f(\xi)}{\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)}{\xi - z} ~ d\xi.\end{align*}\]
If we can differentiate through the integral, we can obtain
\[\begin{align*} {\frac{\partial }{\partial z}\,} f(z) = \frac 1 {2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{(\xi - z)^2} ~d\xi .\end{align*}\]
and thus inductively
\[\begin{align*} (D_z)^n f(z) = \frac{n!}{2\pi i} \int_{{{\partial}}\Omega} \frac{ f(\xi) ~d\xi }{ (\xi - z)^{n+1} } .\end{align*}\]
To prove rigorously, need to write
\[\begin{align*} \Delta_h f(z) = \frac 1 h \qty{ f(z+h) - f(z) } \\ = \frac 1 {2\pi i h} \int_{{{\partial}}\Omega} f(\xi) \qty{ \frac{1}{\xi - (z+h)} - \frac{1}{\xi - z} } ~d\xi = \frac 1 {2\pi i h} \int_{{{\partial}}\Omega} f(\xi) \qty{ \frac{1}{ (\xi - z- h)(\xi - z) } } ~d\xi ,\end{align*}\]
and show the integrand converges uniformly, where \[\begin{align*} \frac{1}{(\xi - z - h)(\xi - z)} \overset{u}\to \frac{1}{(\xi - z)^2} .\end{align*}\]
Continuing inductively yields the integral formula.
Use the parameterization of \(C_r\) given by \(\xi = z + re^{i\theta}\). Then
\[\begin{align*} \frac{1}{2\pi i} \int_{C_r} \frac{f(\xi)}{\xi - z} ~d\xi &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z + re^{i\theta})}{re^{i\theta}} ~ird\theta \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\theta}) ~d\theta \\ &\overset{r \to 0}\to \frac{1}{2\pi} \int_{{{\partial}}\Omega} \frac{f(\xi)} {\xi - z} .\end{align*}\]
where we use the fact that \[\begin{align*}f(z + re^{i\theta}) = f(z) + f'(z)re^{i\theta} + o(r) \overset{r\to 0}\to f(z)\end{align*}\]
Letting \[\begin{align*}F(\xi) = \frac{ f(\xi)}{\xi - z},\end{align*}\] this is holomorphic on \(\Omega\setminus D_r(z)\). Let \(\Omega_r = {{\partial}}\Omega \cup(-C_r)\). Take the following path integral:
Then \[\begin{align*} 0 = \int_{{{\partial}}\Omega_r} F(\xi) ~d\xi = \int_{{{\partial}}\Omega} F(\xi) ~d\xi - \int_{C_r} F(\xi) ~d\xi ,\end{align*}\]
which forces these integrals to be equal.
Fix \(z_0 \in \Omega\) and let \(r = {\left\lvert {z- z_0} \right\rvert}\).
\[\begin{align*} \frac{1}{\xi - z} &= \frac{1}{ \xi - z_0 - (z-z_0) } \\ &= \frac{1}{\xi - z_0} \frac{1}{1 - \qty{ \frac{z-z_0}{\xi - z_0} } } \\ &= \frac{1}{\xi - z_0} \sum_n \qty{ \frac{z - z_0}{\xi - z_0} }^n \quad\text{for } {\left\lvert {z - z_0} \right\rvert} < {\left\lvert {\xi - z_0} \right\rvert} .\end{align*}\]
Note that \(\sum z^n\) converges uniformly for any \({\left\lvert {z} \right\rvert} < \delta < 1\).
Thus
\[\begin{align*} f(z) &= \frac{1}{2\pi i} \int_{\xi \in {{\partial}}\Omega} f(\xi) \sum \frac{(z-z_0)^n}{(\xi - z_0)^{n+1}} ~d\xi \\ &= \sum \qty{ \frac {1}{2\pi i} \int \frac{f(\xi) }{(\xi - z_0)^{n+1}} ~d\xi} (z-z_0)^n \\ &= \sum \frac{f^{(n)} (z_0)}{n!} (z-z_0)^n .\end{align*}\]
Counterexample to keep in mind:
\[\begin{align*} f(x) = \begin{cases} x^2 & x > 0 \\ 0 & x\leq 0 \end{cases} .\end{align*}\]
In the case of \({\mathbb{R}}\), smooth and analytic are very different categories of functions.
Let \(f: \Omega \to {\mathbb{C}}\) be holomorphic, so \(f\in C^1(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu)\). Then for any \(z\in \Omega\),
\[\begin{align*} f(z) = \frac{1}{2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{\xi - z} ~d\xi .\end{align*}\]
In general, \[\begin{align*} f^{(n)}(z) = \frac{n!}{2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{\qty{\xi - z}^{n+1}} ~d\xi .\end{align*}\]
This implies that \(f\) is analytic, i.e. \[\begin{align*} f(z) = \sum a_n (z-z_0)^n \quad\text{where}\quad a_n = \frac{f^{(n)}(z_0) }{n!} .\end{align*}\] Thus \(f\) is holomorphic iff \(f\) is analytic,
and
\[\begin{align*} \int_{{{\partial}}\Omega} f = 0 \implies \int_{{{\partial}}\Omega_\gamma} \frac{f(\xi)}{\xi - z}~d\xi = 0 .\end{align*}\]
where \(\Omega_r = \Omega\setminus D_r(z)\), and \({{\partial}}\Omega_r = {{\partial}}\Omega \cup(-{{\partial}}D_r)\).
We can thus shrink integrals:
\[\begin{align*} \int_{{{\partial}}\Omega} \frac{f(\xi)} {\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)} {\xi - z} ~d\xi .\end{align*}\]
Write \[\begin{align*}\gamma_s(t) = \gamma(s, t): [a, b] \times[0, 1] \to \Omega.\end{align*}\]
We have \(\gamma_s(0) = \gamma_s(1)\) so \({\frac{\partial \gamma}{\partial s}\,}(s, 0) = {\frac{\partial \gamma}{\partial s}\,}(s, 1)\). Then
\[\begin{align*} {\frac{\partial \gamma}{\partial s}\,} &= \int_0^1 \qty{ f'(r(s, t)) {\frac{\partial r}{\partial s}\,} {\frac{\partial r}{\partial t}\,} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial s \partial t} } ~dt \\ &= \int_0^1 \qty{ f'(r(s, t)) {\frac{\partial r}{\partial s}\,} {\frac{\partial r}{\partial t}\,} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial \mathbf{t} \partial \mathbf{s}} } ~dt \\ &= \int_0^1 {\frac{\partial }{\partial t}\,} \qty{ f(\gamma(s, t)) \gamma_s } \\ &= f(\gamma(s, 1))\gamma_s(s, 1) - f(\gamma(s, 0)) \gamma_s(s, 0) \\ &= 0 .\end{align*}\]
where we can just take the paths \(\gamma(s, t) = z_0 \in \Omega\) for all \(s, t\).
Given a compact set \(K \subset \Omega\), pick an \(O\) with smooth boundary such that \(K \subset O \subset \mkern 1.5mu\overline{\mkern-1.5muO\mkern-1.5mu}\mkern 1.5mu \subset \Omega\). We have
\[\begin{align*} f_v(z) &= \frac{1}{2\pi i} \int_{{{\partial}}O} \frac{f_v(\xi)}{\xi - z}~d\xi \\ f_v^{(n)}(z) &= \frac{n!}{2\pi i} \int_{{{\partial}}O} \frac{f_v(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\ .\end{align*}\]
Then on \({{\partial}}O\), we have uniform convergence
\[\begin{align*} \frac{f_v(\xi)}{\qty{\xi - z}^{n+1}} \overset{u}\to \frac{f(\xi)}{\qty{\xi - z}^{n+1}} .\end{align*}\]
By moving the limits inside, we obtain
\[\begin{align*} f(z) &= \frac{1}{2\pi i} \int_{{{\partial}}O} \frac{f(\xi)}{\xi - z}~d\xi \\ f^{(n)}(z) &= \frac{n!}{2\pi i} \int_{{{\partial}}O} \frac{f(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\ .\end{align*}\]
Given \(z_0\in \Omega\), pick the largest disc \(D_R(z_0) \subset \Omega\) and let \(C_R = {{\partial}}D_R\). Using the integral formula, defining \({\left\lVert {f} \right\rVert}_{C_R} = \max_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert}\)
\[\begin{align*} {\left\lvert { f^{(n)}(z_0) } \right\rvert} \leq \frac{n!}{2\pi} \int_0^{2\pi} \frac{{\left\lVert {f} \right\rVert}_{C_R}}{R^{n+1}} R~d\theta = \frac{n! {\left\lVert {f} \right\rVert}_{C_R}}{R^{n}} .\end{align*}\]
Remark: A general proof technique is when proving something for \(f(z)\), consider \(\frac{1}{f(z)}\) and \(f(\frac 1 z)\).
Suppose \(p\) is nonconstant and does not have a root, \(\frac 1 p\) is entire. Assume that \(a_n \neq 0\), then
\[\begin{align*} \frac{p(z)}{z^n} = a_n \qty{ \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n} } \coloneqq a_n + y \end{align*}\]
We can note that \(\lim_{z\to \infty} \frac{a_{n-k}}{z^k} \to 0\), so there exists an \(R>0\) such that
\[\begin{align*} {\left\lvert { \frac{p(z)}{z^n} } \right\rvert} \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} \quad \text{ for } {\left\lvert {z} \right\rvert} > R \\ \implies {\left\lvert {p(z)} \right\rvert} \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} {\left\lvert {z} \right\rvert}^n \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} R^n .\end{align*}\]
Since \(p(z)\) is continuous and has no root in the disc \({\left\lvert {z} \right\rvert} \leq R\), \({\left\lvert {p(z)} \right\rvert}\) is bounded from below in this disc. Since \(p(z)\) is continuous on a compact set, it attains a minimum, and so \({\left\lvert {p(z)} \right\rvert} \geq \min_{{\left\lvert {z} \right\rvert} \leq R} {\left\lvert {p(z)} \right\rvert} = c_2 \neq 0\). Then \({\left\lvert {p(z)} \right\rvert} \geq A = \min(C_2, \frac 1 2 {\left\lvert {a_n} \right\rvert}R^n)\), so \(\frac{1}{p}\) is bounded. Then \(f\) is constant, a contradiction.
Recall that if \(f\) is holomorphic, we have Cauchy’s integral formula.
By induction on the degree of \(P\). From the first corollary, \(P\) has a root \(w_1\), so write \(z = z-w_1 + w_1\). Then
\[\begin{align*} p(z) &= p(z - w_1 + w_1) \\ &= \sum_k^n a_k(z -w_1 + w_1)^k \\ &= \sum_k^n a_k \sum_{j}^k {k\choose j} w_1{k-j} (z-w_1)^j \\ &= \sum_k^n \sum_j^k a_k {k\choose j} w_1^{k-j} (z-w_1)^j \\ &= \sum_j^n \qty{\sum_{k\geq j} a_k {k\choose j}}(z-w_1)^j \\ &= b_0 + b_1(z-w_1) + \cdots + b_n(z-w_1)^n .\end{align*}\]
Since \(P(w_1) = 0\), we must have \(b_0 = 0\), and thus this equals
\[\begin{align*} b_1(z-w_1) + \cdots + b_n(z-w_1)^n &= (z-w_1) \qty{ b_1 + \cdots + b_n (z-w_1)^{n-1} } \\ &\coloneqq(z-w_1) \phi(z) ,\end{align*}\]
where \(\phi(z)\) is degree \(n-1\), which has \(n-1\) roots by induction.
For a sequence \(\left\{{z_n}\right\}\), TFAE
WLOG by restricting to a subsequence, suppose that \(\left\{{w_k}\right\} \in \Omega\) with \(f(w_i) = 0\) for all \(i\) and \(z_0\) is a limit point of \(\left\{{w_i}\right\}\). Let \(U = \left\{{ z\in \Omega {~\mathrel{\Big|}~}f(z) = 0 }\right\}\). Then
Since \(U\) is closed and open, \(U = \Omega\).
We will first show that \(f(z) \equiv 0\) in a disk containing \(z_0\). Choose a disc \(D\) containing \(z_0\) and contained in \(\Omega\). Since \(f\) is holomorphic on \(D\), we can write \[\begin{align*}f(z) = \sum a_nn (z-z_0)^n.\end{align*}\] Since \(f(z_0) = 0\), we have \(a_0 = 0\).
Suppose \(f\not\equiv 0\). Then there exists a smallest \(n\in {\mathbb{Z}}^+\) such that \(a_n \neq 0\), so \(f(z) = a_n(z-z_0)^n + \cdots\). Since \(a_n \neq 0\), we can factor this as \(a_n(z-z_0)^n \qty{ 1 + g(z-z_0) }\) where \[\begin{align*}g(z-z_0) = \sum_{k=n+1}^\infty \frac{a_k}{a_n} (z-z_0)^{k-n}.\end{align*}\] Note that \(g\) is holomorphic, and \(g(z_0 - z_0) = 0\).
Choose some \(w_k\) such that \(f(w_k) = 0\) and \({\left\lvert {g(w_k - z_0)} \right\rvert} \leq \frac 1 2\) by continuity of \(g\). Then \[\begin{align*}{\left\lvert {1 + g(w_k - z_0)} \right\rvert} > 1 - \frac 1 2 = \frac 1 2.\end{align*}\]
So \[\begin{align*}{\left\lvert {f(w_k)} \right\rvert} = {\left\lvert { a_n(w_k - z_0)^n \qty{1 + g(w_k - z_0) } } \right\rvert} > {\left\lvert {a_n} \right\rvert} {\left\lvert {w_k - z_0} \right\rvert}^n \frac 1 2 > 0,\end{align*}\] a contradiction. So \(U\) is open, closed, and nonempty, so \(U = \Omega\).
Let \(z_0\) be a point in \(\Omega\) and \(C_\gamma\) the boundary of \(D_r(z_0)\). Then
\[\begin{align*} f(z_0) &= \frac{1}{2\pi i} \int_{C_\gamma} f(z)/(z-z_0) dz \\ &= \frac {1}{2\pi i} \int_0^{2\pi} f(z_0 + re^{i\theta})/re^{i\theta} r i e^{i\theta} ~d\theta \quad\text{by } z = z_0 + re^{i\theta} \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})~d\theta \\ &= \frac{1}{2\pi r} \int_0^{2\pi} f(z_0 + re^{i\theta})~rd\theta \\ &= \frac{1}{{\left\lvert {C_\gamma} \right\rvert}} \int_0^{2\pi} f(z) ~ds ,\end{align*}\]
which is the average value of \(f\) on the circle.
Note that there is another formula that averages over the disc (see book for derivation?)
\[\begin{align*} f(z_0) &= \frac{1}{D_s(z_0)} \int_{P_s} \int_{D_s} f(z) ~dA .\end{align*}\]
These imply the maximum modulus principle, since the average can not be the max or min unless \(f\) is constant. Note that \({\left\lvert {f(z)} \right\rvert}\) is continuous!
Next time: maximum modulus principle.
Let \(f: \Omega \to {\mathbb{C}}\) be holomorphic where \(\Omega\) is open and connected. Then by Cauchy’s integral formula, we have \(f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta\) for any \(z_0 \in \Omega\).
We can consider \(D_r(z_0)\), in which case we have for all \(0 < s < r\),
\[\begin{align*} f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + se^{i\theta}) ~d\theta \\ \implies s\cdot f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} s\cdot f(z_0 + se^{i\theta}) ~d\theta \\ \implies \cdot f(z_0) \int_0^r s ~ds &= \frac{1}{2\pi} \int_0^{2\pi} \int_0^r f(z_0 + se^{i\theta})\cdot s ~ds~d\theta \\ \implies \frac 1 2 r^2 f(z_0) &= \frac{1}{2\pi} \iint_{D_r(z_0)} f(z) ~dA\\ \implies f(z_0) &= \frac{1}{\pi r^2} \iint_{D_r(z_0)} f(z) ~dA \\ \implies f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta .\end{align*}\]
Let \(f\) be holomorphic on \(\Omega\) be open and connected, and suppose that there is a \(z_0 \in \Omega\) such that \[\begin{align*}{\left\lvert {f(z_0)} \right\rvert} = \sup_{z\in \Omega} {\left\lvert {f(z)} \right\rvert},\end{align*}\] i.e. \(z_0\) is a maximal point of \(f\). Then \(f\) is constant on \(\Omega\).
If \(\Omega\) is additionally bounded, then \(f\) is continuous on \(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu\), then \[\begin{align*}\sup_{z \in \mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu} {\left\lvert {f(z)} \right\rvert} = \max_{z\in\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu} {\left\lvert {f(z)} \right\rvert}.\end{align*}\]
Since \({\left\lvert {f} \right\rvert}\) is continuous and \(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu\) is compact, \({\left\lvert {f} \right\rvert}\) attains a maximum at some point in \(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu\). We want to show that if \({\left\lvert {f(z_0)} \right\rvert} = \sup_{z\in \Omega} {\left\lvert {f(z)} \right\rvert}\), then \(f\) is constant.
Assume that there exists a \(z_0 \in \Omega\) such that \(f(z) = f(z_0)\). Let \(O = \left\{{ \xi\in \Omega {~\mathrel{\Big|}~}f(\xi) = f(z_0) }\right\}\).
Suppose \(\xi_0 \in O\), then there exists a disc \(D_\rho(\xi_0) \subset \Omega\) such that \[\begin{align*}f(\xi_0) = \frac{1}{\pi\rho^2} \int_{D_\rho(\xi_0)} f(z) dA.\end{align*}\] Then (claim) \({\left\lvert {f(\xi_0)} \right\rvert} \geq {\left\lvert {f(z)} \right\rvert}\) for all \(z\in D_\rho(\xi_0)\), which forces \(f(z) = f(\xi_0)\) for all \(z\in D_\rho(\xi_0)\).
Suppose that \(\sup_{a\in \Omega} {\left\lvert {f(z)} \right\rvert} = {\left\lvert {f(\xi_0)} \right\rvert}\) and write \(f(\xi_0) = Be^{i\alpha}\) for \(B>0\) and \(\alpha \in {\mathbb{R}}\). Then define \(g(z) = f(z) e^{-i\alpha}\); then \(g(\xi_0) = B\) is real, and thus
\[\begin{align*} 0 = g(\xi_0) - B = \frac{1}{\pi\rho^2} \iint_{D_\rho(\xi_0)} \Re( g(z) - B ) ~dA .\end{align*}\]
Note that \(\Re(g(z) - B) \leq 0\) implies that \(\Re(g(z) - B) \equiv 0\) on \(D_\rho(z_0)\), so we can write \(g(z) = B + iI(z)\) for some real-valued function \(I\).
But then \({\left\lvert {g(z)} \right\rvert}^2 = B^2 + I(z)^2 = B^2\) by the previous statement, and so \(I(z) = 0\), forcing \(g(z) = B\) and thus \(f(z) = Be^{i\alpha}\). This shows that \(O\) is open, and thus \(O = \Omega\).
Suppose \(f\) is holomorphic on \(D_1(0)\) and \({\left\lvert {f(z)} \right\rvert} \leq 1\) for all \({\left\lvert {z} \right\rvert} < 1\) with \(f(0) = 0\). Then \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) for all \({\left\lvert {z} \right\rvert} < 1\).
Moreover, there is a point \(z_0\in D_1(0)\) such that \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\) iff \(f(z) = c(z)\) for some \(c \in S^1\).
Define
\[\begin{align*} g(z) = \begin{cases} \frac{f(z)}{z} & z\neq 0 \\ f'(0) & z = 0 \end{cases} .\end{align*}\]
Then \(g\) is holomorphic on \(D_1(0)\) and \({\left\lvert {g(z)} \right\rvert} \leq \frac{1}{\rho}\) for all \({\left\lvert {z} \right\rvert} < \rho < 1\). Now apply the maximum principle: since this is true for all \(\rho < 1\), consider the limit \(\rho\to 1^-\).
Then \({\left\lvert {g(z)} \right\rvert} \leq 1\), so \({\left\lvert {\frac{f(z)}{z}} \right\rvert} \leq 1\) and \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\). If \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\) for any point, then \({\left\lvert {g(z_0)} \right\rvert} = 1\) implies \(g(z_0) = c\) and \(c\in S^1\).
Thus \(f(z) = cz\) for some \(c\in S^1\).
Recall that \[\begin{align*}\Phi_a(z) \mathrel{\vcenter{:}}=\frac{z-a}{1-az}.\end{align*}\]
If \(f: D_1(0) \to D_1(0)\) is a biholomorphism, then \[\begin{align*}f(z) = c \Phi_a(z) = e^{i\theta} \Phi_a(z)\end{align*}\] So every such function is a rotated form of \(\Phi_a\).
Let \(\Omega\) be a connected open domain and \(f: \Omega \to {\mathbb{C}}\) holomorphic with \(f\in C^1\). Then \[\begin{align*}\int_\gamma f(z) ~dz = 0\end{align*}\] for every closed curve \(\gamma \subset \Omega\), which implies that \(f^{(k)} (z)\) exists for all \(k\in {\mathbb{N}}\) and \(f\) is smooth/holomorphic.
Suppose \(g: \Omega \to {\mathbb{C}}\) is continuous and \(\int_\gamma g(z)~dz = 0\) whenever \(\gamma = {{\partial}}R\) for some rectangle \(R\subset \Omega\) with sides parallel to the axes:
Then \(g(z)\) is holomorphic in \(\Omega\).
Fix a point \(\alpha = a + ib\) and given \(z = x+iy\), construct a rectangle \(R\) containing \(z\). Then by assumption, \(\int_{{{\partial}}R} g(z) ~dz = 0\). Let \(\gamma_{az}\) be the path given by traversing the bottom edge of \(R\), and \(\sigma_{az}\) by the top path.