\documentclass{article} %\usepackage{teaching} \title{Mathematical Methods: Tutorial sheet 6} \date{21 November 2007} \author{Peter Harrison} \renewcommand{\baselinestretch}{1} % {1,1.25,1.667} = % single/1.5/double line spacing \newcommand{\answerblock}[1]{{\bf Solution:}\nopagebreak[1]#1} \usepackage{mathchars} \newcommand{\nat}{\bbbn} \newcommand{\rat}{\bbbq} \newcommand{\real}{\bbbr} \newcommand{\integers}{\bbbz} \newcommand{\complex}{\ensuremath{\bbbc}} %\renewcommand{\rat}{Q \!\!\!\! O} %\renewcommand{\integers}{Z \!\!\! Z} \renewcommand{\complex}{\cal C} %\renewcommand{\answerblock}[1]{} \newcommand{\bcomment}[1]{} \newcommand{\mtx}[4]{\left(\begin{array}{rr}#1\\#3\end{array}\right)} \newcommand{\cmtx}[4]{\left(\begin{array}{cc}#1\\#3\end{array}\right)} \newcommand{\Bmtx}[9]{\left(\begin{array}{rrr}#1\\#4\\#7 \end{array}\right)} \newcommand{\Bcmtx}[9]{\left(\begin{array}{ccc}#1\\#4\\#7 \end{array}\right)} \newcommand{\vcr}[2]{\left(\begin{array}{r}#1\\#2\end{array}\right)} \newcommand{\cvcr}[2]{\left(\begin{array}{c}#1\\#2\end{array}\right)} \newcommand{\Bvcr}[3]{\left(\begin{array}{r}#1\\#2\\#3\end{array}\right)} \newcommand{\Bcvcr}[3]{\left(\begin{array}{c}#1\\#2\\#3\end{array}\right)} \newcommand{\zeromtx}{\left(\begin{array}{rr}0&0\\0&0\end{array}\right)} \newcommand{\identitymtx}{\left(\begin{array}{rr}1&0\\0&1\end{array}\right)} \newcommand{\determinant}[1]{\left| #1 \right|} \newcommand{\vi}{\vec{i}} \newcommand{\vj}{\vec{j}} \newcommand{\vk}{\vec{k}} \begin{document} \maketitle \noindent {\bf Assessed questions: 3, 4 and 5. Due: Monday 3 December 2007.} \bigskip {\bf Complex numbers} \begin{enumerate} \item Find the complex numbers $z$ that satisfy the following equations: \begin{enumerate} \item $3z + (2-2i) = 1+i$ \item $(1+3i)z - 3(1-2i) = 3+2i - 2(4-i)z$ \end{enumerate} \answerblock{ \begin{enumerate} \item $3z + (2-2i) = 1+i$, so $3z = -1 + 3i, z = -\frac{1}{3} + i $ \item $(1+3i + 8-2i)z = 3(1-2i) + 3+2i = 6 - 4i$, so $$ z = \frac{6-4i}{9+i} = \frac{(6-4i)(9-i)}{82} = \frac{50-42i}{82} = \frac{25}{41}-\frac{21}{41}i $$ \end{enumerate} } \item Write the complex number $i^{4000021}$ in the form $a+ib$. \answerblock{ $i^{4000021} = i^{4(1000005)+1} = (i^4)^{1000005} i = i$ } \item \begin{enumerate} \item Let $X$ be the set of complex numbers $z$ of the form $3e^{i\theta}$, where $-\pi/2 \leq \theta \leq \pi/2$. Plot $X$ on the Argand diagram. \item Let $Y$ be the set of complex numbers $z$ such that $|z-(1+i)| = 1$. Plot $Y$ on the Argand diagram. \end{enumerate} \answerblock{ \begin{enumerate} \item Every $z \in X$ has modulus 3. The Arg ranges from $-\pi/2 $ (negative imaginary axis) to $\pi/2$ (positive imaginary axis). So $X$ is the right half circle, radius 3, centre O. \item $|z-(1+i)|$ is the distance from $z$ to the point representing $1+i$. Hence $Y$ is the circle with centre $1+i$ and radius $1$. \end{enumerate} } \item Let $z=e^{i\theta}$. By expressing $\sin \theta$ in terms of $z$ and $1/z$, expand $(\sin \theta)^4$ to show that $$\cos 4\theta = 8(\sin \theta)^4 + 4 \cos 2\theta -3$$ \answerblock{ $\sin \theta = (z-1/z)/2i$ so \begin{eqnarray*} (\sin \theta)^4 &=& ((z-1/z)/2i)^4 \\ &=& \frac{1}{16}[(z^4+z^{-4}) - 4(z^2+z^{-2}) + 6 z^2 z^{-2}] \\ &=& \frac{1}{8}[\cos 4\theta-4\cos 2\theta + 3] \end{eqnarray*} } \item \begin{enumerate} \item Find the square roots of $-2\sqrt{2}-2\sqrt{2}i$ and plot them on the Argand diagram. \item Find all solutions $z \in {\complex}$ to the equation $ z^5 = -16\sqrt{2}(1+i) $ and plot them on the Argand diagram. \end{enumerate} \answerblock{ \begin{enumerate} \item $-2\sqrt{2}-2\sqrt{2}i = e^{i\pi} 2^{3/2}(1+i) = e^{i\pi} 2^{3/2} 2^{1/2} e^{i(2k\pi + \pi/4)}$ for $k=0,1,\ldots$. So the square root is \begin{eqnarray*} 2[e^{i(2k\pi + 5\pi/4)}]^{1/2} &=& 2 e^{i(k\pi + 5\pi/8)} \\ &=& 2 e^{5\pi i/8}, \;\; 2 e^{13\pi i/8} \end{eqnarray*} Roots are symmetrical in the origin (Args differ by $\pi$). \item $z^5 = 2^5 e^{i(\pi+\pi/4 + 2k\pi)} = 2^5 e^{i(5\pi/4 + 2k\pi)}$ and so $$z = 2 e^{i(\pi/4 + 2k\pi/5)}$$ for $k=0,1,2,3,4$. That is: $$ z = 2 e^{\pi i/4} = \sqrt{2}(1+i), 2 e^{13\pi i/20}, 2 e^{21\pi i/20}, 2 e^{29\pi i/20}, 2 e^{37\pi i/20} $$ Roots lie on the circle, centre O, radius 2, separated by an angle $2\pi/5$, starting at Arg $\pi/4$. \end{enumerate} } \item {\bf (Exam question Q6C1452005)} \begin{enumerate} \item Express the complex number $\sqrt{3}+i$ in polar form, i.e. in the form $re^{i\theta}$ for certain positive real numbers $r$ and $\theta$. \answerblock{ Modulus is $r=\sqrt{3+1}=2$. Argument is $\theta = \arctan 1/\sqrt{3[} = \pi/6$ } \item Using the fact that $\sin(A+B)=\sin A \cos B + \cos A \sin B$ and a corresponding result for $\cos(A+B)$, which you should state, prove that: $$r_1(\cos\theta_1+i\sin\theta_1) r_2(\cos\theta_2+i\sin\theta_2) = r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$$ \answerblock{ $$ r_1(\cos\theta_1+i\sin\theta_1)r_2(\cos\theta_2+i \sin\theta_2) =$$ $$r_1r_2[(\cos\theta_1\cos \theta_2-\sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos \theta_2+\cos\theta_1\sin\theta_2)] = r_1r_2e^{i(\theta_1+\theta_2)}$$ } \item Show that \begin{enumerate} \item $ \cos 3\theta = 4\cos^3 \theta - 3\cos \theta $ \answerblock{ $$\cos 3\theta + i\sin 3\theta = (\cos \theta + i\sin \theta)^3$$ by de Moivre's Theorem. Equating the real parts, $$ \cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta = 4\cos^3 \theta - 3\cos \theta $$ } \item $ 8\cos^4\theta = \cos4\theta + 4\cos2\theta + 3 $ {\bf Note}: You may wish to use the fact that $2\cos\theta = e^{i\theta}+e^{-i\theta}$. \answerblock{ $$ 16\cos^4 \theta = e^{i4\theta} + 4e^{i3\theta}e^{-i\theta} + 6e^{i2\theta}e^{-i2\theta} + 4e^{i\theta}e^{-i3\theta}+e^{-i4\theta} = $$ $$ 2[\cos4\theta + 4\cos2\theta + 3] $$ } \end{enumerate} \item Find {\em all} the roots (for $z$) of the equation $z^4 = 8(\sqrt{3}+i)$, and {\em write them in Cartesian form}, i.e. in the form $a+ib$ for real numbers $a$ and $b$. \answerblock{ In polar form, $z^4 = 16e^{i\pi/6} = 16e^{i\pi(2n+1/6)}$ for $n=0,1,2,\ldots$. Thus, $z = 2e^{i\pi(n/2+1/24)}$ are distinct roots for $n=0,1,2,3$. This gives: \begin{eqnarray*} z_1 &=& 2\cos \pi/24 + i 2 \sin \pi/24 \\ z_2 &=& 2\cos 13\pi/24 + i 2 \sin 13\pi/24 = -2\cos 11\pi/24 + i 2 \sin 11\pi/24 \\ z_3 &=& 2\cos 25\pi/24 + i 2 \sin 25\pi/24 = -2\cos \pi/24 - i 2 \sin \pi/24 \\ z_4 &=& 2\cos 37\pi/24 + i 2 \sin 37\pi/24 = 2\cos 11\pi/24 - i 2 \sin 11\pi/24 \end{eqnarray*} {\bf Alternatively}: {\em any} root multiplied by $1,-1,i,-i$ is OK. } \end{enumerate} \item {\bf (Exam question Q6C1452006)} \begin{enumerate} \item Express the complex number $12+5i$ in polar form, i.e. in the form $re^{i\theta}$ for certain positive real numbers $r$ and $\theta$. \answerblock{ Modulus is $r=\sqrt{144+25}=13$. Argument is $\arctan 5/12 = \phi $, say. Number is then $13e^{i\phi}$ } \item The cube roots of unity, i.e. complex numbers which give result 1 when cubed, are $1, \omega$ and $\omega'$. \begin{enumerate} \item Obtain representations of these three roots in both Cartesian and polar form. \answerblock{ Roots are $e^{2\pi k i/3}$ for $k=0,1,2$, corresponding to $1, \omega, \omega'$. 1 is already in both Cartesian and polar form. $\omega = \cos 2\pi i/3 + i \sin 2\pi i/3 = -1/2 + \sqrt{3}i/2$, $\omega' = \cos 4\pi i/3 + i \sin 4\pi i/3 = -1/2 - \sqrt{3}i/2$. } \item Verify that the roots not equal to 1 are the squares of each other. \answerblock{ The square of the second root ($k=1$) is $(e^{2\pi i/3})^2=e^{4\pi i/3}$, i.e. the third root. The square of the third root ($k=2$) is $(e^{4\pi i/3})^2=e^{8\pi i/3}=e^{2\pi i/3}$, i.e. the second root. } \item Verify that the roots not equal to 1 add up to $-1$. \answerblock{ Either note that $\omega, \omega'$ are complex conjugates with real part $-1/2$ so sum to -1, or observe that the sum of all 3 roots is the coefficient of $z^2$ in the equation $z^3-1=0$, i.e. 0. } \end{enumerate} \item Assuming that $e^{i\theta}=\cos \theta + i \sin \theta$ for real number $\theta$, prove that \begin{enumerate} \item $$(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta $$ for all real $n$. \answerblock{ $\cos n\theta + i \sin n\theta = e^{in\theta} = \left(e^{i\theta} \right)^n= \left(\cos \theta + i \sin \theta \right)^n$ } \item Show that $ \sin 4\theta = 4\cos \theta \sin \theta - 8\cos \theta \sin^3 \theta $ \answerblock{ $$\cos 4\theta + i\sin 4\theta = (\cos \theta + i\sin \theta)^4$$ Equating the imaginary parts, $$ \sin 4\theta = 4\cos^3 \theta \sin \theta - 4\cos \theta \sin^3 \theta = 4\cos \theta \sin \theta (1 - \sin^2 \theta) - 4\cos \theta \sin^3 \theta $$ so the result follows. } \end{enumerate} \item A rock star is worried about the quality of the output signal from one of his amplifiers. Therefore he monitors the signal to look for spurious noise using Fourier analysis. In the course of estimating certain coefficients, he needs to evaluate the integral $$I = \int_0^\pi e^{-\theta} \cos 2n\theta \mbox{d}\theta$$ for integer values $n$. Using the fact that $ \cos 2n\theta$ is the real part of $e^{2in\theta}$, or otherwise, show that $I = \frac{1-e^{-\pi}}{1+4n^2}$. \answerblock{ $I = Re\left( \int_0^\pi e^{(2in-1)\theta}\mbox{d}\theta \right) = Re\left[-\frac{e^{(2in-1)\theta}}{1-2in)}\right]_0^\pi = Re\left(\frac{(1+2in)(1-e^{(2in-1)\pi})}{1+4n^2}\right)$ so result follows. } \end{enumerate} \bigskip \hspace{-0.5cm} {\bf Bounds and Limits} \item What is the {\em least upper bound} ({\bf sup}remum) and {\em greatest lower bound} ({\bf inf}imum) of the following sets of numbers: \begin{enumerate} \item $\{x \in \nat \mid 1 \leq x^2 \leq 29 \}$ \item $\{x \in \rat \mid 1 \leq x^2 \leq 29 \}$ \item $\{x \in \real \mid 1 \leq x^2 \leq 29 \}$ \end{enumerate} in each case, are the infimum and supremum in the given set? \answerblock{ \begin{enumerate} \item The set is $\{1,2,3,4,5\}$ which is finite, so the inf is 1 (minimum element), the sup is 5 (maximum element). \item The set consists of all rationals in $[-\sqrt{29},-1]$ and $[1,\sqrt{29}]$, i.e. it is $\rat \cap ([-\sqrt{29},-1] \cup[1,\sqrt{29}])$. The inf is $-\sqrt{29}$, the sup is $\sqrt{29}$, neither is in the set since they are not rational. \item Same as previous, but both inf and sup are in the set. \end{enumerate} } \item \begin{enumerate} \item Prove {\em rigorously} that, for $\alpha > 0$, the sequence $a_n = n^{-\alpha}$ converges to zero as $n$ tends to infinity. In other words, use $\epsilon$ and $N$, an appropriate value of which should be given as a function of $\epsilon$. \item Hence use the trapping or `sandwich' theorem to show that $$ \frac{n!}{n^n} \rightarrow 0 $$ as $n \rightarrow \infty$ \item What happens to $$ \frac{n!}{n^p} $$ as $n \rightarrow \infty$, where $p$ is a fixed integer? \end{enumerate} \answerblock{ \begin{enumerate} \item Pick $\epsilon > 0$. Then we want an $N$ s.t. $a_N = N^{-\alpha} \leq \epsilon$ so that then all $a_n < \epsilon$ for $n > N$ since $a_n$ is decreasing. Thus we want $N^\alpha \geq 1/\epsilon$ so any integer greater than $\epsilon^{-1/\alpha}$ will do the job, e.g. choose $$N = \lceil \epsilon^{-1/\alpha} \rceil$$ \item $$ \frac{n!}{n^n} = 1 \times (1 - \frac{1}{n}) \times (1 - \frac{2}{n}) \times \ldots \times \frac{1}{n} < \frac{1}{n} $$ But $\frac{1}{n} > 0 \;\; \forall n > 0 $ and the given sequence is trapped. \item Diverges since, for $n>p$, $$ \frac{n!}{n^p} > (n-p)!\frac{(n-p+1)^p}{n^p} = (n-p)! \left( 1 - \frac{p-1}{n} \right)^p > (n-p)! (1/2)^p $$ for $n > 2(p-1)$. \end{enumerate} } \item Investigate the convergence properties (either converges or diverges) of each of the following sequences $a_n, n = 1,2,\ldots$ \begin{enumerate} \item $a_n = \frac{3n^2+2n+4}{5n^2-7n+1}$ \item $a_n = \frac{3n^3+2n+4}{5n^2-7n+1}$ \item $a_n = n + (-1)^n n^2$ \item $a_n = \frac{1}{n^2} + \frac{1}{(n+1)^2} + \ldots + \frac{1}{(2n)^2}$ \end{enumerate} \answerblock{ \begin{enumerate} \item Divide top and bottom by $n^2$ to get: $a_n = \frac{3+2/n+4/n^2}{5-7/n+1/n^2}$. As $n \rightarrow \infty$, $a_n \rightarrow 3/5$ \item Similarly, $a_n = \frac{3n+2/n+4/n^2}{5-7/n+1/n^2} > 3n/5$ so diverges. \item $n^2-n \leq |a_n| \leq n^2 + n$ (because $||x|-|y|| \leq |x \pm y| \leq |x|+|y|$) so $a_n$ diverges as fast as $n^2$, but alternating sign makes it oscillate. \item $a_n < n/n^2 = 1/n$ so $a_n$ converges by trapping. \end{enumerate} } \item Consider the decimal $D = 0.d_1 d_2 d_3\ldots \in [0,1)$, as in lectures, and the sequence of {\em finite decimals} $a_n = 0.d_1 d_2 \ldots d_n, \; n=1,2,\ldots$. (D is also equivalent to a finite decimal if $d_k = 0 \;\; \forall k > K$ for some integer $K>0$.) Show that: \begin{enumerate} \item Each $a_n$ is rational for $n>0$. \item The sequence $a_n$ is increasing and bounded above. \item $D$ is a real number. \end{enumerate} Conversely, let $E$ be a real number in $[0,1)$. Then successively choose the rational decimal numbers $b_n = 0.e_1\ldots e_n, \; n=1,2,\ldots $ such that $b_n \leq E$ and $b_n + 10^{-n} > E$ -- i.e. $b_n$ is the biggest such decimal. Prove that $$\lim_{n \rightarrow \infty} b_n = E$$ and hence that $E$ is a decimal. \answerblock{ \begin{enumerate} \item Obvious since $a_n$ is a finite decimal and so a sum of inverse powers of 10, each of which is rational. \item Obvious since we add a multiple of $10^{-n}$ to get $a_n$ from $a_{n-1}$ ($n>1$). (Strictly, we should say `non-decreasing' as the multiple may be zero.) An upper bound is 1. \item $D$ is real by the fundamental axiom. \end{enumerate} $E$ is clearly the supremum of the sequence $b_n$ (simple proof by contradiction since $E - b_n \leq 10^{-n}$ for arbitrarily large $n$). $b_n$ is increasing and bounded above and therefore has a limit by the fundamental axiom. That limit is $E$ because $\forall \epsilon>0, \exists N \in \nat$ s.t. $|b_N-E| < \epsilon$ since $E$ is the supremum. Hence $|b_n-E| < \epsilon \; \forall n \geq N$. } \end{enumerate} \end{document}