Month: August 2020

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202008190508 Solution to 1973-AL-AMATH-I-1

In calm weather an aeroplane has a speed v and a flying range (out-and-back) R. In a north wind of speed n (n<v) and in a line of flight that makes a constant angle \phi with the north, I wish to know its new flying range.

Assuming that the maximum time of flight is independent of wind conditions, and that the effect of turning round at the end of the outgoing journey can be ignored.

Let \mathbf{v}=v_x\, \hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}} denote the velocity of the aeroplane in calm weather (i.e., zero wind speed).

Then \mathbf{v}=(v_x,\, v_y) = (v\sin\phi , \, v\cos\phi ), where v=|\mathbf{v}| is its speed, and \phi measured anticlockwise from the north.

In a north wind of speed n, the outgoing speed v_{\textrm{out}} of the plane will decrease by (v\sin\phi ,\, v\cos\phi -n), and on its return the speed v_{\textrm{back}} will increase by (v\sin\phi ,\, v\cos\phi +n).

The new range R' is calculated in two parts:

R'_{\textrm{total}}=R'_{\textrm{out}}+R'_{\textrm{back}}.

\because \qquad R'_{\textrm{out}}=R'_{\textrm{back}}

\therefore\qquad R'_{\textrm{out}}=R'_{\textrm{back}}=\displaystyle{\frac{R'}{2}}

If the plane is on schedule, the time of flight t=\displaystyle{\frac{R}{v}} should be kept unchanged.

I.e.,

\begin{aligned} t & = t' \\ t & = t'_{\textrm{out}} + t'_{\textrm{back}} \\ \frac{R}{v} & = \frac{R'_{\textrm{out}}}{v'_{\textrm{out}}} + \frac{R'_{\textrm{back}}}{v'_{\textrm{back}}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi -n)^2}} + \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi +n)^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'}{2}\Bigg\{ \frac{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}\Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}\Bigg\} \\ \end{aligned}

Let a=v^2+n^2 and b=2nv\cos\phi.

\begin{aligned} R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}}{\sqrt{a+b}-\sqrt{a-b}} \Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}(\sqrt{a+b}+\sqrt{a-b})}{(\sqrt{a+b})^2-(\sqrt{a-b})^2} \Bigg\} \\ R' & = \frac{R}{v}\bigg[\frac{(\sqrt{a-b})(a+b)+(\sqrt{a+b})(a-b)}{b}\bigg] \\ R' & = \frac{R}{v}\bigg[ \sqrt{a-b} \bigg( \frac{a}{b} \bigg) +\sqrt{a-b} + \sqrt{a+b} \bigg( \frac{a}{b} \bigg) - \sqrt{a+b} \bigg] \\ \end{aligned}

c^2=a^2+b^2-2ab\cos\theta (The Law of Cosines)

I need a break. To be continued.

After dinner, things are much clearer. Go back to the earlier line:

\displaystyle{\frac{R}{v} = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}}}.

Now invoke its physical meaning:

i. Let \mathbf{n}= -n\,\hat{\mathbf{j}} (n>0) be the north wind velocity vector pointing in the southerly direction.

ii. Let \mathbf{v}= -v_x\,\hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}} (v_x,\, v_y>0) be the velocity vector of the airplane pointing in the northwesterly direction.

iii. Let \mathbf{s}=\mathbf{n}+\mathbf{v} be the side subtending the angle \phi enclosed by the two vector arrows \mathbf{n} and \mathbf{v}. That is, the bottom of vector \mathbf{s} is touching the arrowhead of \mathbf{v}, and the arrowhead of \mathbf{s} is touching the tip of \mathbf{n}.

iv. The sides \mathbf{n}, \mathbf{v}, and \mathbf{s} form a triangle of perimeter given by n+v+s.

v. s^2=|\mathbf{s}|^2= v^2+n^2-2nv\cos\phi.

vi. 2nv\cos\phi =2\mathbf{n}\cdot\mathbf{v}.

(continue)

\begin{aligned} \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \dots & = \frac{R'}{\sqrt{4(v^2 -2nv\cos\phi +n^2)}} + \frac{R'}{\sqrt{4(v^2 +2nv\cos\phi +n^2)}} \\ \dots & = \frac{R'}{\sqrt{4(s^2)}} + \frac{R'}{\sqrt{4(s^2 + 4\mathbf{n}\cdot\mathbf{v})}} \\ \dots & = \bigg( \frac{\sqrt{4(s^2+4\mathbf{n}\cdot \mathbf{v})}+\sqrt{4(s^2)}}{\sqrt{4(s^2)(s^2+4\mathbf{n}\cdot\mathbf{v})}} \bigg) R' \\ \dots & = \bigg( \frac{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s}{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}} \bigg) R' \\ R' & = \bigg( \frac{R}{v} \bigg) \bigg( \frac{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}}{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s} \bigg) \\ \end{aligned}

Roughwork.

\begin{aligned} s^2 & = v^2 + n^2 -2nv\cos\phi \\ s^4 & = (v^2 + n^2 -2nv\cos\phi )^2 \\ \dots & = v^4+n^4+2n^2v^2-4nv^3\cos\phi -4n^3v\cos\phi + 4n^2v^2\cos^2\phi \end{aligned}

Simplifying first the numerator,

\begin{aligned} & \sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}} \\ & = \sqrt{(v^4+n^4+2n^2v^2-4nv^3\cos\phi -4n^3v\cos\phi + 4n^2v^2\cos^2\phi )+4(v^2 + n^2 -2nv\cos\phi )(nv\cos\phi )} \\ & = \sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2} \\ \end{aligned}

then the denominator

\begin{aligned} & \sqrt{s^2+4\mathbf{n}\cdot\mathbf{v}} + s \\ & = \sqrt{v^2+n^2-2nv\cos\phi+4nv\cos\phi } + \sqrt{v^2 + n^2 -2nv\cos\phi } \\ & = \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} \\ \end{aligned}

Putting them together,

\begin{aligned} R' & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{ \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} } \bigg) \\ & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2-n^2)^2+(2nv)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )}} \bigg) \\ \end{aligned}

Unsatisfactory still.

Trimming the numerator again,

\begin{aligned} & \sqrt{(v^2-n^2)^2\bigg( 1 + \frac{(2nv)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}\bigg)} \\ & = (v^2-n^2)\sqrt{\frac{(v^2+n^2)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}} \\ \end{aligned}

and then dividing the denominator by \sqrt{\frac{(v^2+n^2)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}}, or, multiplying the denominator by a factor of

\displaystyle{\frac{v^2-n^2}{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}}

i.e.,

(\sqrt{a+b}+\sqrt{a-b})\frac{v^2-n^2}{\sqrt{a^2-b^2}}.

Note.

\begin{aligned} a - b & = s^2 \\ a + b & = s^2 -4\mathbf{n}\cdot\mathbf{v} \\ \end{aligned}

(to be continued)

Ans.

\displaystyle{\frac{R(v^2-n^2)}{v\sqrt{v^2-n^2\sin^2\phi}}}

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202008190019 Exercise 1.28

In each case, determine all real x and y which satisfy the given relation.

(a) x+\mathrm{i}y=|x-\mathrm{i}y|,

(b) x+\mathrm{i}y=(x-\mathrm{i}y)^2,

(c) \displaystyle{\sum_{k=0}^{100}}\mathrm{i}^k=x+\mathrm{i}y.

Solution.

(a)

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

\therefore y=0 and x\enspace (\geqslant 0)\, \in\mathbb{R} is any non-negative real number.

(b)

\begin{aligned} x + \mathrm{i}y & = (x-\mathrm{i}y)^2 \\ x + \mathrm{i}y & = x^2 - 2xy\mathrm{i} + (\mathrm{i}y)^2 \\ x + \mathrm{i}y & = (x^2-y^2) + \mathrm{i}(-2xy) \end{aligned}

Comparing the imaginary parts:

\begin{aligned} y=-2xy & \Rightarrow (2x+1)y = 0 \\ & \Rightarrow x = -\frac{1}{2}\qquad \textrm{or}\qquad y=0 \end{aligned}

Comparing the real parts:

\begin{aligned} x & = x^2-y^2 \\ x^2 - x - y^2 & = 0 \\ x & = \frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-y^2)}}{2(1)} \\ & = \frac{1\pm \sqrt{1+4y^2}}{2} \end{aligned}

When y=0, x=\displaystyle{\frac{1\pm \sqrt{1+4(0)^2}}{2}}=0,\, 1.

When x=-\frac{1}{2},

\begin{aligned} -\frac{1}{2} & =\frac{1\pm\sqrt{1+4y^2}}{2} \\ \Rightarrow 2 & = \sqrt{1+4y^2} \\ \Rightarrow y & = \pm \displaystyle{\frac{\sqrt{3}}{2}} \end{aligned}

\therefore \begin{pmatrix} x \\ y \end{pmatrix}\in \Bigg\{ \begin{pmatrix} 0 \\ 0 \end{pmatrix},\,\begin{pmatrix} 1 \\ 0 \end{pmatrix},\, \begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix},\, \begin{pmatrix} -\frac{1}{2} \\ -\frac{\sqrt{3}}{2} \end{pmatrix} \Bigg\}.

(c)

\begin{aligned} \sum_{k=0}^{k=100}\mathrm{i}^k & = x+\mathrm{i}y \\ \mathrm{i}^{0} + \mathrm{i}^{1} + \mathrm{i}^{2} + \dots + \mathrm{i}^{100} & = x+\mathrm{i}y \\ \sum_{n=0}^{50} \mathrm{i}^{(2n)} + \sum_{n=0}^{49} \mathrm{i}^{(2n+1)} & = x+\mathrm{i}y \\ \sum_{n=0}^{50} (-1)^n + \mathrm{i}\sum_{n=0}^{49} (-1)^{n} & = x+\mathrm{i}y \\ \end{aligned}

\begin{aligned} \therefore \qquad x & = \sum_{n=0}^{50}(-1)^n \\ & = (-1)^{0} + (-1)^{1} + (-1)^{2} +\dots + (-1)^{50} \\ & = 1-1+1+\dots +1 \\ & = 1 \end{aligned}

\begin{aligned} \therefore \qquad y & = \sum_{n=0}^{49}(-1)^n \\ & = (-1)^{0} + (-1)^{1} + (-1)^{2} +\dots + (-1)^{49} \\ & = 1-1+1+\dots -1 \\ & = 0 \end{aligned}

Sol. \bigg\{\begin{aligned} & x = 1 \\ & y = 0 \end{aligned}\bigg\}.

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202008181129 Homework 1 (Q1)

Solve for z\in \mathbb{C} in the following equations:

(a) z^4+z^3+z^2+z+1=0,

(b) 3z^3+29z^2+497z-169=0.

 Attempts.

(a) Take notice that z\neq 1. Try and see having both sides of the equation multiplied by (1-z),

\begin{aligned} 0 & = (1-z)(z^4+z^3+z^2+z+1) \\ 0 & = 1-z^5 \\ z^5 & = 1 = 1(1+0\,\mathrm{i}) \\ z^5 & = e^{\mathrm{i}(2n\pi)}\qquad \qquad \textrm{where }n=0,1,2,3,4. \\ \end{aligned}

\therefore z=e^{\mathrm{i}(\frac{2n\pi}{5})}.

(n=0 is rejected for z\neq 1=e^{\mathrm{i}(\frac{2(0)\pi}{5})}.)

In polar expression z=e^{\mathrm{i\frac{2\pi}{5}}},\, e^{\mathrm{i\frac{4\pi}{5}}},\, e^{\mathrm{i\frac{6\pi}{5}}},\, e^{\mathrm{i\frac{8\pi}{5}}}.

In trigonometric expression

z=\mathrm{cis}(\frac{2n\pi}{5})=\mathrm{cis}(\frac{2\pi}{5}),\, \mathrm{cis}(\frac{4\pi}{5}),\, \mathrm{cis}(\frac{6\pi}{5}),\, \mathrm{cis}(\frac{8\pi}{5})

I.e.,

\begin{aligned} z^1 & = \cos 72^\circ +\mathrm{i}\sin 72^\circ = \mathrm{cis\,} 72^\circ \\ z^2 & = \cos 144^\circ +\mathrm{i}\sin 144^\circ = \mathrm{cis\,} 144^\circ \\ z^3 & = \cos 216^\circ +\mathrm{i}\sin 216^\circ = \mathrm{cis\,} 216^\circ \\ z^4 & = \cos 288^\circ +\mathrm{i}\sin 288^\circ = \mathrm{cis\,} 288^\circ \\ \end{aligned}

(b)Let f(z)=3z^3+29z^2+497z-169=0. Then (3z-1) is a factor, because \frac{1}{3} is a zero (i.e., f(\frac{1}{3}) = \frac{1}{9}+\frac{29}{9}+\frac{497}{3}-169=0).

\begin{aligned} f(z) & =(3z-1)(z^2+10z+169) \\ & = (3z-1)g(z) \\ \end{aligned}

When g(z)=0, z=\displaystyle{\frac{-10\pm \sqrt{100-4(169)}}{2}}=-5\pm 12\,\mathrm{i}.

The solution to f(z)=0 gives z_1=\frac{1}{3}, z_2=-5+12\,\mathrm{i}, and z_3=-5-12\,\mathrm{i}.