The Physics Language – Page 7

December 13, 2022December 13, 2022

202212131509 Solution to 1985-CE-AMATH-II-6

Find the equations of the two tangents drawn from the point $(-1,0)$ to the parabola $y^2=4x$ .

Roughwork.

$\begin{aligned} y^2 & = 4x \\ 2y\,\mathrm{d}y & = 4\,\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{2}{y} \\ \end{aligned}$

Let $(a_1,b_1)$ and $(a_2,b_2)$ be two points at which the two tangents touches the parabola. Then,

$\displaystyle{\frac{2}{b_i} =\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(a_i,b_i)} =\frac{b_i-(0)}{a_i-(-1)}}$

By solving two equations in two unknowns:

$\begin{cases} b_i^2 = 2a_i+2 \\ b_i^2 = 4a_i \\ \end{cases} \Longrightarrow\quad \begin{pmatrix} a_i\\b_i \end{pmatrix}_{i=1,2} = \Bigg\{ \begin{pmatrix} 1\\ 2 \end{pmatrix} , \begin{pmatrix} 1 \\ -2 \end{pmatrix} \Bigg\}$

This problem is not to be attempted.

December 13, 2022December 13, 2022

202212131316 Solution to 1983-CE-AMATH-II-3

Use the substitution

$u=1+3x^2$

to evaluate

$\displaystyle{\int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }x=\frac{\tan\theta}{\sqrt{3}}\textrm{ so that }\dots \\ \dots &\textrm{ }\mathrm{d}x=\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\textrm{ }\dots \\ \dots &\textrm{ }x=0\Leftrightarrow \theta = 0\textrm{ and }x=1\Leftrightarrow \theta =\frac{\pi}{3}\textrm{ }\dots \\ & = \int_{0}^{\frac{\pi}{3}}\bigg(\frac{\tan\theta}{\sqrt{3}}\bigg)^3\sqrt{1+\tan^2\theta}\,\bigg(\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\bigg) \\ & = \frac{1}{9}\int_{0}^{\frac{\pi}{3}}\sec^3\theta\tan^3\theta\,\mathrm{d}\theta \\ \dots & \textrm{ but }\frac{\mathrm{d}(\sec^3\theta )}{\mathrm{d}\theta}=3\sec^2\theta\cdot\sec\theta\tan\theta\textrm{ so }\dots \\ & = \frac{1}{27}\int_{0}^{\frac{\pi}{3}}\tan^2\theta\,\mathrm{d}(\sec^3\theta )\\ \end{aligned}$

Whoops! Read and follow the instructions.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }u=1+3x^2\textrm{ then }\mathrm{d}x=\frac{\mathrm{d}u}{6x}\textrm{ }\dots \\ \dots &\textrm{ then }x=0\Leftrightarrow u=1\textrm{ and }x=1\Leftrightarrow u=4\textrm{ }\dots \\ & = \int_{1}^{4}x^3\sqrt{u}\,\bigg(\frac{\mathrm{d}u}{6x}\bigg) \\ & = \frac{1}{6}\int_{1}^{4}\bigg(\frac{u-1}{3}\bigg)\sqrt{u}\,\mathrm{d}u \\ & = \frac{1}{18}\int_{1}^{4}\big( u^{\frac{3}{2}}-u^{\frac{1}{2}}\big)\,\mathrm{d}u \\ & = \frac{1}{18}\bigg[ \frac{2u^{\frac{5}{2}}}{5}-\frac{2u^{\frac{3}{2}}}{3}\bigg]\bigg|_{1}^{4} \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 13, 2022

202212121714 Solution to 1974-CE-AMATH-II-X

A bridge $ABC$ is in the form of the parabola given by the equation

$8y=16-x^2$

as shown in the figure.

The slope of the bridge at the point $P$ is $0.5$ . Find the coordinate of $P$ .

Roughwork.

$\begin{aligned} \bigg|\frac{\mathrm{d}y}{\mathrm{d}x}\bigg| & = \bigg|-\frac{1}{4}x\bigg| \\ 0.5 & = \frac{1}{4}|x| \\ x & = \pm 2 \\ 8y & = 16-(\pm 2)^2 \\ y & = 1.5 \\ \end{aligned}$

Answer. $P(2,1.5)$ .

December 12, 2022December 21, 2022

202212121040 Solution to 1968-CE-AMATH-II-X

Find the maximum and minimum values of $y$ on the curve

$y^2=x(1-x)^2$ .

Sketch the curve.

Roughwork. (true-negative/false-positive example)

All computations in the enclosed section below involving higher-order derivatives than the first are wrong owing to the mistaken assumption

$\times :\quad\displaystyle{\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2=\frac{(\mathrm{d}y)^2}{(\mathrm{d}x)^2}}$

which was to be detested. Show

$\begin{aligned} y^2 & = x(1-x)^2 \\ & = x(1-2x+x^2) \\ y^2 & = x^3-2x^2+x \\ \end{aligned}$

$\begin{aligned} 2y\,\mathrm{d}y & = 3x^2\,\mathrm{d}x-4x\,\mathrm{d}x+\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{3x^2-4x+1}{2y} \\ 2(\mathrm{d}y)^2 & = 6x(\mathrm{d}x)^2 - 4(\mathrm{d}x)^2 \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 3x-2 \\ \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ 3 & = 2\bigg(\frac{3x^2-4x+1}{2y}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{3y}{3x^2-4x+1} \\ \end{aligned}$

$\begin{aligned} y'(x) & = 0 \\ 3x^2-4x+1 & = 0 \\ (3x-1)(x-1) & = 0 \\ x & = \frac{1}{3},1 \\ \end{aligned}$

$\begin{aligned} y^2 & = \bigg(\frac{1}{3}\bigg)\bigg(1-\bigg(\frac{1}{3}\bigg)\bigg)^2 \\ y & =\pm\frac{2\sqrt{3}}{9}\\ y^2 & = (1)(1-(1))^2 \\ y & = 0 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \Bigg\{ \begin{pmatrix}\frac{1}{3}\\ -\frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}\frac{1}{3}\\ \frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}\Bigg\}$ are extrema.

$\begin{aligned} y''(x) & = 0 \\ 3y & = 0 \\ \pm\sqrt{x(1-x)^2} & = 0 \\ x(1-x)^2 & = 0 \\ x & = 0,1 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}0\\ 0 \end{pmatrix}, \begin{pmatrix}1\\ 0\end{pmatrix}\bigg\}$ are inflexions.

To answer whether

$\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}\bigg|_{\big( \frac{1}{3},\pm\frac{2\sqrt{3}}{9}\big)}$

is positive or negative is critical to yielding the maximum and the minimum values of the function.

The correction is left a $\textrm{\scriptsize{MUST}}$ for the author.

December 12, 2022December 12, 2022

202212121004 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ if

$ax^2+2hxy+by^2+2gx+2fy+c=0$ .

Roughwork.

$\begin{aligned} ax^2+2hxy+by^2+2gx+2fy+c & =0 \\ 2ax\,\mathrm{d}x+2h(x\,\mathrm{d}y+y\,\mathrm{d}x)+2by\,\mathrm{d}y+2g\,\mathrm{d}x+2f\,\mathrm{d}y & = 0 \\ 2ax + 2h\bigg( x\frac{\mathrm{d}y}{\mathrm{d}x}+y\bigg) +2by\frac{\mathrm{d}y}{\mathrm{d}x}+2g+2f\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ (ax+hy+g) + (hx+by+f)\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{ax+hy+g}{hx+by+f} & = y' \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 12, 2022

202212120958 Solution to 1967-CE-AMATH-II-X

A spherical balloon is of diameter $x$ . By what approximate amount does the volume increases, if the diameter is increased by a very small amount $\Delta x$ ?

Roughwork.

$\begin{aligned} V & = \frac{4}{3}\pi x^3 \\ \Delta V & = 4\pi x^2\,\Delta x \\ \end{aligned}$

December 6, 2022December 6, 2022

202212061712 Solution to 1967-CE-AMATH-I-X

If the population of Hong Kong is $4$ million in $1967$ and it increases by $3\%$ each year, what will be the estimated population in $1987$ ? Give your answer correct to $4$ significant figures.

Roughwork.

$\begin{aligned} & \quad\enspace (4\times 10^6)(1+3\%)^{(1987-1967)} \\ & = (4\times 10^6)(1+0.03)^{20} \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(1)^{20-n}(0.03)^n \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(0.03)^n \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

October 28, 2022November 2, 2022

202210280936 Problem 1.24

Expand $f(t)=\sin^2t\cos^3t$ in Fourier series.

_{Extracted from Hwei Piao Hsu. (1984). HBJ College Outline of Applied Fourier Analysis.}

Roughwork.

One can make use of the identities below:

$\begin{aligned} e^{\pm\mathrm{i}n\theta} & = \cos n\theta\pm\mathrm{i}\sin n\theta \\ \cos n\theta & = \frac{e^{\mathrm{i}n\theta}+e^{-\mathrm{i}n\theta}}{2} \\ \sin n\theta & = \frac{e^{\mathrm{i}n\theta}-e^{-\mathrm{i}n\theta}}{2\mathrm{i}} \\ \end{aligned}$

and thus

$\begin{aligned} & \begin{cases} \sin^2t = \displaystyle{\bigg(\frac{e^{\mathrm{i}t}-e^{-\mathrm{i}t}}{2\mathrm{i}}\bigg)^2 }\\ \cos^3t = \displaystyle{ \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^3 }\\ \end{cases} \\ \Longrightarrow & \begin{cases} \sin^2t = \displaystyle{\frac{e^{2\mathrm{i}t}-2e^{\mathrm{i}t}e^{-\mathrm{i}t}+e^{-2\mathrm{i}t}}{-4}} \\ \cos^3t = \displaystyle{\frac{e^{3\mathrm{i}t}+3e^{\mathrm{i}t}+3e^{-\mathrm{i}t}+e^{-3\mathrm{i}t}}{8}} \\ \end{cases} \\ \end{aligned}$

Or one can make use of the identity

$\sin^2\theta + \cos^2\theta = 1$ .

and hence

$\begin{aligned} f(t) & = (1-\cos^2t)\cos^3t \\ & = \cos^3t-\cos^5t \\ \end{aligned}$

where $\cos^5t$ is as expanded

$\begin{aligned} \cos^5t & = \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^5 \\ & = \frac{e^{5\mathrm{i}t}+5e^{3\mathrm{i}t} +10e^{\mathrm{i}t} +10e^{-\mathrm{i}t}+5e^{-3\mathrm{i}t}+ e^{-5\mathrm{i}t}}{32}\\ \end{aligned}$

so that

$\begin{aligned} f(t) & = -\frac{e^{5\mathrm{i}t}}{32}+\bigg(\frac{1}{8}-\frac{5}{32}\bigg)e^{3\mathrm{i}t} + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{\mathrm{i}t} \\ & \qquad + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{-\mathrm{i}t} + \bigg(\frac{1}{8}-\frac{5}{32}\bigg) e^{-3\mathrm{i}t} - \frac{e^{-5\mathrm{i}t}}{32}\\ & = -\frac{e^{5\mathrm{i}t}}{32}-\frac{e^{3\mathrm{i}t}}{32} + \frac{e^{\mathrm{i}t}}{16} + \frac{e^{-\mathrm{i}t}}{16} - \frac{e^{-3\mathrm{i}t}}{32} -\frac{e^{-5\mathrm{i}t}}{32} \\ & = \frac{1}{16}\Bigg( 2\bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg) - \frac{e^{3\mathrm{i}t}+e^{-3\mathrm{i}t}}{2} - \frac{e^{5\mathrm{i}t}+e^{-5\mathrm{i}t}}{2} \Bigg)\\ & = \cdots \\ \end{aligned}$

Answer. $\displaystyle{\frac{1}{16}(2\cos t-\cos 3t-\cos 5t)}$ .

No slightest disrespect to de Moivre’s and Euler’s formulae but in due respect of Fourier’s analysis, this problem had need be treated again.

Background.

A function $f(t)$ satisfying $f(t+T)=f(t)$ is called a periodic function, and the smallest $T$ its period. A periodic function $f(t)$ can be represented by the trigonometric Fourier series

$\displaystyle{f(t)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos n\omega_0t+b_n\sin n\omega_0t)}$

$\displaystyle{f(t)=C_0+\sum_{n=1}^{\infty}C_n\cos (n\omega_0t-\theta_n)}$

where $\omega_0=2\pi /T$ . The coefficients of the Fourier series are found by using the orthogonality properties of sine and cosine functions over a period:

$\begin{aligned} \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\begin{Bmatrix}\cos n\omega_0t \\ \sin n\omega_0t \end{Bmatrix}\,\mathrm{d}t \\ C_0 & = \frac{1}{2}a_0 \\ C_n & = \sqrt{a_n^2+b_n^2} \\ \theta_n & = \tan^{-1}\bigg(\frac{b_n}{a_n}\bigg) \\ \end{aligned}$

_{Text on pg. 11}

For $f(t)=\sin^2t\cos^3t$ ,

$\begin{aligned} T & = 2\pi \\ \omega_0 & = 1 \\ \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\begin{Bmatrix}\cos nt \\ \sin nt\end{Bmatrix}\,\mathrm{d}t\\ \end{aligned}$

When $n=0$ :

$\begin{aligned} a_0 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\,\mathrm{d}t \\ \cdots\enspace\textrm{Let }& u=\sin t\textrm{ s.t. }\mathrm{d}t=\frac{\mathrm{d}u}{\cos t}\enspace\cdots \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2\cos^2t\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2(1-u^2)\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}(u^2-u^4)\,\mathrm{d}u \\ & = \frac{1}{\pi}\bigg[ \frac{u^3}{3}-\frac{u^5}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = \frac{1}{\pi}\bigg[ \frac{\sin^3t}{3}-\frac{\sin^5t}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = 0 \\ \end{aligned}$

When $n=1$ :

$\begin{aligned} a_1 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^4t\,\mathrm{d}t \\ \dots\enspace &\textrm{By WolframAlpha}\enspace\dots \\ & = \frac{1}{\pi}\bigg[\frac{1}{192}(12t+3\sin (2t)-3\sin (4t)-\sin (6t))\bigg]\bigg|_{-\pi}^{\pi} \\ & = \frac{1}{\pi}\bigg(\frac{24\pi}{192}\bigg) \\ & = \frac{1}{8} \\ \end{aligned}$

a bit tiring an exercise.

(to be continued)

October 10, 2022October 24, 2022

202210101009 Exercise 3.9

(a) Prove that the sequence $\{ z_n\}$ converges, and give its limit, when $z_n$ is given by
i. $\displaystyle{\frac{1}{n}\mathrm{i}^n}$ ,
ii. $(1+\mathrm{i})^{-n}$ ,
iii. $\displaystyle{\frac{n^2+\mathrm{i}n}{n^2+\mathrm{i}}}$ .
(b) Prove that the sequence $\{ z_n\}$ does not converge when $z_n$ is given by
i. $\mathrm{i}^n$ ,
ii. $(1+\mathrm{i})^n$ ,
iii. $\displaystyle{(-1)^n\frac{n}{n+\mathrm{i}}}$ .

_{Extracted from H. A. Priestley. (2003). Introduction to Complex Analysis.}

Background.

Definition. (geometric series)
A series of complex number $\displaystyle{\sum_{n=0}^{\infty}z_n}$ is called a geometric series if

$\exists\, t\in\mathbb{C}\textrm{ s.t. }z_{n+1}=tz_n\enspace\forall\, n\in\mathbb{N}$ .

N.b. Such geometric series as $z_0\neq 0$ is divergent whenever $|t|\geqslant 1$ .

Proposition. (ratio test)
Let $\displaystyle{\sum_{n=0}^{\infty}z_n}$ be a series of complex numbers, $z_n\neq 0\enspace\forall\, n\in\mathbb{N}$ . Define $\displaystyle{r_n:=\bigg|\frac{z_{n+1}}{z_n}\bigg|}$ for $n\in\mathbb{N}$ . Assume that $\displaystyle{\lim_{n\to\infty}r_n=r}$ exists, $r\in [0,\infty)\cup \{\infty\}$ . The series is absolutely convergent if $r<1$ , divergent if $r>1$ , and undetermined if $r=1$ .

Proposition. (root test)
Let $\displaystyle{\sum_{n=0}^{\infty}z_n}$ be a series of complex numbers. For $n\geqslant 1$ define $\rho_n =|z_n|^{\frac{1}{n}}$ . Assume that $\displaystyle{\lim_{n\to\infty} \rho_n :=A}$ exists, $A\in [0,\infty )\cup \{\infty\}$ . The series is absolutely convergent if $A<1$ , divergent if $A>1$ , and undetermined if $A=1$ .

(a) i.

From $\displaystyle{z_n=\frac{1}{n}\mathrm{i}^n}$ observe that

$\displaystyle{z_{n+1}=\bigg(\frac{n}{n+1}\mathrm{i}}\bigg)\cdot z_n$

is a geometric series of common ratio $t$ :

$\displaystyle{t:=\frac{n}{n+1}\mathrm{i}}}$ .

$\begin{aligned} |t| & =\bigg| \frac{n}{n+1} \mathrm{i}\bigg| \\ & = \sqrt{(0)^2+\bigg(\frac{n}{n+1}\bigg)^2} \\ & = \frac{n}{n+1} < 1\enspace \forall\, n\in\mathbb{N} \\ \end{aligned}$

$\therefore$ $\{ z_n\}$ converges to the limit $1$ .

(a) ii.

$\begin{aligned} z_n & = (1+\mathrm{i})^{-n} \\ z_{n+1} & = (1+\mathrm{i})^{-(n+1)} \\ & = (1+\mathrm{i})^{-1}(1+\mathrm{i})^{-n} \\ & = \bigg(\frac{1}{1+\mathrm{i}}\bigg) \cdot z_n \\ t & := \frac{1}{1+\mathrm{i}} \\ & = \frac{1}{1+\mathrm{i}} \cdot \frac{1-\mathrm{i}}{1-\mathrm{i}} \\\ & = \frac{1-\mathrm{i}}{1-\mathrm{i}^2} \\ & = \frac{1-\mathrm{i}}{1-(-1)} \\ & = \frac{1}{2} - \frac{1}{2}\mathrm{i} \\ |t| & = \sqrt{\bigg(\frac{1}{2}\bigg)^2 + \bigg( -\frac{1}{2}\bigg)^2} \\ & = \cdots \\ & = \frac{\sqrt{2}}{2}<1 \\ \end{aligned}$

$\therefore$ $\{ z_n\}$ converges to the limit $0$ .

(a) iii.

$\begin{aligned} z_n & = \frac{n^2+\mathrm{i}n}{n^2+\mathrm{i}} \\ z_{n+1} & = \frac{(n+1)^2+\mathrm{i}(n+1)}{(n+1)^2+\mathrm{i}} \\ \frac{z_{n+1}}{z_n} & = \frac{(n+1)^2+\mathrm{i}(n+1)}{(n+1)^2+\mathrm{i}}\bigg/ \frac{n^2+\mathrm{i}n}{n^2+\mathrm{i}} \\ & = \frac{(n+1)^2+\mathrm{i}(n+1)}{(n+1)^2+\mathrm{i}}\cdot \frac{n^2+\mathrm{i}}{n^2+\mathrm{i}n} \\ & =: r_n \\ \end{aligned}$

By ratio test or root test for convergence the series being undetermined, we had rather adopt another approach.

$\begin{aligned} \frac{n^2+\mathrm{i}n}{n^2+\mathrm{i}} & = \frac{(n^2+\mathrm{i}n)(n^2-\mathrm{i})}{(n^2+\mathrm{i})(n^2-\mathrm{i})} \\ z_n & = \frac{(n^4+n)+\mathrm{i}(n^3-n^2)}{n^4+1} \\ \textrm{Re}\, z_n & = \frac{n^4+n}{n^4+1} \\ \textrm{Im}\, z_n & = \frac{n^3-n^2}{n^4+1} \\ \lim_{n\to\infty} \textrm{Re}\, z_n & = \cdots = 1 \\ \lim_{n\to\infty} \textrm{Im}\, z_n & = \cdots = 0 \\ \end{aligned}$

$\therefore$ $\{ z_n\}$ converges to the limit $1$ .

(b) i., ii., and iii. are left the reader as an exercise.

October 5, 2022October 24, 2022

202210051329 Exercise 3.1

(a) Prove that the following are open sets:
i. $\{ z\in\mathbb{C}:|z-1|<|z+\mathrm{i}|\}$ ;
ii. $\mathbb{C}\backslash [0,1]$ .
(b) Prove that the following are not open:
i. $\{ z\in\mathbb{C}:\mathrm{Re}\, z\geqslant 0\}$ ;
ii. $\{ z\in\mathbb{C}:|z|\leqslant 2,\mathrm{Re}\, z>1\}$ .

_{Extracted from H. A. Priestley. (2003). Introduction to Complex Analysis.}

Background.

Definition. (open set)
A set $S\subseteq \mathbb{C}$ is open if, given $z\in S$ , there exists $r>0$ (depending on $z$ ) such that $\mathrm{D}(z;r)\subseteq S$ .

Definition. (open disc)
The open disc centre $a\in\mathbb{C}$ and radius $r>0$ is defined to be

$\mathrm{D}(a;r):=\{ z\in\mathbb{C}: |z-a|<r \}$ .

(a) i.

Warm-up.

$\begin{aligned} |z-1| & = |(x+\mathrm{i}y)-1| \\ & = \sqrt{(x-1)^2+(y)^2} \\ |z+\mathrm{i}| & = |(x+\mathrm{i}y)+\mathrm{i}| \\ & = \sqrt{(x)^2+(y+1)^2} \\ \end{aligned}$

No, just let $r=|z+\mathrm{i}|$ is done.

(a) ii.

Set-up.

In set notation a complex interval number may be represented in the form

$\mathcal{A}=[a,b]+[c,d]\mathrm{i}=\{ x+\mathrm{i}y\, |\, a\leqslant x\leqslant b,c\leqslant y \leqslant d\}$ .

_{Extracted from R. Boche. (1966). Complex Interval Arithmetic with Some Applications.}

Thus $\mathbb{C}\backslash [0,1]$ is equivalent to

$\{z=a+\mathrm{i}b\textrm{ where }a,b\notin [0,1]\}$ ,

as shown in the figure below:

Abortive attempt.

Lemma.

Let $S\subset X$ be a non-empty subset and $U\subset S$ . Then $U$ is open in $S$ if and only if $U=V\cap S$ for some $V\subset X$ which is open in $X$ .

Proof. Necessity. $\forall\, z\in U,\,\exists\, r>0$ s.t. $\mathrm{D}_S(z;r)\subset U$ . Let $\displaystyle{V=\bigcup_{z\in U}\mathrm{D}_X(z;r)}$ . Then $U=V\cap S$ . Note that $V$ is open in $X$ . Sufficiency. $\forall\, z\in U\subset V,\,\exists\, r>0$ s.t. $\mathrm{D}_X(z;r)\subset V$ . Thus $\mathrm{D}_S(z;r)=\mathrm{D}_X(z;r)\cap S\subset V\cap S\subset U$ . $\blacksquare$

Now that $U=\mathbb{C}\backslash [0,1]$ , $S=\mathbb{C}$ , and $X=\mathbb{C}$ … quit this circular reasoning of no use.

Make use of the following

Lemma.

The union of any collection of open sets is open.

Proof.

Let $\displaystyle{S=\bigcup_{\lambda\in I}S_\lambda}$ where $S_\lambda$ is open in $\mathbb{C}$ and $I$ any index set. Then $\forall\, z\in S$ , $z\in S_\lambda$ for some $\lambda$ and so $\exists\, r>0$ s.t. $\mathrm{D}(z;r)\subset S_\lambda\subset S$ . $\blacksquare$

Thereby

$\mathbb{C}\backslash [0,1]=\mathbb{C}(-\infty ,0)\cup\mathbb{C}(1,\infty)$

is open.

QED

(b)

A set $S$ is not open whenever $S^0$ (/ $\textrm{int}\, S$ ), the interior of $S$ , falls short of its whole, i.e., $S\supset S^0\neq S$

For i. and ii. the unbelonging boundaries $\partial S$ ‘s to either are $\{z\in\mathbb{C}:\textrm{Re }z=0\}$ and $\{z\in\mathbb{C}:|z|=2,\mathrm{Re}\, z>1\}$ .