\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

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2011.3.8 Question 8

Since \(w = u + iv\), \(z = x + iy\), we have \begin {align*} u + iv & = w \\ & = \frac {1 + iz}{i + z} \\ & = \frac {1 + i(x + iy)}{i + (x + iy)} \\ & = \frac {(1 - y) + xi}{x + (y + 1)i} \\ & = \frac {(1 - y) + xi}{x + (y + 1)i} \cdot \frac {x - (y + 1)i}{x - (y + 1)i} \\ & = \frac {\left [(1 - y) + xi\right ] \left [x - (y + 1)i\right ]}{x^2 + (y + 1)^2} \\ & = \frac {(1 - y)x + x(y + 1)}{x^2 + (y + 1)^2} + \frac {x^2 - (1 - y) \cdot (y + 1)}{x^2 + (y + 1)^2} \cdot i \\ & = \frac {2x}{x^2 + (y + 1)^2} + \frac {x^2 + y^2 - 1}{x^2 + (y + 1)^2} \cdot i, \end {align*}

and hence \[ (u, v) = \left (\frac {2x}{x^2 + (y + 1)^2}, \frac {x^2 + y^2 - 1}{x^2 + (y + 1)^2}\right ). \]

  1. When \(y = 0\), we have \[ (u, v) = \left (\frac {2x}{x^2 + 1}, \frac {x^2 - 1}{x^2 + 1}\right ). \]

    Let \(x = \tan \left (\frac {\theta }{2}\right )\). The tangent half-angle substitution also gives that \(u = \sin \theta \) and \(v = -\cos \theta \), and hence \(u^2 + v^2 = 1\).

    For the range of \(\theta \), we have \(-\frac {\pi }{2} < \frac {\theta }{2} < \frac {\pi }{2}\), which means \(-\pi < \theta < \pi \).

    This represents the unit circle without the point \((\sin \pi , - \cos \pi ) = (0, 1)\) corresponding to \(\theta = \pi ( + 2k\pi )\) for some integer \(k\).

  2. When \(-1 < x < 1\), we have \(-\frac {\pi }{4} < \frac {\theta }{2} < \frac {\pi }{4}\), which means \(-\frac {\pi }{2} < \theta < \frac {\pi }{2}\). This is the unit circle with only the part below the \(u\) axis (exclusive).

  3. When \(x = 0\), we have \[ (u, v) = \left (0, \frac {y^2 - 1}{(y + 1)^2}\right ). \]

    Notice that \[ v = \frac {y^2 - 1}{(y + 1)^2} = \frac {(y + 1)(y - 1)}{(y + 1)^2} = \frac {y - 1}{y + 1} = 1 - \frac {2}{y + 1}, \] and hence \(-1 < v < 1\).

    This means the locus of \(w\) is the line segment \(u = 0, -1 < v < 1\).

  4. When \(y = 1\), we have \[ (u, v) = \left (\frac {2x}{x^2 + 4}, \frac {x^2}{x^2 + 4}\right ). \]

    First, let \(x = 2t\), and we have \[ (u, v) = \left (\frac {4t}{4t^2 + 4}, \frac {4t^2}{4t^2 + 4}\right ) = \left (\frac {t}{t^2 + 1}, \frac {t^2}{t^2 + 1}\right ). \]

    Let \(t = \tan \left (\frac {\theta }{2}\right )\), and we have \(-\pi < \theta < \pi \). Notice that \[ u = \frac {1}{2} \cdot \frac {2t}{t^2 + 1} = \frac {1}{2} \sin \theta , \] and \[ v - \frac {1}{2} = \frac {1}{2} \cdot \frac {t^2 - 1}{t^2 + 1} = - \frac {1}{2} \cos \theta . \]

    This means the loci is a subset of the circle centred at \(\left (0, \frac {1}{2}\right )\) with radius \(\frac {1}{2}\), with the point \[ (u, v) = \left (\frac {1}{2} \sin \pi , \frac {1}{2} - \frac {1}{2} \cos \pi \right ) = (0, 1) \] missing, which corresponds to \(\theta = \pi (+ 2k\pi )\) for some integer \(k\).

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