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Add script to extract question/answer/solution triplets from given examples
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‎wizard/OandW_sample_PS1_Complete.tex

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\documentclass[12pt]{article}
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\input{../../Shared/header}
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\pagestyle{empty}
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\setlength{\textheight}{9.6in} \setlength{\textwidth}{6.5in}
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\setlength{\oddsidemargin}{-0.15in} \setlength{\evensidemargin}{0.2in}
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\setlength{\topmargin}{-50pt}
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\setlength{\arraycolsep}{2pt}
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\setlength{\parskip}{0pt}
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\setlength{\parindent}{0pt}
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\begin{document}
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\centerline{\Large Complex Numbers, Functions and Ordinary Differential Equations}
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\bigskip\bigskip
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\noindent
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Problem Sheet 1\hfill 2019
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\vskip-9pt\hrulefill
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\bigskip
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%\subsubsection*{Exercises}
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\begin{enumerate}
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\item Find the real and imaginary parts of:
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$
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\begin{array}[h!]{lll}
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{\rm (a)}\hskip5pt 8+3\,i\hskip24pt&
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{\rm (b)}\hskip5pt 4-15\,i\hskip24pt&
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{\rm (c)}\hskip5pt \cos\theta-i\,\sin\theta\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt i^2&
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{\rm (e)}\hskip5pt i\,(2-5\,i)&
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{\rm (f)}\hskip5pt (1+2\,i)(2-3\,i)
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\end{array}
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$
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\item Write each of the following expressions as a complex number in the form $x+i\,y$:
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$
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\begin{array}[h!]{lll}
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{\rm (a)}\hskip5pt (5-i)(2+3\,i)\hskip24pt&
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{\rm (b)}\hskip5pt (3-4\,i)(3+4\,i)\hskip24pt&
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{\rm (c)}\hskip5pt (1+2\,i)^2\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt \displaystyle{10\over4-2\,i}&
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{\rm (e)}\hskip5pt \displaystyle{3-i\over4+3\,i}&
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{\rm (f)}\hskip5pt \displaystyle{1\over i}
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\end{array}
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$
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\item Define $z=(5+7\,i)(5+b\,i)$.
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$
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\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt {\rm If}\ b\ {\rm and}\ z\ {\rm are\ both\ real,\ find}\ b.\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt {\rm If}\ {\rm Im}(b)={4\over5},\ {\rm and}\ z\ {\rm is\ pure\ imaginary,\ find}\
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{\rm Re}(b).
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\end{array}
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$
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\item Plot the following complex numbers in the complex plane:
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$\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt 2\,i\hskip24pt
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{\rm (b)}\hskip5pt -3+i\,2\hskip24pt
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{\rm (c)}\hskip5pt (-3+i\,2)^\ast\hskip24pt
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{\rm (d)}\hskip5pt \displaystyle{1+i\over\sqrt{2}}\hskip24pt
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\end{array}
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$
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\item Write the following numbers in polar form:
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$\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt i\hskip24pt
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{\rm (b)}\hskip5pt -i\hskip24pt
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{\rm (c)}\hskip5pt 1+i\hskip24pt
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{\rm (d)}\hskip5pt 1-i\,\sqrt{3}\hskip24pt
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\end{array}
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$
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\item Write the following complex numbers in Cartesian coordinates:
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$\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt e^{-3\pi i/4}\hskip24pt
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{\rm (b)}\hskip5pt e^{5\pi i/4}\hskip24pt
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{\rm (c)}\hskip5pt 3\,e^{i}\hskip24pt
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{\rm (d)}\hskip5pt \displaystyle{1\over\sqrt{3}\,e^{\pi i/3}}\hskip24pt
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\end{array}
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$
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\item Consider the polar form of a complex number $z=r\,e^{i\,\theta}$. Show that
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$(z^2)^\ast=(z^\ast)^2$.
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\item
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\begin{enumerate}
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\item Consider the polynomial $az^2+bz+c$, where $a$, $b$,
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and $c$ may be complex and $a\ne 0$. Given that $z_0$ and $z_0^\ast$ are distinct solutions to
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$az^2 + bz + c = 0$ and thus are also solutions to $z^2+(b/a)z+(c/a)=z^2+b'z+c'=0$, show that $b'$ and $c'$ must be real.
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Conversely, if $a$, $b$ and $c$ are all real, and $z_0$ is a solution of $az^2 + bz + c = 0$, deduce that $z_0^\ast$ is also a solution.
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\item Let $b=c=1$ and $a=t$, $t>\frac{1}{4}$. Make a graph of the
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pattern that the solutions to this equation traces out in the
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complex plane (hint: work out $|z+1|$).
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\item Derive functions $a(t)$, $b(t)$ and $c(t)$ for which
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the solutions to the equation traces out a pattern which is rotated
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by $90^\circ$ counterclockwise around the origin with respect to the previous
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pattern (n.b. it is no longer the case that $a$, $b$ and $c$ are all real. Why?).
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How can you achieve an arbitrary rotation?
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\end{enumerate}
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\end{enumerate}
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\end{document}
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\documentclass[12pt]{article}
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\input{../../Shared/header}
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\pagestyle{empty}
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\setlength{\textheight}{9.6in} \setlength{\textwidth}{6.5in}
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\setlength{\oddsidemargin}{-0.15in} \setlength{\evensidemargin}{0.2in}
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\setlength{\topmargin}{-50pt}
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\setlength{\arraycolsep}{2pt}
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\setlength{\parskip}{0pt}
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\setlength{\parindent}{0pt}
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\begin{document}
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\centerline{\Large Complex Numbers, Functions and Ordinary Differential Equations}
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\bigskip\bigskip
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\noindent
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Solutions to Problem Sheet 1\hfill 2023
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\vskip-9pt\hrulefill
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%\subsubsection*{Homework}
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\begin{enumerate}
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\item For a complex number $a+i\,b$, in which $a$ and $b$ are real, the real and imaginary parts
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are given by ${\rm Re}(a+i\,b)=a$ and ${\rm Im}(a+i\,b)=b$, respectively. Thus,
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$
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\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt {\rm Re}(8+3\,i)=8\, ,\qquad{\rm Im}(8+3\,i)=3\, .\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt {\rm Re}(4-15\,i)=4\, ,\qquad {\rm Im}(4-15\,i)=-15\, .\\
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\noalign{\vskip12pt}
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{\rm (c)}\hskip5pt {\rm Re}(\cos\theta-i\,\sin\theta)=\cos\theta\, ,\qquad {\rm
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Im}(\cos\theta-i\,\sin\theta)=-\sin\theta\, .\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt i^2=-1.\quad {\rm Re}(i^2)=-1\, ,\qquad {\rm Im}(i^2)=0\, .\\
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\noalign{\vskip12pt}
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{\rm (e)}\hskip5pt i\,(2-5\,i)=5+2\,i.\qquad {\rm Re}(5+2\,i)=5\, ,\qquad {\rm Im}(5+2\,i)=2\, .\\
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\noalign{\vskip12pt}
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{\rm (f)}\hskip5pt (1+2\,i)(2-3\,i)=2-3\,i+4\,i+6=8+i\, .\qquad {\rm Re}(8+i)=8\, ,\qquad {\rm
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Im}(8+i)=1\, .
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\end{array}
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$
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\item Applying the rules for the multiplication and division of
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complex numbers yields:
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$
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\begin{array}[h!]{lll}
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{\rm (a)}\hskip5pt (5-i)(2+3\,i)=10+15\,i-2\,i+3=13+13\,i\, .\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt (3-4\,i)(3+4\,i)=9+12\,i-12\,i+16=25\, .\\
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\noalign{\vskip12pt}
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{\rm (c)}\hskip5pt (1+2\,i)^2=(1+2\,i)(1+2\,i)=1+2\,i+2\,i-4=-3+4\,i\, .\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt \displaystyle{{10\over4-2\,i}={10\over4-2\,i}\times{4+2\,i\over4+2\,i}=
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{40+20\,i\over16+8\,i-8\,i+4}={40+20\,i\over20}=2+i\, .}\\
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\noalign{\vskip12pt}
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{\rm (e)}\hskip5pt \displaystyle{{3-i\over4+3\,i}={3-i\over4+3\,i}\times{4-3\,i\over4-3\,i}=
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{12-9\,i-4\,i-3\over16-12\,i+12\,i+9}={9-13\,i\over25}={9\over25}-i\,{13\over25}\, .}\\
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\noalign{\vskip12pt}
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{\rm (f)}\hskip5pt \displaystyle{{1\over i}={1\over i}\times{-i\over-i}=-i\, .}
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\end{array}
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$
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\item We have that $z=(5+7\,i)(5+b\,i)=25+5b\,i+35\,i-7b$.
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$
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\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt {\rm If}\ b\ {\rm and}\ z\ {\rm are\ both\ real},\ {\rm then\ the \ imaginary\
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parts\ of\ both\ quantities\ vanish.}\ {\rm Thus,}\\
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\noalign{\vskip12pt}
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\hskip20pt {\rm Im}(z)=35+5b=0, {\rm so}\ b=-7.\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt {\rm If}\ {\rm Im}(b)={4\over5},\ {\rm and}\ z\ {\rm is\ pure\ imaginary,}\
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{\rm then\ the\ real\ part\ of}\ z\ {\rm vanishes\!:}\\
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\noalign{\vskip12pt}
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\hskip20pt{\rm Re}(z)=25+5\bigl[i\,{\rm Im}(b)\bigr]i-7\,{\rm Re}(b)=25-4-7\,{\rm Re}(b)=21-7\,{\rm
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Re}(b)=0,\\
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\noalign{\vskip12pt}
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\hskip20pt{\rm so}\ {\rm Re}(b)=3.
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\end{array}
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$
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\item The graphical representation of the complex number $z=x+i\,y$ is
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the point $(x,y)$ on a set of axes where the $x$-axis corresponds to
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the real part of the complex number and the $y$-axis the imaginary
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part. The required points are
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\begin{center}
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\includegraphics[width=8cm]{PS1-2}
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\end{center}
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\item For a complex number $z=x+i\,y$, the polar form is $z=r\,e^{i\theta}$, where
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\begin{equation*}
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r=\sqrt{x^2+y^2}\, ,\qquad
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\theta={\rm tan}^{-1}\biggl({y\over x}\biggr)\, .
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\end{equation*}
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$\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt z=i.\ {\rm We\ have\ that}\ x=0\ {\rm and}\ y=1,\ {\rm so}\ r=1,\ {\rm and}\
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\theta={1\over2}\pi.\ {\rm Thus}\ i=e^{i\,\pi/2}.\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt z=-i.\ {\rm We\ have\ that}\ x=0\ {\rm and}\ y=-1,\ {\rm so}\ r=1,\ {\rm and}\
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\theta={3\over2}\pi.\ {\rm Thus}\ -i=e^{3i\,\pi/2}.\\
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\noalign{\vskip12pt}
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{\rm (c)}\hskip5pt z=1+i.\ {\rm We\ have\ that}\ x=1\ {\rm and}\ y=1,\ {\rm so}\ r=\sqrt{2},\ {\rm
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and}\ \theta={1\over4}\pi.\\
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\noalign{\vskip12pt}
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\hskip20pt{\rm Thus}\ 1+i=\sqrt{2}\,e^{i\,\pi/4}.\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt z=1-i\,\sqrt{3}.\ {\rm We\ have\ that}\ x=1\ {\rm and}\ y=-\sqrt{3},\ {\rm so}\
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r=2,\ {\rm and}\\
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\noalign{\vskip12pt}
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\hskip20pt\theta={\rm tan}^{-1}(-\sqrt{3})=-{1\over3}\pi.\ {\rm Thus}\
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1-i\,\sqrt{3}=2\,e^{-i\,\pi/3}.
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\end{array}
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$
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\item A complex number $z=r\,e^{i\,\theta}$ can be written as
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$z=r\cos\theta+i\,r\sin\theta$. Thus,
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$\begin{array}[h!]{l}
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{\rm (a)}\hskip5pt e^{-3\pi i/4}=\cos\bigl({3\over4}\pi\bigr)-i\,\sin\bigl({3\over4}\pi\bigr)=
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\displaystyle{-{1+i\over\sqrt{2}}}=-{1\over2}\sqrt{2}-i\,{1\over2}\sqrt{2}\, .\\
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\noalign{\vskip12pt}
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{\rm (b)}\hskip5pt e^{5\pi i/4}=\cos\bigl({5\over4}\pi\bigr)+i\,\sin\bigl({5\over4}\pi\bigr)=
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\displaystyle{-{1+i\over\sqrt{2}}}=-{1\over2}\sqrt{2}-i\,{1\over2}\sqrt{2}\, .\\
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\noalign{\vskip12pt}
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{\rm (c)}\hskip5pt 3\,e^{i}=3\cos 1+i\,\sin 1\, .\\
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\noalign{\vskip12pt}
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{\rm (d)}\hskip5pt \displaystyle{{1\over\sqrt{3}\,e^{\pi i/3}}=
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{\sqrt{3}\over3}\,e^{-\pi i/3}=
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{\sqrt{3}\over3}\cos\bigl({\textstyle{1\over3}}\pi\bigr)-
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{i\,\sqrt{3}\over3}\sin\bigl({\textstyle{1\over3}}\pi\bigr)=
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{\sqrt{3}\over6}-{i\over2}}\, .
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\end{array}
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$
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%\end{enumerate}
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%\subsubsection*{Tutorial}
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\item Given that $z=r\,e^{i\,\theta}$, we have
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\begin{eqnarray*}
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(z^2)^\ast=\bigl[\bigl(r\,e^{i\,\theta}\bigr)^2\bigr]^\ast=\bigl(r^2\,e^{2i\,\theta}\bigr)^\ast=r^2\,e^{-2i\,\theta}=\bigl(r\,e^{-i\,\theta}\bigr)^2=
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(z^\ast)^2\, .
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\end{eqnarray*}
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\item
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\begin{enumerate}
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\item Suppose $z_0$ is a solution of $az^2+bz+c=0$, and therefore also of $z^2+b'z+c'=0$ where $b'=b/a$ and $c'=c/a$ and $a\ne0$. Given that $z_0^\ast$ is also a solution implies that
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\[
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(z-z_0)(z-z_0^\ast)=0~,
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\]
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or
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\[
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z^2-(z_0+z_0^\ast)z+|z_0|^2=0~,
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\]
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showing that $b'$ and $c'$ must be real. Conversely, if $a$, $b$, and $c$ are real, and $z_0$ is a solution of $az^2+bz+c=0$, then $az_0^2+bz_0+c=0$, and taking the conjugate yields
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\[
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a(z_0^2)^\ast+bz_0^\ast+c=0~,
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\]
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or
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\[
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a(z_0^\ast)^2+bz_0^\ast+c=0~,
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\]
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showing that $z_0^\ast$ is also a solution.
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\item The roots of the polynomial are
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\[
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z_{\pm} = \frac{1}{2t} \left( -1 \pm \sqrt{1 - 4t} \right)
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\]
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which for $t>\frac{1}{4}$ has complex roots,
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\[
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z_{\pm} = - \frac{1}{2t} \pm i\, \frac{1}{2t}\sqrt{4t - 1} \, .
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\]
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The solutions will map out a circle centered at $z=-1$ in the complex plane.
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(To see this either show that $x={\rm Re}(z), y={\rm Im}(z)$ satisfies $(x+1)^2+y^2=1$,
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or that $|z+1|^2=1$.)
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\begin{center}
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\includegraphics[width=8cm]{PS1-1}
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\end{center}
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\item Rotating the figure through $90^\circ$ is equivalent to multiplying both
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solutions by $e^{i\pi/2} = i$, thus
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\begin{align}
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z_{\pm} & = i \left( - \frac{1}{2t} \pm i\, \frac{1}{2t}\sqrt{4t - 1} \right)
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\nonumber \\
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& = \pm \frac{1}{2t}\sqrt{4t - 1} - \frac{i}{2t} \, . \nonumber
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\end{align}
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Forming the polynomial by multiplying together the two solutions gives
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\begin{align}
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p & =\left(z-z_+\right)\left(z-z_-\right)\nonumber\\
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&=
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\left( z - \frac{1}{2t}\sqrt{4t - 1} + \frac{i}{2t} \right)
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\left( z + \frac{1}{2t}\sqrt{4t - 1} + \frac{i}{2t} \right)
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\nonumber \\
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& = -\frac{1}{t}(-t z^2 - i\,z + 1) \nonumber \, ,
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\end{align}
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thus reducing $p=0$ to $-t z^2 - i\,z + 1=0$.
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%It can be seen that this is just the original polynomial with $z$ substituted with $iz=e^{i\pi/2}z$.
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So one solution is $a(t)=-t$, $b=-i$ and $c=1$. All other solutions are multiples of this set. The solutions that were conjugate pairs for the original equation are not conjugate pairs for the new equation as the coefficients of the latter are no longer real.
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An arbitrary rotation of angle $\theta$ can be made by substituting $z_{\pm}$ with $e^{i\theta}z_{\pm}$.
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\end{enumerate}
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\end{enumerate}
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\end{document}

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