October 2022 – Page 2

October 19, 2022October 19, 2022

202210191436 Solution to 1969-CE-AMATH-II-XX

Calculate the gradient of the curve $x=3y+y^2$ at the point $(0,-3)$ .

Roughwork.

Differentiating $x(y)$ wrt to $y$ and taking reciprocal,

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

the gradient $\nabla y(x)$ at point $(x,y) = (0,-3)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(0,-3)} = \frac{1}{3+2(-3)} = -\frac{1}{3}}$ .

October 19, 2022October 19, 2022

202210191139 Solution to 1973-CE-AMATH-II-XX

Find the area bounded by the curve $y=4x-x^2$ and the straight line $x-y+2=0$ .

Roughwork.

Rewriting,

$\begin{cases} C:y=-x^2+4x \\ L:y=x+2 \\ \end{cases}$

Solving,

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

$\begin{aligned} x=1 \Rightarrow y=(1)+2=3 \\ x=2 \Rightarrow y=(2)+2=4 \\ \end{aligned}$

Now that the points of intersection are $(1,3)$ and $(2,4)$ , plot a graph below.

By either definite integral,

$\textrm{Area} & = \begin{cases} \displaystyle{\int_{1}^{2}\big(y_C(x)-y_L(x)\big)\,\mathrm{d}x }\\ \displaystyle{\int_{3}^{4}\big(x_L(y)-x_C(y)\big)\,\mathrm{d}y }\\ \end{cases}$

The upper integral is easier solvable than the lower one. Exercise.

October 19, 2022October 19, 2022

202210191050 Solution to 1977-CE-AMATH-II-XX

Prove that

$\displaystyle{\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)}$ .

Hence solve the equation

$\cos 4\theta + 4\cos 2\theta +1=0$

for $0^\circ\leqslant\theta\leqslant 360^\circ$ .

Roughwork.

$\begin{aligned} \textrm{RHS} & = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \\ & = \frac{1}{8}\big(\cos 2(2\theta ) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}\big( (2\cos^22\theta -1) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}(2\cos^22\theta +4\cos 2\theta +2) \\ \dots\,\because\enspace & \cos2\theta = 2\cos^2\theta -1\,\dots \\ & = \frac{1}{8}\big( 2(2\cos^2\theta -1)^2+4(2\cos^2\theta -1)+2\big) \\ & = \dots \\ & = \cos^4\theta \\ & = \textrm{LHS} \\ \end{aligned}$

The missing steps are left the reader.

October 18, 2022October 18, 2022

202210181552 Solution to 1971-CE-AMATH-I-XX

Show that the straight line $x+2y+4=0$ touches the curve $y^2=4x$ .

Roughwork.

If there exists some point $(a,b)$ of intersection of the straight line and the curve, we have

$\begin{cases} a+2b+4 = 0 \\ b^2 = 4a \\ \end{cases}$

Substituting $-2b-4$ for $a$ in the second equation, we have

$\begin{aligned} & b^2 = 4(-2b-4) \\ \Leftrightarrow\quad & b^2 + 8b +16 = 0 \\ \Leftrightarrow\quad & (b+4)^2 = 0 \\ \Leftrightarrow\quad & b = -4\quad\textrm{(rep.)} \\ \Rightarrow\quad & a = -2(-4)-4= 4 \\ \end{aligned}$

$\begin{cases} a = 4 \\ b = -4 \\ \end{cases}$

Such point as $(4,-4)$ exists.

$\because$ The intersection of $\begin{cases} L: x+2y+4=0 \\ C: y^2 = 4x \\ \end{cases}$ is non-empty.

$\therefore$ The straight line touches the curve.

October 14, 2022October 20, 2022

202210141629 Solution to 1976-CE-AMATH-II-XX

By using

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x = \int_{a}^{c}f(x)\,\mathrm{d}x + \int_{c}^{b}f(x)\,\mathrm{d}x}$

or otherwise, evaluate

$\displaystyle{\int_{0}^{\pi}\cos^{7}x\,\mathrm{d}x}$ .

Warm-up.

$\begin{aligned} f(x) & =\cos^7x \\ f(-x) & = \cos^7(-x) \\ & = \cos^7(x) \\ & = f(x) \\ \end{aligned}$

$\therefore$ $f(x)$ is an even function such that

$\displaystyle{\int_{-a}^{a}f(x)\,\mathrm{d}x = 2\int_{0}^{a}f(x)\,\mathrm{d}x=2\int_{-a}^{0}f(x)\,\mathrm{d}x}$ .

Assume $f(x+T)=f(x)$ is a periodic function for some $T\in (0,2\pi ]$ .

$\begin{aligned} & x\overset{f} \longrightarrow \cos^7x \\ \sim\enspace & x\overset{g}\longrightarrow \cos x\overset{h}\longrightarrow \cos^7x \\ \end{aligned}$

where $g(x)=\cos x$ , $h(x)=x^7$ , and $f=h\circ g$ .

$g:[0,\pi ]\to [-1,1]$ by $x\mapsto \cos x$ is one-to-one and onto.

$h:[-1,1]\to [-1,1]$ by $x\mapsto x^7$ is injective and surjective as well.

Hence $f$ , the composition $h\circ g$ of bijections $g$ and $h$ , is also bijective. Besides $f:[0,\pi ]\to [-1,1]$ is a continuously differentiable function that $f(0)=1$ , $f(\frac{\pi}{2}) =0$ , and $f(\pi )=-1$ .

How do you evaluate the integral below:

$\displaystyle{F(\theta )=\int_{0}^{\theta}f(x)\,\mathrm{d}x}$

where $F$ is an odd function such that

$\begin{aligned} F(0) & = F(2\pi ) = 0 \\ F(\theta ) & = -F(-\theta ) \\ \end{aligned}$

The curve of function $f$ is flat at point(s) whose slope $f'$ is zero, i.e.,

$\begin{aligned} f'(x) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos^7x) & = 0 \\ -7\cos^6x\sin x & = 0 \\ x & = 0,\frac{\pi}{2}, \pi \\ \end{aligned}$

Roughwork. Show

Integrating by parts,

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}\cos^6x\,\mathrm{d}(\sin x) \\ & = \big[\cos^6x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^6x)\\ & = 0 + \int_{0}^{\pi}6\sin^2x\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6(1-\cos^2x)\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x - \int_{0}^{\pi}6\cos^7x\,\mathrm{d}x \\ \int_{0}^{\pi}7\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x \\ & = \frac{6}{7}\int_{0}^{\pi}\cos^4x\,\mathrm{d}(\sin x) \\ &= \frac{6}{7}\bigg\{ \big[\cos^4x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^4x)\bigg\} \\ &= \frac{6}{7}\bigg\{ 0 + \int_{0}^{\pi}4\sin^2 x\cos^3x\,\mathrm{d}x\bigg\} \\ & = \frac{24}{7}\int_{0}^{\pi} (1-\cos^2x)\cos^3x\,\mathrm{d}x \\ \frac{30}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{24}{7} \int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \end{aligned}$

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\bigg\{ \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \bigg\} \\ & = \frac{24}{35} \int_{0}^{\pi} \cos^2x\,\mathrm{d}(\sin x) \\ & = \frac{24}{35}\bigg\{ \big[ \cos^2x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^2x) \bigg\} \\ & = \frac{24}{35} \bigg\{ 0 + \int_{0}^{\pi}2\sin^2x\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{48}{35} \int_{0}^{\pi}(1-\cos^2x)\cos x\,\mathrm{d}x \\ \bigg(\frac{24}{35}+\frac{48}{35}\bigg) \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{48}{35}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{24}{35}\bigg\{ \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{16}{35}\big[ \sin x\big]\big|_{0}^{\pi} \\ & = 0 \\ \end{aligned}$

Solution.

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{\frac{\pi}{2}}^{\pi}\cos^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{0}^{\frac{\pi}{2}}\cos^7(x-\pi /2)\,\mathrm{d}(x-\pi /2) \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x - \int_{0}^{\frac{\pi}{2}}\sin^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}(\cos^7x-\sin^7x)\,\mathrm{d}x \\ \end{aligned}$

The rest is left as an exercise to the reader.

October 14, 2022November 10, 2022

202210141016 Solution to 1975-CE-AMATH-II-XX

In the Figure below, $OABC$ is a square. $D$ and $E$ are the mid-points of $AB$ and $BC$ respectively. Let the vector $\overrightarrow{OA}$ be represented by $\mathbf{i}$ and $\overrightarrow{OC}$ by $\mathbf{j}$ .

(a) Express $\overrightarrow{OD}$ in terms of $\mathbf{i}$ and $\mathbf{j}$ .
(b) Express $\overrightarrow{OE}$ in terms of $\mathbf{i}$ and $\mathbf{j}$ .
(c) Evaluate $\displaystyle{\frac{\overrightarrow{OD}\cdot\overrightarrow{OE}}{\left|\overrightarrow{OD}\right|\left|\overrightarrow{OE}\right|}}$ , where $\left|\overrightarrow{OD}\right|$ and $\left|\overrightarrow{OE}\right|$ are the magnitudes of $\overrightarrow{OD}$ and $\overrightarrow{OE}$ respectively.
(d) Hence calculate $\angle DOE$ .

Notation.

In what follow vectors will be typographically boldfaced, i.e.,

$\begin{aligned} \mathbf{OA} & := \overrightarrow{OA} \\ \mathbf{OC} & := \overrightarrow{OC} \\ \mathbf{OD} & := \overrightarrow{OD} \\ \mathbf{OE} & := \overrightarrow{OE} \\ \end{aligned}$

etc., in place of an overhead arrow $\overrightarrow{[\cdot ]}$ more than usually handwritten, the direction of which (e.g. $\overrightarrow{AB}$ ) is intended an initial point (e.g. $A$ ) make to a terminal point (e.g. $B$ ).

(a)

$\begin{aligned} \mathbf{OD} & = \mathbf{OA} + \mathbf{AD} \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{AB} \\ \dots\enspace\because\enspace & \mathbf{AB}=\mathbf{OC}\enspace\dots \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{OC} \\ & = \mathbf{i} + \frac{1}{2}\mathbf{j} \\ \end{aligned}$

(b)

$\begin{aligned} \mathbf{OE} & = \mathbf{OC} + \mathbf{CE} \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{CB} \\ \dots\enspace\because\enspace & \mathbf{CB}=\mathbf{OA}\enspace\dots \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{OA} \\ & = \mathbf{j} + \frac{1}{2}\mathbf{i} \\ \end{aligned}$

(c) and (d) are not chosen.

October 13, 2022October 18, 2022

202210131233 Solution to 1969-CE-AMATH-I-XX

Find $y$ in terms of $x$ , if $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=8x-1}$ and $y=-1$ when $x=-1$ .

Roughwork.

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 8x-1 \\ y(x) & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (8x-1)\,\mathrm{d}x \\ & = 4x^2-x+C\qquad\textrm{for some constant }C \\ \because\enspace -1 & = y(-1) \\ -1 & = 4(-1)^2-(-1)+C \\ \therefore\enspace C & = -6 \\ \end{aligned}$

In conclusion, $y(x) = 4x^2-x-6$ .

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 $\{z\in\mathbb{C}:|z|=2,\mathrm{Re}\, z>1\}$ .