# Geometry, Complex Arithmetic

## Some Geometry

Let $$z_{k}(k=1, \cdots, n)$$ be complex numbers lying on the same side of a straight line passing through the origin. Show that

\begin{align*} z_{1}+z_{2}+\cdots+z_{n} \neq 0, \quad 1 / z_{1}+1 / z_{2}+\cdots+1 / z_{n} \neq 0 \end{align*}

Hint: Consider a special situation first.

## Images of circles

Let $$f(z)=z+1 / z$$. Describe the images of both the circle $$|z|=r$$ of radius $$r(r \neq 0)$$ and the ray $$\arg z=\theta_{0}$$ under $$f$$ in terms of well known curves.

## Geometric Identities

Prove that $$\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}=2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$$ for any two complex numbers $$z_{1}, z_{2}$$, and explain the geometric meaning of this identity

## Geometric Identities

Use $$n$$-th roots of unity (i.e. solutions of $$z^{n}-1=0$$ ) to show that

\begin{align*} \cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2(n-1) \pi}{n}+=-1 \text { and } \\ \sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\sin \frac{6 \pi}{n}+\cdots \frac{2(n-1) \pi}{n}=0 \end{align*}

Hint: If $$z^{n}+c_{1} z^{n-1}+\cdots+c_{n-1} z+c_{n}=0$$ has roots $$z_{1}, z_{2}, \ldots, z_{n}$$, then \begin{align*} z_{1}+z_{2}+\cdots+z_{n}=-c_{1} \\ z_{1} z_{2} \cdots z_{n}=(-1)^{n} c_{n} \text { (not used) } \end{align*}

## Geometry from equations

Describe each set in the $$z$$-plane in (a) and (b) below, where $$\alpha$$ is a complex number and $$k$$ is a positive number such that $$2|\alpha|<k$$.

• $$|z-\alpha|+|z+\alpha|=k$$;

• $$|z-\alpha|+|z+\alpha| \leq k$$.

## Spring 2020.1, Spring 2020 HW 1.4 #complex/qual/completed

• Prove that if $$c>0$$, \begin{align*} {\left\lvert {w_1} \right\rvert} = c{\left\lvert {w_2} \right\rvert} \implies {\left\lvert {w_1 - c^2 w_2} \right\rvert} = c{\left\lvert {w_1 - w_2} \right\rvert} .\end{align*}

• Prove that if $$c>0$$ and $$c\neq 1$$, with $$z_1\neq z_2$$, then the following equation represents a circle: \begin{align*} {\left\lvert {z-z_1 \over z-z_2} \right\rvert} = c .\end{align*} Find its center and radius.

Hint: use part (a)

\begin{align*} {\left\lvert {w_1 - c^2 w_2} \right\rvert}^2 &= (w_1 - c^2 w_2) ( \mkern 1.5mu\overline{\mkern-1.5muw_1\mkern-1.5mu}\mkern 1.5mu - c^2 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu ) \\ &= {\left\lvert {w_1} \right\rvert}^2 + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= {\color{green} c^2 {\left\lvert {w_2} \right\rvert}^2 } + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_2} \right\rvert}^2 + {\color{green} c^2 {\left\lvert {w_1} \right\rvert}^2 } - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_1 - w_2} \right\rvert} ,\end{align*} where we’ve applied the assumption $${\left\lvert {w_1} \right\rvert} = c{\left\lvert {w_2} \right\rvert}$$ twice.

Using part 1: \begin{align*} w_1\coloneqq z-z_1, w_2 \coloneqq z-z_2 \implies {\left\lvert {w_1} \right\rvert} &= c{\left\lvert {w_2} \right\rvert} \\ \implies {\left\lvert {w_1 - c^2 w_2} \right\rvert} &= c {\left\lvert {w_1 - w_2} \right\rvert} \\ \implies {\left\lvert { z-z_1 - c^2 (z-z_2) } \right\rvert} &= {\left\lvert {(z-z_1) - (z-z_2)} \right\rvert} \\ \implies {\left\lvert {(1-c^2) z - z_3} \right\rvert} &= {\left\lvert { z_2 - z_1 } \right\rvert} \\ \implies {\left\lvert {z-z_4} \right\rvert} &= r ,\end{align*} where the $$z_i$$ and $$r$$ are all constant, so this is the equation of a circle.

## Spring 2020 HW 1.1 #complex/exercise/completed

Geometrically describe the following subsets of $${\mathbb{C}}$$:

• $${\left\lvert {z-1} \right\rvert} = 1$$
• $${\left\lvert {z-1} \right\rvert} = 2{\left\lvert {z-2} \right\rvert}$$
• $$1/z = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$
• $$\Re(z) = 3$$
• $$\Im(z) = a$$ with $$a\in {\mathbb{R}}$$.
• $$\Re(z) > a$$ with $$a\in {\mathbb{R}}$$.
• $${\left\lvert {z-1} \right\rvert} < 2{\left\lvert {z-2} \right\rvert}$$

• A circle of radius 1 about $$z=1$$.

• 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: \begin{align*} {\left\lvert {z-1} \right\rvert}^2 &= 4{\left\lvert {z-2} \right\rvert}^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 ,\end{align*} which is a circle of radius $$2/3$$ with center $$\qty{{14\over 6}, 0}$$. To avoid the calculation, use \begin{align*} Ax^2 + Bxy + Cy + \cdots = 0,\quad A=1, B=0, C=1 \implies \Delta \coloneqq B^2 - 4AC < 0 ,\end{align*} which is an ellipse, and since $$A=C$$ it is in fact a circle.

• $$S^1$$, using that $${1\over z} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\over {\left\lvert {z} \right\rvert}^2}$$ and if this equals $$\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$, then $${\left\lvert {z} \right\rvert}^2=1$$. Alternatively, $$1 = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5muz = {\left\lvert {z} \right\rvert}^2$$.

• Vertical line through $$z=3$$.

• Horizontal line through $$z=ia$$.

• Region to the right of the vertical line through $$z=a$$.

• 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.

## Fixed argument exercise #complex/exercise/completed

Fix $$a,b\in {\mathbb{C}}$$ and $$\theta$$, and describe the locus \begin{align*} \left\{{z{~\mathrel{\Big\vert}~}\operatorname{Arg}\qty{z-a\over z-b} = \theta}\right\} .\end{align*}

The geometry at hand:

By the inscribed angle theorem, this locus is an arc of a circle whose center $$O$$ is the point for which the angle $$aOb$$ is $$2\theta$$:

## Fall 2019.2, Spring 2020 HW 1.11 #complex/qual/completed

Prove that the distinct complex numbers $$z_1, z_2, z_3$$ are the vertices of an equilateral triangle if and only if \begin{align*} z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1} .\end{align*}

$$\implies$$: Write the vertices as $$z_1, z_2, z_3$$ and the sides as

• $$s_1 \coloneqq z_2-z_1$$
• $$s_2 \coloneqq z_3 - z_2$$
• $$s_1 \coloneqq z_1 -z_3$$

Note that $$s_i = \pm \zeta_3 s_{i-1}$$, dividing yields \begin{align*} {s_2 \over s_3} &= {s_1\over s_2} \\ &\iff s_2^2 - s_1 s_3 = 0 \\ &\iff \left(z_{2}-z_{3}\right)^{2}-\left(z_{2}-z_{1}\right)\left(z_{1}-z_{3}\right)=0 \\ &\iff \left(z_{2}^{2}+z_{3}^{2}-2 z_{2} z_{3}\right)-\left(z_{2} z_{1}-z_{2} z_{3}-z_{1}^{2}+z_{1} z_{3}\right)=0 \\ &\iff z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-\left(z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right)=0 .\end{align*}

$$\impliedby$$: We still have $$s_i = \theta_i s_{i-1}$$ for some angles $$\theta_i$$ We have

and \begin{align*} {s_1\over s_2} &= {\theta_1 \over \theta_2} \cdot {s_3\over s_1} \\ {s_2\over s_3} &= {\theta_2 \over \theta_3} \cdot {s_1\over s_2} \\ {s_3\over s_1} &= {\theta_3 \over \theta_1} \cdot {s_2\over s_3} .\end{align*}

Running the above calculation backward yields $$s_2/s_3 = s_1/s_2$$, and by the 2nd equality above, this forces $$\theta_2 = \theta_3$$. Similar arguments show $$\theta_1=\theta_2 = \theta_3$$ which forces $$s_1=s_2 = s_3$$.

# Complex Arithmetic

## Sum of Sines

Use de Moivre’s theorem (i.e. $$\left(e^{i \theta}\right)^{n}==\cos n \theta+i \sin n \theta$$, or $$\left.(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\right)$$ to find the sum

\begin{align*} \sin x+\sin 2 x+\cdots+\sin n x \end{align*}

## Solving Equations

Characterize positive integers $$n$$ such that $$(1+i)^{n}=(1-i)^{n}$$

## Characters

Let $$n$$ be a natural number. Show that

\begin{align*} [1 / 2(-1+\sqrt{3} i)]^{n}+[1 / 2(-1-\sqrt{3} i)]^{n} \end{align*}

is equal to 2 if $$n$$ is a multiple of 3 , and it is equal to $$-1$$ otherwise.

## Spring 2019.3 #complex/qual/stuck

Let $$R>0$$. Suppose $$f$$ is holomorphic on $$\left\{{z{~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} < 3R}\right\}$$. Let \begin{align*} M_{R}:=\sup _{|z| \leq R}|f(z)|, \quad N_{R}:=\sup _{|z| \leq R}\left|f^{\prime}(z)\right| \end{align*}

• Estimate $$M_{R}$$ in terms of $$N_{R}$$ from above.

• Estimate $$N_{R}$$ in terms of $$M_{2 R}$$ from above.

First note that by the maximum modulus principal, it suffices to consider sups on the boundary, i.e. \begin{align*} M_R = \sup_{{\left\lvert {z} \right\rvert} = R}{\left\lvert {f(z)} \right\rvert}, \qquad N_R = \sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f'(z)} \right\rvert} .\end{align*}

The first estimate: stuck!

The second estimate: suppose $$z_0 \in {\mathbb{D}}_R(0)$$, then any $$D_R(z_0)$$ is contained in $$D_{2R}(0)$$, So for any such $$z_0$$, apply Cauchy’s integral formula centered at $$z_0$$: \begin{align*} f^{(1)}(z_0) &= {1\over 2\pi i }\oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {f(\xi)\over (\xi-z_0)^2 }\,d\xi\\ \implies {\left\lvert { f^{(1)}(z_0)} \right\rvert} &\leq {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {\left\lvert {f(\xi)\over (\xi-z_0)^2 } \right\rvert}\,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi-z_0} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over R^2 } \,d\xi\\ &\leq {1\over 2\pi} R^{-2} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { \sup_{z\in {{\partial}}{\mathbb{D}}_{R}(z_0) } {\left\lvert {f(z)} \right\rvert} } \,d\xi\\ &= {1\over 2\pi} R^{-2} \sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \cdot 2\pi R \\ &= R^{-1}\sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \\ &\leq R^{-1}M_{2R} ,\end{align*} where we’ve used in the last step that $${\mathbb{D}}_R(z_0) \subseteq {\mathbb{D}}_{2R}(0)$$, and sups can only get larger when taken over larger sets. Since this was an arbitrary $$z_0\in {\mathbb{D}}_R(0)$$, this holds for all $$z$$ with $${\left\lvert {z} \right\rvert} \leq R$$. Since taking sups preserves inequalities, we have \begin{align*} {\left\lvert {f'(z_0)} \right\rvert} \leq R^{-1}M_{2R}\, \forall {\left\lvert {z} \right\rvert} \leq R \implies N_R\coloneqq\sup_{{\left\lvert {z} \right\rvert} \leq R}{\left\lvert {f'(z)} \right\rvert} \leq R^{-1}M_{2R} .\end{align*}

## Spring 2021.1 #complex/qual/completed

The question as written on the original qual has several errors. What is below is the correct version of the inequality.

• Let $$z_{1}$$ and $$z_{2}$$ be two complex numbers.
• Show that \begin{align*} \left|1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_{1} z_{2}\right|^{2}-\left|z_{1}-z_{2}\right|^{2}=\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^2\right) \end{align*}

• Show that if $$\left|z_{1}\right|<1$$ and $$\left|z_{2}\right|<1$$, then $$\left|\frac{z_{1}-z_{2}}{1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_{1} z_{2}}\right|<1 .$$

• Assume that $$z_{1} \neq z_{2}$$. Show that $$\left|\frac{z_{1}-z_{2}}{1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_{1} z_{2}}\right|=1$$ if only if $$\left|z_{1}\right|=1$$ or $$\left|z_{2}\right|=1$$.

Part 1: For ease of notation, let $$z=z_1$$ and $$w=z_2$$ We want to show \begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 = \qty{{1 - {\left\lvert {z} \right\rvert}^2 }} \qty{{1 - {\left\lvert {w} \right\rvert}^2}} .\end{align*}

So write \begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 &= (1-z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu)(1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w) - (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= 1 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 + w\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \\ &= 1 - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 \\ &=(1-{\left\lvert {z} \right\rvert}^2)(1-{\left\lvert {w} \right\rvert}^2) .\end{align*}

Part 2 and 3: \begin{align*} {\left\lvert {z-w \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 \leq 1 &\iff 0 \leq {\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 - {\left\lvert {z-w} \right\rvert}^2 \\ &\iff 0 \leq (1 - {\left\lvert {z} \right\rvert}^2 )(1 - {\left\lvert {w} \right\rvert}^2) ,\end{align*} where we’ve used part 1. But this is clearly true when $${\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1$$, so the RHS is positive. Moreover if $${\left\lvert {z} \right\rvert} = {\left\lvert {w} \right\rvert} = 1$$, the RHS is zero, yielding equalities at every step instead.

## Spring 2020 HW 1.5 #complex/qual/completed

• Let $$z, w \in {\mathbb{C}}$$ with $$\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w \neq 1$$. Prove that \begin{align*} {\left\lvert {w-z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} < 1 \quad\text{ if } {\left\lvert {z} \right\rvert}<1,~ {\left\lvert {w} \right\rvert} < 1 \end{align*} with equality when $${\left\lvert {z} \right\rvert} = 1$$ or $${\left\lvert {w} \right\rvert} = 1$$.

• Prove that for a fixed $$w\in {\mathbb{D}}$$, the mapping $$F: z\mapsto {w-z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z}$$ satisfies

• $$F$$ maps $${\mathbb{D}}$$ to itself and is holomorphic.
• $$F(0) = w$$ and $$F(w) = 0$$.
• $${\left\lvert {z} \right\rvert} = 1$$ implies $${\left\lvert {F(z)} \right\rvert} = 1$$.
• $$F$$ is a bijection.

Part 1: See Spring 2021.1 above.

Part 2, holomorphicity: This is clearly meromorphic, as it’s a rational function, and has a singularity only at $$z$$ such that $$\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1$$. This can only happen if $$z, w \in S^1$$: taking the modulus yields \begin{align*} \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1 \implies {\left\lvert {w} \right\rvert}^2{\left\lvert {z} \right\rvert}^2 = 1 ,\end{align*} and moreover since $${\left\lvert {w} \right\rvert}^2 \leq 1$$ and $${\left\lvert {z} \right\rvert}^2\leq 1$$, the only way this product can be one is when $${\left\lvert {w} \right\rvert}^2 = {\left\lvert {z} \right\rvert}^2 = 1$$. This also forces $$z=1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$$.

The claim is that the singularity $$1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$$ is removable. Note that $$1\over w = \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$$ on the circle, so $$1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\mkern-1.5mu}\mkern 1.5mu = 2$$, so \begin{align*} \qty{ z- \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} &= \qty{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z - 1 \over \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}(w-z) \\ &= w(w-z) \\ &\overset{z\to \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}=w }\to 0 .\end{align*}

Part 2, being a bijection: This follows from finding an explicit inverse, using that $$F^2 = \operatorname{id}$$: \begin{align*} F(F(z)) &= \frac{w- \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } }{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } } \\ &= \frac{w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z)-(w-z)}{q-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu(w-z)} \\ &= \frac{w-|w|^{2} z-w+z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-|w|^{2}+\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \frac{z\left(1-|w|^{2}\right)}{1-|w|^{2}} \\ &= z .\end{align*}

Part 2, being an involution: A direct check shows that $$F(w) = 0$$, since the numerator vanishes, and $$F(0) = {w - 0 \over 1 - 0} = w$$.

Part 3, preserving the circle: Follows from the estimate in part 1.

## Spring 2020 HW 1.2 #complex/qual/completed

Prove the following inequality, and explain when equality holds: \begin{align*} {\left\lvert {z-w} \right\rvert} \geq {\left\lvert { {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} } \right\rvert} .\end{align*}

\begin{align*} {\left\lvert {z-w} \right\rvert}^2 &= (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ &\geq {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2{\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu } \right\rvert}{\left\lvert {z} \right\rvert} \\ &\geq \qty{ {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} }^2 ,\end{align*} and taking square roots introduces an absolute value on the final term. Here we’ve used the basic estimate \begin{align*} \Re(z) \leq {\left\lvert {z} \right\rvert} \implies -\Re(z) \geq -{\left\lvert {z} \right\rvert} .\end{align*}

## Fall 2020.1, Spring 2020 HW 1.6 #complex/qual/completed

Use $$n$$th roots of unity to show that \begin{align*} 2^{n-1} \sin\qty{\pi \over n} \sin\qty{2\pi \over n} \cdots \sin\qty{(n-1)\pi \over n} = n .\end{align*}

Hint: \begin{align*} 1 - \cos(2\theta) &= 2\sin^2(\theta) \\ 2 \sin(2\theta) &= 2\sin(\theta) \cos(\theta) .\end{align*}

• $$\zeta_n \coloneqq e^{2\pi i \over n}$$
• $$\Phi_n(z) \coloneqq\prod_{1\leq j \leq n-1}(z-\zeta_n^j)$$
• $$\Phi_n(1) = n$$, since $$\Phi_n(z) = {z^n-1\over z-1} = \sum_{0\leq j\leq n-1} z^j$$.
• $$\sin(z) = \qty{e^{iz} - e^{-iz} \over 2i}$$.
• $$\prod_k \exp(c_k) = \exp\qty{\sum_k c_k}$$.

\begin{align*} \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}\Big|_{z=1} \\ &= 2^{1-n} \qty{ \sum_{0\leq k \leq n-1} z^k}\Big|_{z=1} \\ &= n2^{1-n} .\end{align*}

\begin{align*} \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}} .\end{align*}

## Spring 2020 HW 1.5 #complex/qual/completed

• Let $$z, w \in {\mathbb{C}}$$ with $$\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w \neq 1$$. Prove that \begin{align*} {\left\lvert {w-z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} < 1 \quad\text{ if } {\left\lvert {z} \right\rvert}<1,~ {\left\lvert {w} \right\rvert} < 1 \end{align*} with equality when $${\left\lvert {z} \right\rvert} = 1$$ or $${\left\lvert {w} \right\rvert} = 1$$.

• Prove that for a fixed $$w\in {\mathbb{D}}$$, the mapping $$F: z\mapsto {w-z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z}$$ satisfies

• $$F$$ maps $${\mathbb{D}}$$ to itself and is holomorphic.
• $$F(0) = w$$ and $$F(w) = 0$$.
• $${\left\lvert {z} \right\rvert} = 1$$ implies $${\left\lvert {F(z)} \right\rvert} = 1$$.

\begin{align*} 0 &\leq (1 - {\left\lvert {w} \right\rvert}^2)(1-{\left\lvert {z} \right\rvert}^2) \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ \implies {\left\lvert {w-z} \right\rvert}^2 &\leq {\left\lvert {1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5muz} \right\rvert}^2 .\end{align*} Note that if either $${\left\lvert {w} \right\rvert}^2 = 1$$ or $${\left\lvert {z} \right\rvert}^2 = 1$$ then the first line is an equality, yielding equality in the final line.

• That $$F: {\mathbb{D}}\to {\mathbb{D}}$$: follows from the inequality, since $${\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1$$ for $$z,w\in {\mathbb{D}}$$. Holomorphicity: follows from the fact that rational expressions of holomorphic functions are holomorphic away from where the denominators vanish.

• Then just note that $${\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} \lleq {\left\lvert {w} \right\rvert}{\left\lvert {z} \right\rvert} < 1$$, so $${\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} > 0$$.

• $$F(0) = {w-0 \over 1-0} = w$$

• $$F(w) = {w-w\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu w} = 0$$

• $${\left\lvert {z} \right\rvert} = 1$$ yields equality in part 1.

Other notes: $$F$$ is bijective on $${\mathbb{D}}$$: \begin{align*} F(F(z)) &= {w - \qty{w-z\over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\qty{w-z\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } \\ &= {w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - (w-z) \over (1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu (w-z)} \\ &= {z-{\left\lvert {w} \right\rvert}^2 z \over 1 - {\left\lvert {w} \right\rvert}^2 }\\ &= z .\end{align*}

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