\(% 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}}\)

[next] [prev] [prev-tail] [tail] [up]

2023.2.7 Question 7

  1. Let \(z = a + ib\) and \(\abs *{z} = \sqrt {a^2 + b^2}\). Let \(w = c + id\) and \(\abs *{w} = \sqrt {c^2 + d^2}\).

    We have \(zw = (ac - bd) + (bc + ad) i\), and hence \begin {align*} \abs *{zw} & = \sqrt {(ac - bd)^2 + (bc + ad)^2} \\ & = \sqrt {a^2 c^2 + b^2 d^2 - 2abcd + b^2 c^2 + a^2 d^2 + 2abcd} \\ & = \sqrt {a^2 c^2 + b^2 d^2 + b^2 c^2 + a^2 d^2} \\ & = \sqrt {\left (a^2 + b^2\right ) \left (c^2 + d^2\right )} \\ & = \sqrt {a^2 + b^2} \sqrt {c^2 + d^2} \\ & = \abs *{z} \abs *{w} \end {align*}

    as desired.

  2. Let \(z = 2 + i\) and \(w = 10 + 11i\), we have \(\abs *{z} = \sqrt {5}\) and \(\abs *{w} = \sqrt {221}\).

    Multiplying them gives us \(zw = (2 \times 10 - 1 \times 11) + (10 \times 1 + 2 \times 11) i = 9 + 32 i\).

    We have \(\abs *{zw} = \abs *{z} \abs *{w}\), and hence \(\sqrt {9^2 + 32^2} = \sqrt {5 \times 221}\).

    This means \(9^2 + 32^2 = 5 \times 221\), and hence a possible pair is \((h, k) = (9, 32)\).

  3. We have \(8045 = 5 \times 1609 = \left (1^2 + 2^2\right ) \left (40^2 + 3^2 \right )\).

    Let \(z = 2 + i, w = 3 + 40 i\), \(zw = (2 \times 3 - 1 \times 40) + (2 \times 40 + 3 \times 1) i = -34 + 83i\).

    Since \(\abs *{zw} = \abs *{z} \abs *{w}\), we must have \begin {align*} 34^2 + 83^2 & = \left (1^2 + 2^2\right ) \times \left (40^3 + 3^2\right ) \\ & = 5 \times 1609 \\ & = 8045, \end {align*}

    and hence \((m, n) = (34, 83)\) is a possible pair of solution.

  4. We notice that \(36\) is a square number, and \begin {align*} 36 \times 50805 & = 6^2 \left (102^2 + 201^2\right ) \\ & = 6^2 \cdot 102^2 + 6^2 \cdot 201^2 \\ & = (6 \times 102)^2 + (6 \times 201)^2 \\ & = 612^2 + 1206^2. \end {align*}

    Hence, \((p, q) = (612, 1206)\) is a possible pair of solution.

  5. First, we observe that \(1002082 = 1002001 + 81 = 1001^2 + 9^2\), and hence similar to the previous part, we have \begin {align*} 25 \times 1002082 & = 5^2 \left (9^2 + 1001^2\right ) \\ & = \left (5 \times 9\right )^2 + \left (5 \times 1001\right )^2 \\ & = 45^2 + 5005^2, \end {align*}

    and \((r, s) = (45, 5005)\) is a possible pair of solution.

    Furthermore, since \(1002082 = 1001^2 + 9^2\), and \(5^2 = 4^2 + 3^2\), consider \(z = 3 + 4i\), \(w = 1001 + 9i\), we have \begin {align*} zw & = (3 \times 1001 - 4 \times 9) + (4 \times 1001 + 3 \times 9) i \\ & = (3003 - 36) + (4004 + 27) i \\ & = 2967 + 4031 i, \end {align*}

    and \((r, s) = (2967, 4031)\) is a possible pair of solution since \(\abs *{zw} = \abs *{z} \abs *{w}\).

    Similarly, \(z = 4 + 3i\) and \(w = 1001 + 9i\) gives \begin {align*} zw & = (4 \times 1001 - 3 \times 9) + (3 \times 1001 + 4 \times 9) i \\ & = (4004 - 27) + (3003 + 36) i \\ & = 3977 + 3039 i, \end {align*}

    and therefore \((R, s) = (3039, 3977)\) is another possible pair of solution.

  6. We have \(109 = 100 + 9 = 10^2 + 3^2\), and let \(z = 10 + 3i\), \(w = t + ui\), we examine the linear system of equations \[ \left \{ \begin {aligned} 10t - 3u & = 1001, \\ 3t + 10u & = 6. \end {aligned} \right . \]

    This solves to \(t = 92\) and \(u = -27\). But since \((-27)^2 = 27^2\), we must have \((t. u) = (92, 27)\) satisfies the desired equation.

[next] [prev] [prev-tail] [front] [up]