Complex arithmetic

1. A circle of radius 1 about \(z=1\).

2. A circle, using that Apollonius circles are characterized as the locus of distances whose ratios to some fixed points \(A, B\) are constant. To actually compute this: \[ \abs{z-1}^2 &= 4\abs{z-2}^2 \\ \implies (x-1)^2 + y^2 - 4\qty{(x-2)^2 + y^2 } &=0 \\ -3x^2 + 14x - 3y^2 - 15 &= 0 && \star\\ \implies x^2 - {14\over 3}x + y^2 + 5 &= 0 \\ \implies (x- {14\over 2\cdot 3})^2 - {14\over 2\cdot 3}^2 + y^2 + 5 &= 0 \\ \implies (x-{14\over 6})^2 + y^2 = \qty{2\over 3}^2 ,\] which is a circle of radius \(2/3\) with center \(\qty{{14\over 6}, 0}\). To avoid the calculation, use \[ Ax^2 + Bxy + Cy + \cdots = 0,\quad A=1, B=0, C=1 \implies \Delta \da B^2 - 4AC < 0 ,\] which is an ellipse, and since \(A=C\) it is in fact a circle.

3. \(S^1\), using that \({1\over z} = {\bar{z} \over z\bar{z}} = {\bar z\over \abs{z}^2}\) and if this equals \(\bar{z}\), then \(\abs{z}^2=1\). Alternatively, \(1 = \bar{z}z = \abs{z}^2\).

4. Vertical line through \(z=3\).

5. Horizontal line through \(z=ia\).

6. Region to the right of the vertical line through \(z=a\).

7. Exterior of a circle: same calculation is (2), replacing \(=0\) with \(<0\). Note that the line marked \(\star\) involves dividing by a negative, so this flips the sign, and we get \(\cdots > \qty{2\over 3}^2\) at the end.

Goal: Prove \(\abs{z + w} \ge \abs{\abs{z} - \abs{w}}\) and explain when equality holds.

<1>1. \(\abs{z} \le \abs{z + w} + \abs{w}\).

<1>2. \(\abs{w} \le \abs{z + w} + \abs{z}\).

<1>3. \(\abs{\abs{z} - \abs{w}} \le \abs{z + w}\).

<1>4. Equality characterization.

<1>5. Q.E.D.

Goal: Prove that all roots of \(p(z) = z^3 + 2z + 4\) lie outside the unit circle.

<1>1. On \(\abs{z} = 1\), the modulus of the first two terms is at most \(3\).

<1>2. On \(\abs{z} = 1\), \(\abs{z^3 + 2z} < \abs{4}\).

<1>3. \(p(z) = z^3 + 2z + 4\) has the same number of zeros in \(\abs{z} < 1\) as the constant \(4\).

<1>4. \(p\) has no zeros in \(\abs{z} < 1\) and none on \(\abs{z} = 1\).

<1>5. Q.E.D.

\[ \abs{w_1 - c^2 w_2}^2 &= (w_1 - c^2 w_2) ( \bar{w_1} - c^2 \bar{w_2} ) \\ &= \abs{w_1}^2 + c^4 \abs{w_2}^2 - 2c^2 \Re(w_1 \bar{w_2}) \\ &= {\color{green} c^2 \abs{w_2}^2 } + c^4 \abs{w_2}^2 - 2c^2 \Re(w_1 \bar{w_2}) \\ &= c^2 \abs{w_2}^2 + {\color{green} c^2 \abs{w_1}^2 } - 2c^2 \Re(w_1 \bar{w_2}) \\ &= c^2 \abs{w_1 - w_2} ,\] where we’ve applied the assumption \(\abs{w_1} = c\abs{w_2}\) twice.

Using part 1: \[ w_1\da z-z_1, w_2 \da z-z_2 \implies \abs{w_1} &= c\abs{w_2} \\ \implies \abs{w_1 - c^2 w_2} &= c \abs{w_1 - w_2} \\ \implies \abs{ z-z_1 - c^2 (z-z_2) } &= \abs{(z-z_1) - (z-z_2)} \\ \implies \abs{(1-c^2) z - z_3} &= \abs{ z_2 - z_1 } \\ \implies \abs{z-z_4} &= r ,\] where the \(z_i\) and \(r\) are all constant, so this is the equation of a circle.

Goal: (a) Show \(\abs{\frac{w - z}{1 - \bar w z}} < 1\) for \(\abs z, \abs w < 1\) (with equality when \(\abs z = 1\) or \(\abs w = 1\)), for \(\bar z w \neq 1\); (b) show \(F(z) = \frac{w - z}{1 - \bar w z}\) maps \(\DD\) to itself, is holomorphic, sends \(0 \mapsto w\), \(w \mapsto 0\), the circle to the circle, and is bijective.

<1>1. Compute \(\abs{w - z}^2 - \abs{1 - \bar w z}^2 = (\abs w^2 - 1)(1 - \abs z^2)\).

<1>2. (a): For \(\abs z, \abs w < 1\), \(\abs{\frac{w - z}{1 - \bar w z}} < 1\).

<1>3. (a): If \(\abs z = 1\) or \(\abs w = 1\), then \(\abs{\frac{w - z}{1 - \bar w z}} = 1\).

The case \(\abs w = 1\) is symmetric: \(\abs{1 - \bar w z} = \abs{w - z\abs w^2} = \abs{w - z}\).

<1>4. (b): \(F\) is holomorphic on \(\DD\).

<1>5. (b): \(F(0) = w\) and \(F(w) = 0\).

<1>6. (b): \(F\) maps \(\DD\) into itself and \(\abs z = 1 \implies \abs{F(z)} = 1\).

<1>7. (b): \(F: \DD \to \DD\) is bijective.

Clearing the inner denominator \(1 - \bar w z\) gives \[ F(F(z)) = \frac{w(1 - \bar w z) - (w - z)}{(1 - \bar w z) - \bar w(w - z)} = \frac{z(1 - \abs w^2)}{1 - \abs w^2} = z \] for \(\abs w < 1\). Thus \(F \circ F = \mathrm{id}_{\DD}\), so \(F\) is a bijection.

<1>8. Q.E.D.

\[ \prod_{1\leq k \leq n-1} \sin\qty{k\pi\over n} &= \prod_{1\leq k \leq n-1} {\omega_n^k - \omega_n^{-k} \over 2i} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \omega_n^{k} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \exp\qty{i\pi k\over n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \exp\qty{ {i\pi\over n} \displaystyle\sum_{1\leq k \leq n-1} k } \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} e^{i\pi n(n-1)\over 2n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2}^{n-1}\qty{1\over i}^{n-1} \qty{ e^{i\pi \over 2} }^{n-1} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{k}} \\ &= 2^{1-n} { \Phi_n(z) \over z-1}\evalfrom_{z=1} \\ &= 2^{1-n} \qty{ \sum_{0\leq k \leq n-1} z^k}\evalfrom_{z=1} \\ &= n2^{1-n} .\]

\[ \prod_{1\leq j\leq n-1} \sin \left(\frac{j \pi}{n}\right) &=\prod_{1\leq j\leq n-1} \frac{1}{2 i} \qty{\zeta_n^{j\over 2} - \zeta_n^{- {j \over 2} }} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \zeta_n^{j\over 2} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \exp\qty{ij\pi \over n} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\sum_{1\leq j\leq n-1} \frac{i j \pi}{n}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\frac{(n-1) i \pi}{2}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1}\left(e^{i \pi / 2}\right)^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} i^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2}\right)^{n-1} \Phi_{n}(1)\\ &=\frac{n}{2^{n-1}} .\]

Translate so that \(z_1=0\), and put \[ a=z_2-z_1,\qquad b=z_3-z_1. \] The displayed equation is translation-invariant and becomes \[ a^2+b^2=ab. \] Since the three points are distinct, \(a\ne0\). Dividing by \(a^2\) and writing \(t=b/a\) gives \[ t^2-t+1=0, \] so \[ t=e^{i\pi/3}\quad\text{or}\quad t=e^{-i\pi/3}. \] Thus \(|b|=|a|\) and the angle between \(a\) and \(b\) is \(\pi/3\). Consequently \[ |a-b|=|a|=|b|, \] so the three vertices form an equilateral triangle.

Conversely, if the triangle is equilateral, after the same translation and division by \(a\ne0\) we have \(b/a=e^{\pm i\pi/3}\), which satisfies \(t^2-t+1=0\). Reversing the algebra gives the displayed identity.

The easy check: \(f\) is differentiable iff \(\delbar_z f = 0\), but \[ \delbar_z \abs{z}^2 = \delbar_z z\bar{z} = z \neq 0 ,\] unless of course \(z=0\).

A more explicit check: check the limits. \[ {f(z) - f(0) \over z-0} = { \abs{z}^2 \over z } = {z\bar z \over z} = \bar{z} \converges{z\to 0}\too 0 ,\] so \(f\) is differentiable at \(w=0\). Now taking \(w = Re^{i\theta} \neq 0\), \[ {f(z) -f(w) \over z-w} = {\abs{z}^2 - \abs{w}^2 \over z - w} = {\qty{\abs{z} + \abs{w} } \qty{\abs{z} - \abs{w}} \over z-w } = {\abs z - \abs w \over z-w}\cdot \qty{\abs{z} + \abs{w}} .\] First let \(z\to w\) along \(\bd \DD_{R'}(0)\) where \(R' \da \abs{w}\), so that the numerator vanishes and the limit is zero. Then let \(z\to w\) along the curve \(\ts{tw\st t\in [0, 1]}\), then \(\abs{z} = t \abs{w}\), so the ratio becomes \[ {\abs z - \abs w \over z-w}\cdot \qty{\abs{z} + \abs{w}} &= {t\abs{w} - \abs w \over tw-w}\cdot \qty{t\abs{w} + \abs{w}} \\ &= {\abs{w}\qty{t-1 } \over w(t-1)} \cdot \abs{w}(t+1) \\ &= { \abs{w}^2(t+1) \over w} \\ &= \bar{w}(t+1) \\ &\converges{t\to 1}\to 2\bar{w} ,\] which is nonzero is \(w\neq 0\).

Write \(f=u+iv\) on the connected domain.

<1>1. If \(|f|\) is constant, then either \(f\equiv0\), or \(f\) is nonzero and its image lies in a circle. A nonconstant holomorphic map is open, but a circle has empty interior. Hence \(f\) is constant.

<1>2. If \(\Re f=u\) is constant, then \(u_x=u_y=0\). By the Cauchy–Riemann equations, \[ v_y=u_x=0,\qquad v_x=-u_y=0, \] so \(v\) is constant and therefore \(f\) is constant.

<1>3. If \(\arg f\equiv\theta_0\), then \(f\) never vanishes and \[ e^{-i\theta_0}f \] is real-valued. It is holomorphic, so by part (2) it is constant. Hence \(f\) is constant.

<1>4. If \(\overline f=u-iv\) is also holomorphic, then applying the Cauchy–Riemann equations to both \(u+iv\) and \(u-iv\) gives \[ u_x=v_y=-v_y, \qquad u_y=-v_x=v_x. \] Thus \(u_x=u_y=v_x=v_y=0\), so \(f\) is constant.

Goal: Prove that if \(z \mapsto f(z)\) is analytic on a domain \(\Omega\), then \(z \mapsto \overline{f(\bar z)}\) is analytic on \(\overline{\Omega} \definedas \theset{\bar z \suchthat z \in \Omega}\).

<1>1. Fix \(z_0 \in \Omega\); near \(z_0\), \(f\) has a convergent power series \(f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n\).

<1>2. Compute \(\overline{f(\bar z)}\) for \(z\) near \(\bar z_0\). <2>1. \(\overline{f(\bar z)} = \sum_{n=0}^{\infty} \bar a_n (z - \bar z_0)^n\).

<2>2. The series \(\sum \bar a_n (z - \bar z_0)^n\) converges for \(\abs{z - \bar z_0} < r\), where \(r\) is the radius of convergence of <1>1.

<1>3. \(g(z) := \overline{f(\bar z)}\) is analytic at \(\bar z_0\).

<1>4. Q.E.D.

No: polynomials are holomorphic and the uniform limit of holomorphic functions is holomorphic. However, \(f(z) \da \bar{z}\) is continuous on \(\bar\DD\) but not holomorphic.

The claim is that there is some \(N\) such that \(\dd{^N f}{z^N} \equiv 0\), making \(f\) a polynomial. If no derivative is identically zero, the following set is countable: \[ \Union_{n\in \NN} \inverseof{\dd{^n f}{z^n}}(0) .\] This is because each derivative is analytic and the zeros of an analytic function are countable. But by hypothesis this set is uncountable.

Let the pole at \(z_0\) have order \(m\). Its principal part can be written \[ Q(z)=\sum_{j=1}^m\frac{c_j}{(z_0-z)^j}, \qquad c_m\ne0. \] After subtracting \(Q\), the function \(g=f-Q\) is holomorphic on a neighborhood of the closed unit disk. Hence there is some \(R>1\) such that \(g\) is holomorphic on \(D_R(0)\).

<1>1. For \(|z|<1\), \[ \frac1{(z_0-z)^j} =\sum_{n=0}^\infty \binom{n+j-1}{j-1}z_0^{-(n+j)}z^n. \] Thus, if \(q_n\) denotes the \(n\)th Taylor coefficient of \(Q\), \[ q_n=\sum_{j=1}^m c_j \binom{n+j-1}{j-1}z_0^{-(n+j)}. \] The term \(j=m\) has highest polynomial degree in \(n\), so \[ q_n= \frac{c_m}{(m-1)!}n^{m-1}z_0^{-(n+m)} \left(1+O(n^{-1})\right). \]

<1>2. Write \(g(z)=\sum b_nz^n\). Cauchy’s estimates on any circle of radius \(r\) with \(1<r<R\) give \(b_n=O(r^{-n})\). Hence the Taylor coefficients \(a_n=q_n+b_n\) of \(f\) satisfy \[ a_n= \frac{c_m}{(m-1)!}n^{m-1}z_0^{-(n+m)} \left(1+O(n^{-1})\right), \] because the exponentially decaying \(b_n\) is negligible compared with the nonzero polynomial-size leading term.

Therefore, for all sufficiently large \(n\), \(a_{n+1}\ne0\), and \[ \frac{a_n}{a_{n+1}} =z_0\left(\frac{n}{n+1}\right)^{m-1} \frac{1+O(n^{-1})}{1+O((n+1)^{-1})} \longrightarrow z_0. \]

<1>1. Put \(r_n = \abs{a_{n+1}}/\abs{a_n}\). Then \(r_n>0\) and \(r_n \to L\).

<1>2. For any \(\varepsilon > 0\), put \(\ell=\max\{L-\varepsilon,0\}\). There is \(N\) such that \(\ell \le r_n < L + \varepsilon\) for all \(n \ge N\).

<1>3. For \(n > N\), \[\abs{a_N}\,\ell^{n-N} \le \abs{a_n} < \abs{a_N}(L + \varepsilon)^{n-N}.\]

<1>4. One has \[L - \varepsilon \le \ell \le \liminf_{n\to\infty} \abs{a_n}^{1/n} \le \limsup_{n\to\infty} \abs{a_n}^{1/n} \le L + \varepsilon.\]

<1>5. One has \(\lim_{n\to\infty} \abs{a_n}^{1/n} = \boxed{L}\).

<1>6. Q.E.D.

Suppose \[ f(z)=\sum_{n=0}^\infty a_nz^n \] has radius of convergence \(R>0\), and let \(z_0\) satisfy \(|z_0|<R\). Then the sum \(f\) is holomorphic on \(D(0,R)\). Choose \[ 0<\rho<R-|z_0|. \] By Cauchy’s formula, for \(|z-z_0|<\rho\), \[ f(z)=\sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k. \] Indeed, taking any circle \(|\zeta-z_0|=s\) with \(|z-z_0|<s<R-|z_0|\), expand \[ \frac1{\zeta-z} =\frac1{\zeta-z_0} \sum_{k=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^k \] uniformly on the circle and insert this into Cauchy’s integral formula. Thus \(f\) has a power-series expansion about every point of its original disk of convergence.

Goal: Prove: (a) \(\sum_{n} nz^n\) does not converge at any point of \(S^1\); (b) \(\sum_n z^n/n^2\) converges at every point of \(S^1\); (c) \(\sum_n z^n/n\) converges at every point of \(S^1\) except \(z = 1\).

<1>1. Part (a): for \(\abs z = 1\), the terms do not tend to \(0\).

<1>2. Part (b): \(\sum_n \frac{z^n}{n^2}\) converges absolutely at every point of \(S^1\). <2>1. \(\abs{\frac{z^n}{n^2}} = \frac{1}{n^2}\) for \(\abs z = 1\).

<2>2. \(\sum_n \frac{1}{n^2}\) converges.

<2>3. Hence the series converges absolutely, in particular converges, at every \(z \in S^1\).

<1>3. Part (c) at \(z = 1\): \(\sum_n \frac{1}{n}\) diverges.

<1>4. Part (c) at \(z \in S^1 \setminus \theset{1}\): \(\sum_n \frac{z^n}{n}\) converges. <2>1. The partial sums \(A_N := \sum_{n=1}^{N} z^n\) are bounded.

<2>2. The sequence \(\frac{1}{n}\) decreases to \(0\).

<2>3. Dirichlet’s test applies: \(\sum \frac{z^n}{n}\) converges.

<1>5. Q.E.D.

Parametrize the vertical segment by \[ z=x+it, \qquad 0\le t\le b, \qquad dz=i\,dt. \] Then \[ \int_{\gamma_x}f(z)\,dz =i\int_0^b f(x+it)\,dt. \] Since \(f(x+iy)\to A\) uniformly for \(0\le y\le b\), \[ \sup_{0\le t\le b}|f(x+it)-A|\longrightarrow0. \] Therefore \[ \begin{aligned} \left|\int_{\gamma_x}f(z)\,dz-iAb\right| &=\left|i\int_0^b\bigl(f(x+it)-A\bigr)\,dt\right|\\ &\le b\sup_{0\le t\le b}|f(x+it)-A|\longrightarrow0. \end{aligned} \] Hence \[ \lim_{x\to\infty}\int_{\gamma_x}f(z)\,dz=iAb. \]

First find a principal root: \(z^4=i = e^{i\pi\over 2} \implies z_0 = e^{i\pi \over 8}\). Now all of the roots are \(\ts{z_k = z_0 \zeta_4^k \st k=0,1,2,3}\) where \(\zeta_4=e^{2\pi i \over 4} = e^{i\pi \over 2}\), so

• \(z_0 = e^{i\pi \over 8} e^{0i\pi \over 4} = e^{i\pi \over 8}\) corresponding to \({1\over 8} + {0\over 4}\),

• \(z_1 = e^{i\pi \over 8} e^{1i\pi \over 4} = e^{3i\pi \over 8}\) corresponding to \({1\over 8} + {1\over 4} = {3\over 8}\),

• \(z_2 = e^{i\pi \over 8} e^{2i\pi \over 4} = e^{5i\pi \over 8}\) corresponding to \({1\over 8} + {2\over 4} = {5\over 8}\),

• \(z_3 = e^{i\pi \over 8} e^{3i\pi \over 4} = e^{7i\pi \over 8}\) corresponding to \({1\over 8} + {3\over 4} = {7\over 8}\).

\[ {z+2\over z-1}\cdot{\bar{z-1} \over \bar{z-1}} &= {(z+2)\bar{z-1} \over \abs{z-1}^2} \\ &= {((z-1) + 3)\bar{z-1} \over \abs{z-1}^2} \\ &= {\abs{z-1}^2 + 3\bar{z} -3 \over \abs{z-1}^2} \\ &= {(x-1)^2 + y^2 + 3(x-iy) - 3 \over (x-1)^2 + y^2} \\ &= {(x-1)^2 + 3x - 3 + y^2\over (x-1)^2 + y^2} + i{-3y\over (x-1)^2 + y^2} \\ &= {(x-1)(x+2) + y^2\over (x-1)^2 + y^2} + i{-3y\over (x-1)^2 + y^2} .\]

Write \(x=\cos(\theta), y= \sin(\theta)\), so \(e^{i\theta} = x+iy\). Then \[ \cos(4\theta) + i\sin(4\theta) &= e^{4i\theta } \\ &= (x+iy)^4 \\ &= \sum_{0\leq k \leq 4} {4\choose k} x^k (iy)^{4-k} \\ &= x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4 \\ &= (x^4 - 6x^2y^2 + y^4) + i(4x^3y - 4xy^3) .\] So \[ \cos(4\theta) &= \cos^4(\theta) - 6\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta)\\ \sin(4\theta) &= 4\cos^3(\theta)\sin(\theta) - 4\cos(\theta)\sin^3(\theta) .\]

Write \(z=x+iy\), so \(\Re(z) = x\). Then \[ \abs{e^z} = \abs{e^{x+iy}} = \abs{e^x}\abs{e^{iy}} = e^x ,\] using that \(\abs{e^{iy}}=\abs{\cos y+i\sin y}=1\) and \(e^x>0\) for all \(x\in \RR\).

Write \(w=(z-1)/(z+1)\). Then \[ w\in i\RR &\iff w=-\overline w \\ &\iff \frac{z-1}{z+1}=\frac{1-\bar z}{1+\bar z} \\ &\iff (z-1)(1+\bar z)=(1-\bar z)(1+z) \\ &\iff 2\abs z^2-2=0 \\ &\iff \abs z=1. \]