The Physics Language – Page 5

February 20, 2023February 20, 2023

202302200931 Review Problem 6.5.7

Let $z\in\mathbb{C}$ . Recall that $z=x+iy$ for some $x,y\in\mathbb{R}$ , and we can form the complex conjugate of $z$ by taking $\overline{z}=x-iy$ . The function $c:\mathbb{R}^2\to\mathbb{R}^2$ which sends $(x,y)\mapsto (x,-y)$ agrees with complex conjugation.

(a) Show that $c$ is a linear map over $\mathbb{R}$ (i.e., scalars in $\mathbb{R}$ ).
(b) Show that $\overline{z}$ is not linear over $\mathbb{C}$ .

_{Extracted from D. Cherney, et al. (2013). Linear Algebra.}

Roughwork.

A function $L:V\to W$ is linear if $V$ and $W$ are vector spaces and

$L(ru+sv)=rL(u)+sL(v)$

for all $u,v\in V$ and $r,s\in\mathbb{R}$ .

(a)

$\begin{aligned} c(az_1+bz_2) & = c(a(x_1+iy_1)+b(x_2+iy_2)) \\ & = c((ax_1+bx_2)+i(ay_1+by_2)) \\ & = (ax_1+bx_2)+i(-(ay_1+by_2)) \\ & = (ax_1+bx_2)-i(ay_1+by_2) \\ & = a(x_1-iy_1) + b(x_2-iy_2) \\ & = a(x_1+i(-y_1)) + b(x_2+i(-y_2)) \\ & = a(c(x_1+iy_1)) + b(c(x_2+iy_2)) \\ & = ac(z_1) + bc(z_2) \\ \end{aligned}$

(b)

Not to be attempted.

February 14, 2023February 14, 2023

202302140910 Exercises 2.1

What do the following equations represent geometrically? Give sketches.

i. $|z+2|=6$ ,
ii. $|z-3\,\mathrm{i}|=|z+\mathrm{i}|$ ,
iii. $|\mathrm{i}z-1|=|\mathrm{i}z+1|$ ,
iv. $|z-\omega |=|z-1|$ ( $\omega=e^{2\pi\,\mathrm{i}/3}$ ).

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

Roughwork.

i.

$\begin{aligned} |z+2|&=6 \\ |(x+\mathrm{i}y)+2| & = 6 \\ |(x+2)+\mathrm{i}y| & = 6 \\ \sqrt{(x+2)^2+y^2} & = 6 \\ (x+2)^2 + y^2 & = 6^2 \\ \end{aligned}$

i.e., a circle centred at $(-2,0)$ with $6$ units in radius.

ii.

$\begin{aligned} |z-3\,\mathrm{i}| & = |z+\mathrm{i}| \\ |(x+\mathrm{i}y)-3\,\mathrm{i}| & = |(x+\mathrm{i}y)+\mathrm{i}| \\ |x+\mathrm{i}(y-3)| & = |x+\mathrm{i}(y+1)| \\ \sqrt{x^2+(y-3)^2} & = \sqrt{x^2+(y+1)^2} \\ x^2+(y-3)^2 & = x^2 + (y+1)^2 \\ y^2-6y+9 & = y^2+2y+1 \\ y & = 1 \\ \end{aligned}$

i.e., a horizontal line with $y$ -intercept $1$ unit.

iii.

$\begin{aligned} |\mathrm{i}z-1| & = |\mathrm{i}z+1| \\ |\mathrm{i}(x+\mathrm{i}y)-1| & = |\mathrm{i}(x+\mathrm{i}y)+1| \\ |(-y+\mathrm{i}x)-1| & = |(-y+\mathrm{i}x+1|\\ |(-y-1)+\mathrm{i}x| & = |(-y+1)+\mathrm{i}x| \\ \sqrt{(-y-1)^2+x^2} & = \sqrt{(-y+1)^2+x^2} \\ y^2+2y+1 & = y^2-2y+1 \\ y & = 0 \\ \end{aligned}$

i.e., the $x$ -axis.

iv.

$\begin{aligned} |z-\omega | & = |z-1| \\ |(x+\mathrm{i}y)-\mathrm{cis}(2\pi/3)|& = |(x+\mathrm{i}y)-1| \\ \bigg|\bigg(x+\frac{1}{2}\bigg)+\mathrm{i}\bigg(y-\frac{\sqrt{3}}{2}\bigg)\bigg| & = |(x-1)+\mathrm{i}y| \\ \end{aligned}$

Not to be completed.

January 17, 2023May 4, 2023

202301170910 Exercise 3.2

Solve these simultaneous equations:

1. $\begin{cases} x+y=5 \\ xy=6 \\ \end{cases}$

Roughwork. Read More

Substituting $x=5-y$ for $x$ :

$\begin{aligned} (5-y)y & = 6 \\ y^2-5y+6 & = 0 \\ (y-2)(y-3) & = 0 \\ y & = 2,3 \\ \begin{pmatrix}x \\ y \end{pmatrix} = \bigg\{ \begin{pmatrix} 2 \\ 3 \end{pmatrix},\begin{pmatrix} 3 \\ 2 \end{pmatrix} \bigg\} & \\ \end{aligned}$

2. $\begin{cases} x-2y=1\\ x^2+y^2=29 \\ \end{cases}$

Roughwork. Read More

Substituting $x=1+2y$ for $x$ :

$\begin{aligned} (1+2y)^2 + y^2 & =29 \\ (1+4y+4y^2)+y^2 & = 29 \\ 5y^2 +4y -28 & = 0 \\ (5y+14)(y-2) & = 0 \\ y & = -\frac{14}{5},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}5\\2\end{pmatrix}, \begin{pmatrix}-23/5\\-14/5\end{pmatrix} \bigg\} & \\ \end{aligned}$

3. $\begin{cases} 2x+y=5 \\ x^2-xy=12 \\ \end{cases}$

Roughwork. Read More

Substituting $y=5-2x$ for $y$ :

$\begin{aligned} x^2-x(5-2x) & = 12 \\ 3x^2 -5x -12 & = 0 \\ (3x+4)(x-3) & = 0 \\ x & = -\frac{4}{3},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-4/3\\7/3\end{pmatrix},\begin{pmatrix}3\\-1\end{pmatrix}\bigg\} &\\ \end{aligned}$

4. $\begin{cases} 3x-y=7 \\ x^2-xy+y^2=7\\ \end{cases}$

Roughwork. Read More

Substituting $y=3x-7$ for $y$ :

$\begin{aligned} x^2-x(3x-7)+(3x-7)^2 & = 7 \\ 7x^2 -35x +42 & = 0 \\ (7x+7)(x-6) & = 0 \\ x & = -1,6 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-1\\-10\end{pmatrix} , \begin{pmatrix}6\\11\end{pmatrix}\bigg\} & \\ \end{aligned}$

5. $\begin{cases} 5x+y=9 \\ 3xy+y^2=-5 \\ \end{cases}$

Roughwork. Read More

Substituting $y=9-5x$ for $y$ :

$\begin{aligned} 3x(9-5x)+(9-5x)^2 & = -5 \\ 10x^2-63x +86 & = 0 \\ (10x-43)(x-2) & = 0 \\ x & = \frac{43}{10},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}43/10\\-25/2\end{pmatrix} & , \begin{pmatrix}2\\-1\end{pmatrix}\bigg\} \\ \end{aligned}$

6. $\begin{cases} 3x+2y=13 \\ 3x^2+y^2=31 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y = \frac{13-3x}{2}}$ for $y$ :

$\begin{aligned} 3x^2 + \bigg(\frac{13-3x}{2}\bigg)^2 = 31 \\ 7x^2 - 26x+15 & = 0 \\ (7x-5)(x-3) & = 0 \\ x & = \frac{7}{5},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}7/5\\22/5\end{pmatrix},\begin{pmatrix}3\\-13\end{pmatrix} \bigg\} & \\ \end{aligned}$

7. $\begin{cases} 2x-3y+11=0 \\ 2x^2-xy=36 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x+11}{3}}$ for $y$ :

$\begin{aligned} 2x^2-x\bigg( \frac{2x+11}{3}\bigg) & =36 \\ 4x^2-11x-108 &= 0 \\ (4x-27)(x+4) & = 0 \\ x & = -4,\frac{27}{4} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}-4\\1\end{pmatrix},\begin{pmatrix}27/4\\49/6\end{pmatrix}\bigg\} \\ \end{aligned}$

8. $\begin{cases} 2x-5y=1\\ x^2-xy+3y^2=9 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x-1}{5}}$ for $y$ :

$\begin{aligned} x^2-x\bigg(\frac{2x-1}{5}\bigg)+3\bigg(\frac{2x-1}{5}\bigg)^2 & = 9 \\ 27x^2 -7x -222 & = 0 \\ (27x+74)(x-3) & = 0 \\ x & = -\frac{74}{27},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-74/27\\-35/27\end{pmatrix} , \begin{pmatrix}3\\1\end{pmatrix} \bigg\} & \\ \end{aligned}$

9. $\begin{cases} 2x+3y=7 \\ y^2=26-x^2 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{7-3y}{2}}$ for $x$ :

$\begin{aligned} y^2 & = 26-\bigg(\frac{7-3y}{2}\bigg)^2 \\ 0 & = 13y^2 -42y -55 \\ 0 & = (13y-55)(y+1) \\ y & = -1,\frac{55}{13} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}5\\-1\end{pmatrix} , \begin{pmatrix}74/26\\55/13\end{pmatrix}\bigg\} \\ \end{aligned}$

10. $\begin{cases} 5x+3y+7=0 \\ 3y^2=x^2-4y+3 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{-7-3y}{5}}$ for $x$ :

$\begin{aligned} 3y^2 & = \bigg(\frac{-7-3y}{5}\bigg)^2-4y+3 \\ 0 & = 33y^2+29y-62 \\ 0 & = (33y+62)(y-1) \\ y & = 1,-\frac{62}{33}\\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{\begin{pmatrix} -2 \\1\end{pmatrix},\begin{pmatrix} -3/11\\-62/33\end{pmatrix}\bigg\} \\ \end{aligned}$

_{Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.}

This problem is not to be attempted.

January 16, 2023January 16, 2023

202301160933 Exercise 3.4.B (Q29)

A ball is thrown vertically upward from the ground. Its displacement at time $t$ is given by $20t-5t^2$ . Find the maximum distance of the ball from the ground and the time this occurs.

_{Extracted from Y. F. So & M. S. Wong. (1988). New Way Additional Mathematics.}

Roughwork.

By differentiation wrt time

$\begin{aligned} s(t) & = 20t-5t^2 \\ v(t) & = 20 - 10t \\ a(t) & = -10 \\ \end{aligned}$

Its velocity is zero at some time $t'$ , i.e.,

$\begin{aligned} 0=v(t') & =20-10t' \\ t' & = 2\,\mathrm{s} \\ \end{aligned}$

once it reaches the maximum height

$\begin{aligned} s(t') & = 20(2)-5(2)^2 \\ & = 20\,\mathrm{m} \\ \end{aligned}$

This problem is not to be attempted.

December 31, 2022January 7, 2023

202212311431 Problem 1.4

Let $g:\Omega\to\mathbb{R}$ be such that $\displaystyle{\frac{\partial g}{\partial x}}$ and $\displaystyle{\frac{\partial g}{\partial y}}$ exist at $(x_0,y_0)\in\Omega$ , and suppose that one of these partials exists in a neighbourhood of $(x_0,y_0)$ and is continuous at $(x_0,y_0)$ . Show that $g$ is real-differentiable at $(x_0,y_0)$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Granted that

$\begin{aligned} w=f(z)&=u(x,y)+iv(x,y) \\ \textrm{s.t. }A & =u_0+iv_0 \\ z_0 & =x_0+iy_0 \\ \end{aligned}$

the concepts of existence of limit

$\begin{aligned} &\qquad\enspace \lim_{z\to z_0}f(z) = A \\ & \Longleftrightarrow \begin{cases} \lim_{(x,y)\to (x_0,y_0)}u(x,y)=u_0 \\ \lim_{(x,y)\to (x_0,y_0)}v(x,y)=v_0 \\ \end{cases} \textrm{exist} \end{aligned}$

and continuity at a point

$\begin{aligned} &\qquad\enspace w=f(z)\textrm{ continuous at }z_0 \\ &\Longleftrightarrow \textrm{ both }u(x,y)\textrm{ and }v(x,y)\textrm{ continuous at }z_0 \\ \end{aligned}$

are held by definition.

This problem is not to be attempted.

December 31, 2022December 31, 2022

202212311359 Problem 1.3

Let $z_1$ and $z_2$ be nonzero complex numbers, and let $\theta$ ( $0\leqslant \theta\leqslant \pi$ ) be the angle between them. Show that

(a) $\mathrm{Re}z_1\overline{z}_2=|z_1||z_2|\cos\theta$ , $\mathrm{Im}z_1\overline{z}_2=\pm |z_1||z_2|\sin\theta$ , and consequently
(b) The area of the triangle formed by $z_1$ , $z_2$ , and $z_2-z_1$ is $|\mathrm{Im}z_1\overline{z}_2|/2$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

The area of a triangle $\triangle ABC$ , constructed by any two sides $\mathbf{AB}$ and $\mathbf{AC}$ with an included angle $\theta$ , is

$\begin{aligned} \textrm{Area} & = \frac{1}{2}\mathbf{AB}\times \mathbf{AC} \\ \bigg( & =\frac{1}{2}(AB)(AC)\sin\theta\bigg) \\ \end{aligned}$

i.e., half the area of the parallelogram spanned by these two vectors.

Hence, $\mathtt{(a)} \Longrightarrow \mathtt{(b)}$ .

This problem is not to be attempted.

December 31, 2022January 1, 2023

202212311329 Problem 1.2

Show that $|z_1+z_2|=|z_1|+|z_2|$ iff $z_1$ and $z_2$ lie on a common ray from $0$ iff one of $z_1$ or $z_2$ is a nonnegative multiple of the other.

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Let $z_1=r_1e^{i\theta_{1}}$ and $z_2=r_2e^{i\theta_{2}}$ , rewrite

$\begin{aligned} |r_1e^{i\theta_{1}}+r_2e^{i\theta_{2}}| & = |r_1e^{i\theta_{1}}|+|r_2e^{i\theta_{2}}| \\ ? = r_3 & = r_1 + r_2 \\ \end{aligned}$

by Lemma.

$\begin{aligned} r_3e^{i\theta_{3}} & = r_1e^{i\theta_{1}}+r_2e^{i\theta_{2}} \\ & = r_1(\cos\theta_1+i\sin\theta_1) + r_2(\cos\theta_2+i\sin\theta_2) \\ & = (r_1\cos\theta_1+r_2\cos\theta_2) + i(r_1\sin\theta_1+r_2\sin\theta_2) \\ r_3^2 & = r_1^2+r_2^2+2r_1r_2\cos (\theta_1-\theta_2) \\ r_3 & = \sqrt{r_1^2+r_2^2+2r_1r_2\cos (\theta_1-\theta_2)} \\ \end{aligned}$

we have

$\begin{aligned} \cos (\theta_1-\theta_2) & = 1 \\ \theta_1 - \theta_2 & = 0 \\ \theta_1 & = \theta_2 \\ \end{aligned}$

The rest is left an exercise for the reader.

December 30, 2022December 31, 2022

202212301726 Solution to 1987-CE-AMATH-I-3

For any complex number $z$ , let $\overline{z}$ , $|z|$ , and $\mathrm{Re}(z)$ be its conjugate, modulus, and real part respectively. Show that

$z+\overline{z}=2\mathrm{Re}(z)$ and $|z|\geqslant \mathrm{Re}(z)$ .

Hence, or otherwise, show that for any complex numbers $z_1$ and $z_2$ ,

$z_1z_2+\overline{z_1z_2}\leqslant 2|z_1||z_2|$ .

Roughwork.

In the field $\mathbb{C}=\{ x+iy:x,y\in\mathbb{R}\}$ of complex numbers, $x$ is called the real part of $z$ and $y$ the imaginary part, i.e.,

$z=x+iy=\mathrm{Re}(z)+i\,\mathrm{Im}(z)$ ,

its trigonometric form being $z=r(\cos\theta +i\sin\theta )$ with $r$ the modulus and $\theta$ the argument, and its exponential form, $z=re^{i\theta}$ .

This problem is not to be attempted.

December 28, 2022January 6, 2023

202212281153 Solution to 1999-CE-AMATH-II-2

Find $\displaystyle{\int x(x+2)^{99}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\enspace \int x(x+2)^{99}\,\mathrm{d}x \\ \dots\, &\textrm{let }u=x+2\textrm{ s.t. }\mathrm{d}u=\mathrm{d}x\,\dots \\ & = \int (u-2)(u)^{99}\,\mathrm{d}u \\ & = \int \big( u^{100}-2u^{99}\big)\,\mathrm{d}u \\ & = \frac{u^{101}}{101} - \frac{u^{100}}{50} + C\textrm{ for some constant} \\ & = \frac{(x+2)^{101}}{101} - \frac{(x+2)^{100}}{50} + C \\ \end{aligned}$

This problem is not to be attempted.

December 28, 2022December 28, 2022

202212281117 Problem 1.18

Show that an equation for the circle $C(z_0,r)$ is

$z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0=r^2$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Writing

$\begin{aligned} |z-z_0| & = r \\ |(a+bi)-(a_0+b_0i)| & = r \\ |(a-a_0)+(b-b_0)i| & = r \\ (a-a_0)^2 + (b-b_0)^2 & = r^2 \\ \end{aligned}$

and

$\begin{aligned} &\quad\enspace z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0 \\ & = (a+bi)(a-bi) - (a_0-b_0i)(a+bi) \\ & \qquad\qquad -(a_0+b_0i)(a-bi)+(a_0+b_0i)(a_0-b_0i) \\ & = \cdots \\ \end{aligned}$

is to let ends meet, without use of properties of conjugates:

$\begin{aligned} |\overline{z}| & = |z| \\ \mathrm{arg}\overline{z} & = -\mathrm{arg}z \\ \overline{z_1+z_2} & = \overline{z}_1+\overline{z}_2 \\ \overline{z_1-z_2} & = \overline{z}_1-\overline{z}_2 \\ \overline{z_1z_2} & = \overline{z}_1\overline{z}_2 \\ \mathrm{Re}z & = (z+\overline{z})/2 \\ \mathrm{Im}z & = (z-\overline{z})/2i \\ z\overline{z} & = |z|^2 \\ \end{aligned}$

(to be continued)