{
  "index": "1958-A-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. If \\( a_{1}, a_{2}, \\ldots, a_{n} \\) are complex numbers such that\n\\[\n\\left|a_{1}\\right|=\\left|a_{2}\\right|=\\cdots=\\left|a_{n}\\right|=r \\neq 0\n\\]\nand if \\( { }_{n} T_{s} \\) denotes the sum of all products of these \\( n \\) numbers taken \\( s \\) at a time, prove that\n\\[\n\\left|\\frac{{ }_{n} T_{s}}{{ }_{n} T_{n-s}}\\right|=r^{2 s-n}\n\\]\nwhenever the denominator of the left-hand side is different from zero.",
  "solution": "Solution. For any non-zero complex number, \\( z \\), we have \\( z \\quad 1=\\bar{z} /|z|^{2} \\) where the bar denotes complex conjugation. If \\( J \\) is a set of \\( s \\) indices selected from \\( \\{1,2, \\ldots, n\\} \\), then\n\\[\n\\begin{aligned}\n\\prod_{i £ J} a_{i} & =a_{1} a_{2} \\cdots a_{n} \\prod_{i \\in J} a_{i}^{-1}=a_{1} a_{2} \\cdots a_{n} \\prod_{i \\in J} \\frac{\\bar{a}_{i}}{r^{2}} \\\\\n& =\\frac{a_{1} a_{2} \\cdots a_{n}}{r^{2 s}} \\prod_{i \\in J} \\bar{a}_{i}\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\n{ }_{n} T_{n-s} & =\\sum_{J} \\prod_{i \\notin J} a_{i}=\\frac{a_{1} a_{2} \\cdots a_{n}}{r^{2 s}} \\sum_{J} \\prod_{i \\in J} \\overline{a_{1}} \\\\\n& =\\frac{a_{1} a_{2} \\cdots a_{n}}{r^{2 s}} \\bar{T}_{s} .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left.\\left|{ }_{n} T_{n-s}\\right|=\\left.\\frac{r^{n}}{r^{2 s}}\\right|_{n} \\overline{T_{s}}\\left|=r^{n-2 s}\\right|_{n} T_{s} \\right\\rvert\\,,\n\\]\nwhence the required formula follows at once.",
  "vars": [
    "a_1",
    "a_2",
    "a_n",
    "a_i",
    "i",
    "z",
    "J"
  ],
  "params": [
    "n",
    "r",
    "s"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_1": "alphaone",
        "a_2": "alphatwo",
        "a_n": "alphanlast",
        "a_i": "alphavar",
        "i": "indexer",
        "z": "complexz",
        "J": "indexset",
        "n": "totalnum",
        "r": "modulus",
        "s": "subsetsize"
      },
      "question": "4. If \\( alphaone, alphatwo, \\ldots, alphanlast \\) are complex numbers such that\n\\[\n\\left|alphaone\\right|=\\left|alphatwo\\right|=\\cdots=\\left|alphanlast\\right|=modulus \\neq 0\n\\]\nand if \\( { }_{totalnum} T_{subsetsize} \\) denotes the sum of all products of these \\( totalnum \\) numbers taken \\( subsetsize \\) at a time, prove that\n\\[\n\\left|\\frac{{ }_{totalnum} T_{subsetsize}}{{ }_{totalnum} T_{totalnum-subsetsize}}\\right|=modulus^{2 subsetsize-totalnum}\n\\]\nwhenever the denominator of the left-hand side is different from zero.",
      "solution": "Solution. For any non-zero complex number, \\( complexz \\), we have \\( complexz \\quad 1=\\bar{complexz} /|complexz|^{2} \\) where the bar denotes complex conjugation. If \\( indexset \\) is a set of \\( subsetsize \\) indices selected from \\( \\{1,2, \\ldots, totalnum\\} \\), then\n\\[\n\\begin{aligned}\n\\prod_{indexer \\in indexset} alphavar & =alphaone alphatwo \\cdots alphanlast \\prod_{indexer \\in indexset} alphavar^{-1}=alphaone alphatwo \\cdots alphanlast \\prod_{indexer \\in indexset} \\frac{\\bar{alphavar}}{modulus^{2}} \\\\\n& =\\frac{alphaone alphatwo \\cdots alphanlast}{modulus^{2 subsetsize}} \\prod_{indexer \\in indexset} \\bar{alphavar}\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\n{ }_{totalnum} T_{totalnum-subsetsize} & =\\sum_{indexset} \\prod_{indexer \\notin indexset} alphavar=\\frac{alphaone alphatwo \\cdots alphanlast}{modulus^{2 subsetsize}} \\sum_{indexset} \\prod_{indexer \\in indexset} \\overline{alphavar} \\\\\n& =\\frac{alphaone alphatwo \\cdots alphanlast}{modulus^{2 subsetsize}} \\bar{T}_{subsetsize} .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left.\\left|{ }_{totalnum} T_{totalnum-subsetsize}\\right|=\\left.\\frac{modulus^{totalnum}}{modulus^{2 subsetsize}}\\right|_{totalnum} \\overline{T_{subsetsize}}\\left|=modulus^{totalnum-2 subsetsize}\\right|_{totalnum} T_{subsetsize} \\right\\rvert\\,,\n\\]\nwhence the required formula follows at once."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_1": "pineapple",
        "a_2": "toothbrush",
        "a_n": "raincloud",
        "a_i": "snowflake",
        "i": "lanternfly",
        "z": "buttercup",
        "J": "scarecrow",
        "n": "tortoise",
        "r": "pendulum",
        "s": "goldfish"
      },
      "question": "4. If \\( pineapple, toothbrush, \\ldots, raincloud \\) are complex numbers such that\n\\[\n\\left|pineapple\\right|=\\left|toothbrush\\right|=\\cdots=\\left|raincloud\\right|=pendulum \\neq 0\n\\]\nand if \\( { }_{tortoise} T_{goldfish} \\) denotes the sum of all products of these \\( tortoise \\) numbers taken \\( goldfish \\) at a time, prove that\n\\[\n\\left|\\frac{{ }_{tortoise} T_{goldfish}}{{ }_{tortoise} T_{tortoise-goldfish}}\\right|=pendulum^{2 goldfish-tortoise}\n\\]\nwhenever the denominator of the left-hand side is different from zero.",
      "solution": "Solution. For any non-zero complex number, \\( buttercup \\), we have \\( buttercup \\quad 1=\\bar{buttercup} /|buttercup|^{2} \\) where the bar denotes complex conjugation. If \\( scarecrow \\) is a set of \\( goldfish \\) indices selected from \\( \\{1,2, \\ldots, tortoise\\} \\), then\n\\[\n\\begin{aligned}\n\\prod_{lanternfly £ scarecrow} snowflake & =pineapple toothbrush \\cdots raincloud \\prod_{lanternfly \\in scarecrow} snowflake^{-1}=pineapple toothbrush \\cdots raincloud \\prod_{lanternfly \\in scarecrow} \\frac{\\bar{snowflake}}{pendulum^{2}} \\\\\n& =\\frac{pineapple toothbrush \\cdots raincloud}{pendulum^{2 goldfish}} \\prod_{lanternfly \\in scarecrow} \\bar{snowflake}\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\n{ }_{tortoise} T_{tortoise-goldfish} & =\\sum_{scarecrow} \\prod_{lanternfly \\notin scarecrow} snowflake=\\frac{pineapple toothbrush \\cdots raincloud}{pendulum^{2 goldfish}} \\sum_{scarecrow} \\prod_{lanternfly \\in scarecrow} \\overline{pineapple} \\\\\n& =\\frac{pineapple toothbrush \\cdots raincloud}{pendulum^{2 goldfish}} \\bar{T}_{goldfish} .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left.\\left|{ }_{tortoise} T_{tortoise-goldfish}\\right|=\\left.\\frac{pendulum^{tortoise}}{pendulum^{2 goldfish}}\\right|_{tortoise} \\overline{T_{goldfish}}\\left|=pendulum^{tortoise-2 goldfish}\\right|_{tortoise} T_{goldfish} \\right\\rvert\\,,\n\\]\nwhence the required formula follows at once."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_1": "finaldatum",
        "a_2": "ultimatepiece",
        "a_n": "firstdatum",
        "a_i": "staticunit",
        "i": "constantidx",
        "z": "realzero",
        "J": "singularset",
        "n": "emptiness",
        "r": "flatness",
        "s": "entirety"
      },
      "question": "4. If \\( finaldatum, ultimatepiece, \\ldots, firstdatum \\) are complex numbers such that\n\\[\n\\left|finaldatum\\right|=\\left|ultimatepiece\\right|=\\cdots=\\left|firstdatum\\right|=flatness \\neq 0\n\\]\nand if \\( { }_{emptiness} T_{entirety} \\) denotes the sum of all products of these \\( emptiness \\) numbers taken \\( entirety \\) at a time, prove that\n\\[\n\\left|\\frac{{ }_{emptiness} T_{entirety}}{{ }_{emptiness} T_{emptiness-entirety}}\\right|=flatness^{2 entirety-emptiness}\n\\]\nwhenever the denominator of the left-hand side is different from zero.",
      "solution": "Solution. For any non-zero complex number, \\( realzero \\), we have \\( realzero \\quad 1=\\bar{realzero} /|realzero|^{2} \\) where the bar denotes complex conjugation. If \\( singularset \\) is a set of \\( entirety \\) indices selected from \\{1,2, \\ldots, emptiness\\}, then\n\\[\n\\begin{aligned}\n\\prod_{constantidx £ singularset} staticunit & =finaldatum ultimatepiece \\cdots firstdatum \\prod_{constantidx \\in singularset} staticunit^{-1}=finaldatum ultimatepiece \\cdots firstdatum \\prod_{constantidx \\in singularset} \\frac{\\bar{staticunit}}{flatness^{2}} \\\\\n& =\\frac{finaldatum ultimatepiece \\cdots firstdatum}{flatness^{2 entirety}} \\prod_{constantidx \\in singularset} \\bar{staticunit}\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\n{ }_{emptiness} T_{emptiness-entirety} & =\\sum_{singularset} \\prod_{constantidx \\notin singularset} staticunit=\\frac{finaldatum ultimatepiece \\cdots firstdatum}{flatness^{2 entirety}} \\sum_{singularset} \\prod_{constantidx \\in singularset} \\overline{finaldatum} \\\\\n& =\\frac{finaldatum ultimatepiece \\cdots firstdatum}{flatness^{2 entirety}} \\bar{T}_{entirety} .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left.\\left|{ }_{emptiness} T_{emptiness-entirety}\\right|=\\left.\\frac{flatness^{emptiness}}{flatness^{2 entirety}}\\right|_{emptiness} \\overline{T_{entirety}}\\left|=flatness^{emptiness-2 entirety}\\right|_{emptiness} T_{entirety} \\right\\rvert\\,,\n\\]\nwhence the required formula follows at once."
    },
    "garbled_string": {
      "map": {
        "a_1": "qzxwvtnp",
        "a_2": "hjgrksla",
        "a_n": "mnbvcxzr",
        "a_i": "plmoknji",
        "i": "qazwsxed",
        "z": "edcrfvtg",
        "J": "yhnujmik",
        "n": "lkjhgfdsa",
        "r": "poiuytre",
        "s": "mikojnuh"
      },
      "question": "4. If \\( qzxwvtnp, hjgrksla, \\ldots, mnbvcxzr \\) are complex numbers such that\n\\[\n\\left|qzxwvtnp\\right|=\\left|hjgrksla\\right|=\\cdots=\\left|mnbvcxzr\\right|=poiuytre \\neq 0\n\\]\nand if \\( { }_{lkjhgfdsa} T_{mikojnuh} \\) denotes the sum of all products of these \\( lkjhgfdsa \\) numbers taken \\( mikojnuh \\) at a time, prove that\n\\[\n\\left|\\frac{{ }_{lkjhgfdsa} T_{mikojnuh}}{{ }_{lkjhgfdsa} T_{lkjhgfdsa-mikojnuh}}\\right|=poiuytre^{2 mikojnuh-lkjhgfdsa}\n\\]\nwhenever the denominator of the left-hand side is different from zero.",
      "solution": "Solution. For any non-zero complex number, \\( edcrfvtg \\), we have \\( edcrfvtg \\quad 1=\\overline{edcrfvtg} /|edcrfvtg|^{2} \\) where the bar denotes complex conjugation. If \\( yhnujmik \\) is a set of \\( mikojnuh \\) indices selected from \\( \\{1,2, \\ldots, lkjhgfdsa\\} \\), then\n\\[\n\\begin{aligned}\n\\prod_{qazwsxed £ yhnujmik} plmoknji & =qzxwvtnp\\, hjgrksla \\cdots mnbvcxzr \\prod_{qazwsxed \\in yhnujmik} plmoknji^{-1}=qzxwvtnp\\, hjgrksla \\cdots mnbvcxzr \\prod_{qazwsxed \\in yhnujmik} \\frac{\\overline{plmoknji}}{poiuytre^{2}} \\\\\n& =\\frac{qzxwvtnp\\, hjgrksla \\cdots mnbvcxzr}{poiuytre^{2 mikojnuh}} \\prod_{qazwsxed \\in yhnujmik} \\overline{plmoknji}\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\n{ }_{lkjhgfdsa} T_{lkjhgfdsa-mikojnuh} & =\\sum_{yhnujmik} \\prod_{qazwsxed \\notin yhnujmik} plmoknji=\\frac{qzxwvtnp\\, hjgrksla \\cdots mnbvcxzr}{poiuytre^{2 mikojnuh}} \\sum_{yhnujmik} \\prod_{qazwsxed \\in yhnujmik} \\overline{qzxwvtnp} \\\\\n& =\\frac{qzxwvtnp\\, hjgrksla \\cdots mnbvcxzr}{poiuytre^{2 mikojnuh}} \\overline{T}_{mikojnuh} .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left.\\left|{ }_{lkjhgfdsa} T_{lkjhgfdsa-mikojnuh}\\right|=\\left.\\frac{poiuytre^{lkjhgfdsa}}{poiuytre^{2 mikojnuh}}\\right|_{lkjhgfdsa} \\overline{T_{mikojnuh}}\\left|=poiuytre^{lkjhgfdsa-2 mikojnuh}\\right|_{lkjhgfdsa} T_{mikojnuh} \\right\\rvert\\,,\n\\]\nwhence the required formula follows at once."
    },
    "kernel_variant": {
      "question": "Let $m\\ge 2$ be an integer and let $b_{1},\\dots ,b_{m}$ be non–zero complex numbers.  \nFor $0\\le k\\le m$ put  \n\\[\n\\sigma_{k}\\;=\\;\\sum_{1\\le i_{1}<\\dots <i_{k}\\le m} b_{i_{1}}\\dots b_{i_{k}},\n\\qquad \n\\sigma_{0}:=1 .\n\\]\n\nFix a real number $\\rho>0$ and introduce the anti–linear involution  \n\\[\n\\varphi_{\\rho}: \\Bbb C\\setminus\\{0\\}\\longrightarrow\\Bbb C\\setminus\\{0\\},\n\\qquad \n\\varphi_{\\rho}(z)=\\dfrac{\\rho^{2}}{\\overline z}.\n\\tag{1}\n\\]\n\nDefine the monic polynomial  \n\\[\nP(z)=\\sum_{k=0}^{m} (-1)^{k}\\sigma_{k}\\,z^{\\,m-k}\n        =\\prod_{j=1}^{m}(z-b_{j}),\n\\tag{2}\n\\]\nand its $\\rho$–conjugate–reciprocal  \n\\[\nP^{\\sharp}(z):=z^{\\,m}\\,\\overline{P\\!\\left(\\dfrac{\\rho^{2}}{\\overline z}\\right)} .\n\\tag{3}\n\\]\n\nThroughout we shall meet two different coefficient relations\n\\[\n(R_{\\rho})\\;:\\; \n\\sigma_{m-k}\\neq 0\\;\\Longrightarrow\\;\n\\left|\\frac{\\sigma_{k}}{\\sigma_{m-k}}\\right|\n          =\\rho^{\\,2k-m}\n\\quad (0\\le k\\le m),\n\\tag{4}\n\\]\n\n\\[\n(A_{\\rho})\\;:\\;\\;\n\\text{there exists } \\lambda\\in\\Bbb C,\\;|\\lambda|=1\n\\text{ such that }\\;\n\\overline{\\sigma_{k}}\n      =\\lambda\\,(-1)^{m}\\rho^{\\,2k-m}\\,\\sigma_{m-k}\n\\quad (0\\le k\\le m).\n\\tag{5}\n\\]\n(Thus $(A_{\\rho})$ refines $(R_{\\rho})$ by prescribing the phases.)\n\nQuestions  \n\na)  Prove that $(A_{\\rho})$ is equivalent to  \n\\[\nP^{\\sharp}(z)=\\lambda\\,\\rho^{\\,m}\\,P(z)\\qquad(\\forall z\\in\\Bbb C);\n\\tag{6}\n\\]\npolynomials satisfying (6) are called $\\rho$–self–reciprocal.\n\nb)  Show that $\\rho$–self–reciprocity is equivalent to invariance of the\nroot multiset under $\\varphi_{\\rho}$; namely,\n\\[\n\\varphi_{\\rho}(B)=B,\n\\qquad B=\\{b_{1},\\dots ,b_{m}\\}.\n\\tag{7}\n\\]\nHence the roots can be reordered as  \n\\[\nB=\\{w_{1},\\varphi_{\\rho}(w_{1}),\\dots ,w_{s},\\varphi_{\\rho}(w_{s}),\n      u_{1},\\dots ,u_{t}\\},\n\\tag{8}\n\\]\nwith $|w_{j}|\\neq\\rho$ and $|u_{i}|=\\rho$, $2s+t=m$.  \nConversely every multiset of the form (8) yields a $\\rho$–self–reciprocal\npolynomial.\n\nc)  Suppose $P$ is $\\rho$–self–reciprocal.  \nProve that the following are equivalent:\n\n(i)  $|b_{j}|=\\rho$ for every $j$;\n\n(ii) $B$ contains no distinct $\\varphi_{\\rho}$–pair, that is,\n$\\varphi_{\\rho}(b_{j})=b_{j}$ for all $j$.\n\nIn particular, under $(A_{\\rho})$ the condition “no distinct\n$\\varphi_{\\rho}$–pair’’ is equivalent to $|b_{j}|=\\rho\\;(\\forall j)$.\n\nd)  Show that $(R_{\\rho})$ by itself does \\emph{not} imply\n$\\rho$–self–reciprocity.  More precisely, let $m=2$.\n\ni)   Prove that $(R_{\\rho})$ is equivalent to $|b_{1}b_{2}|=\\rho^{2}$.\n\nii)  Prove that $(A_{\\rho})$ additionally holds iff, with the unimodular\nconstant \n\\[\n\\lambda=\\frac{\\overline{b_{1}b_{2}}}{|b_{1}b_{2}|},\n\\]\none has  \n\\[\n\\overline{b_{1}+b_{2}}=\\lambda\\,(b_{1}+b_{2}).\n\\tag{9}\n\\]\n\niii) Exhibit concrete numbers\n\\[\nb_{1}=2i,\\qquad b_{2}=1,\\qquad \\rho=\\sqrt 2,\n\\]\nfor which $(R_{\\rho})$ is valid while (9) fails, thereby giving a\nquadratic example with $(R_{\\rho})$ but \\emph{not} $(A_{\\rho})$.\n\n\n\n----------------------------------------------------------------------",
      "solution": "We keep the notation of the statement throughout.\n\n0.  Coefficients of $P^{\\sharp}$.  \nExpanding (3) with (2) gives  \n\\[\n\\begin{aligned}\nP^{\\sharp}(z)\n      &=z^{\\,m}\\sum_{k=0}^{m}(-1)^{k}\\,\n           \\overline{\\sigma_{k}}\\,\n           \\left(\\dfrac{\\rho^{2}}{\\overline z}\\right)^{m-k}\n      =\\sum_{k=0}^{m}(-1)^{k}\\rho^{\\,2(m-k)}\n           \\overline{\\sigma_{k}}\\,\n           z^{\\,k}.\n\\end{aligned}\n\\tag{10}\n\\]\n\nHence, for $0\\le k\\le m$,  \n\\[\np_{k}=(-1)^{\\,m-k}\\sigma_{m-k}= \\operatorname{coeff}_{z^{\\,k}}P,\n\\qquad\nq_{k}=(-1)^{k}\\rho^{\\,2(m-k)}\n      \\overline{\\sigma_{k}}\n      =\\operatorname{coeff}_{z^{\\,k}}P^{\\sharp}.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nPart (a) $(A_{\\rho})\\Longleftrightarrow$ (6)\n--------------------------------------------------------------------\n\n(i)  $(A_{\\rho})\\Rightarrow$ (6).  \nInsert (5) into $q_{k}$:\n\\[\n\\begin{aligned}\nq_{k}\n   &=(-1)^{k}\\rho^{\\,2(m-k)}\n       \\lambda\\,(-1)^{m}\\rho^{\\,2k-m}\\,\\sigma_{m-k}  \\\\\n   &=\\lambda\\,(-1)^{m-k}\\rho^{\\,m}\\sigma_{m-k}\n   =\\lambda\\,\\rho^{\\,m}\\,p_{k}\\qquad(0\\le k\\le m).\n\\end{aligned}\n\\]\nThus $P^{\\sharp}=\\lambda\\rho^{m}P$.\n\n(ii)  (6)$\\Rightarrow(A_{\\rho})$.  \nAssume $P^{\\sharp}=cP$ with $c\\in\\Bbb C^{\\times}$.  \nComparing the coefficients of $z^{m}$ in $P^{\\sharp}$ and $P$ (see (2),\n(10)) gives  \n\\[\nc=(-1)^{m}\\overline{\\sigma_{m}},\\qquad |c|=|\\sigma_{m}|.\n\\tag{12}\n\\]\nComparing the constant terms yields  \n\\[\nc=\\frac{\\rho^{2m}}{(-1)^{m}\\sigma_{m}},\\qquad\n|c|=\\frac{\\rho^{2m}}{|\\sigma_{m}|}.\n\\tag{13}\n\\]\nEquating the moduli in (12) and (13) gives $|\\sigma_{m}|^{2}=\\rho^{2m}$,\nhence $|\\sigma_{m}|=\\rho^{m}$ and $|c|=\\rho^{m}$.\nWrite $c=\\lambda\\rho^{m}$ with $|\\lambda|=1$.  \nBecause $P^{\\sharp}=cP$ we have $q_{k}=c\\,p_{k}$ for \\emph{every}\n$k=0,\\dots ,m$.  Substituting the explicit forms (11) gives\n\\[\n(-1)^{k}\\rho^{\\,2(m-k)}\\overline{\\sigma_{k}}\n      =\\lambda\\rho^{m}(-1)^{\\,m-k}\\sigma_{m-k},\n\\]\nwhich rearranges exactly to (5) for all $k$.  \nThus (6) implies $(A_{\\rho})$.\n\n--------------------------------------------------------------------\nPart (b) $\\rho$–self–reciprocity $\\Longleftrightarrow$\n          $\\varphi_{\\rho}$–invariance\n--------------------------------------------------------------------\n\nSuppose $P$ satisfies (6).  If $b\\in B$ then\n\\[\n0=P(b)=\\lambda^{-1}\\rho^{-m}P^{\\sharp}(b)\n      =\\lambda^{-1}\\rho^{-m}b^{\\,m}\\,\n        \\overline{P\\!\\bigl(\\varphi_{\\rho}(b)\\bigr)} ,\n\\]\nso $P\\!\\bigl(\\varphi_{\\rho}(b)\\bigr)=0$ and $\\varphi_{\\rho}(b)\\in B$.\nCounting multiplicities proves (7).\n\nConversely, assume (7).  \nThen $P$ and $P^{\\sharp}$ share the same zeros, hence\n$P^{\\sharp}=cP$ for some $c\\neq 0$.\nUsing the argument of part (a) we already know $|c|=\\rho^{m}$, so\n$c=\\lambda\\rho^{m}$ with $|\\lambda|=1$ and (6) holds.\n\nRoot decomposition.  \nBecause $\\varphi_{\\rho}$ is an involution, each orbit in $B$ is either a\ntwo–cycle $\\{w,\\varphi_{\\rho}(w)\\}$ with $|w|\\neq\\rho$ or a fixed point\n$u$ with $|u|=\\rho$.  Re–ordering the roots yields (8).  \nConversely any multiset of the form (8) is $\\varphi_{\\rho}$–stable, so\nits monic polynomial is $\\rho$–self–reciprocal by the first part of (b).\n\n--------------------------------------------------------------------\nPart (c) All roots on the circle\n--------------------------------------------------------------------\n\nAssume $P$ is $\\rho$–self–reciprocal.  \n\n(i)$\\Rightarrow$(ii).  \nIf $|b_{j}|=\\rho$ then\n$\\varphi_{\\rho}(b_{j})=\\rho^{2}/\\overline{b_{j}}=b_{j}$, so $B$ contains\nno distinct $\\varphi_{\\rho}$–pair.\n\n(ii)$\\Rightarrow$(i).  \nConversely, if every $\\varphi_{\\rho}$–orbit is trivial then (8) has\n$s=0$, hence every root is fixed by $\\varphi_{\\rho}$ and consequently\nhas modulus $\\rho$.  \n\nSince $\\rho$–self–reciprocity is equivalent to $(A_{\\rho})$ by (a), the\nlast sentence follows immediately.\n\n--------------------------------------------------------------------\nPart (d) $(R_{\\rho})\\nRightarrow(A_{\\rho})$\n          in the quadratic case\n--------------------------------------------------------------------\n\nLet $m=2$.  Then $\\sigma_{1}=b_{1}+b_{2}$ and $\\sigma_{2}=b_{1}b_{2}$.\n\n(i)  From $k=0$ or $k=2$ in (4) we obtain  \n\\[\n|\\sigma_{2}|=\\rho^{2}\\quad\\Longleftrightarrow\\quad\n|b_{1}b_{2}|=\\rho^{2},\n\\]\nwhile the $k=1$ condition is automatically satisfied.\nThus $(R_{\\rho})\\iff |b_{1}b_{2}|=\\rho^{2}$.\n\n(ii)  Assume $(R_{\\rho})$.  \nSet \n\\[\n\\lambda=\\frac{\\overline{\\sigma_{2}}}{|\\sigma_{2}|}\\quad(|\\lambda|=1).\n\\]\nCondition (5) for $k=2$ is then automatically fulfilled, and for $k=0$\nis trivial.  The remaining case $k=1$ yields precisely (9).  \nHence $(A_{\\rho})$ holds iff (9) is satisfied.\n\n(iii)  Take  \n\\[\nb_{1}=2i,\\qquad b_{2}=1,\\qquad \\rho=\\sqrt 2 .\n\\]\nThen $\\sigma_{2}=2i$ so $|\\sigma_{2}|=2=\\rho^{2}$, i.e.\\ $(R_{\\rho})$\nholds.  With \n\\[\n\\lambda=\\frac{\\overline{\\sigma_{2}}}{|\\sigma_{2}|}\n        =\\frac{-2i}{2}=-i\n\\]\nwe have  \n\\[\n\\overline{b_{1}+b_{2}}=1-2i,\n\\qquad\n\\lambda\\,(b_{1}+b_{2})=-i(1+2i)=2-i,\n\\]\nwhich are different; hence (9) fails and $(A_{\\rho})$ is not satisfied.\nThis exhibits a quadratic polynomial for which $(R_{\\rho})$ holds\nwhereas $(A_{\\rho})$ (and therefore $\\rho$–self–reciprocity) fails.\n\n\\hfill$\\square$\n\n\n----------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.505188",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra directions: the original problem only asks for the forward implication “equal moduli ⇒ ratio of symmetric sums”.  The enhanced variant additionally demands  \n   • the converse (deducing equal moduli from the ratios),  \n   • a functional identity between a polynomial and its reciprocal, and  \n   • a spectral characterisation of the location of its zeros.  \n   Each new claim introduces a fresh layer of algebraic and analytic reasoning.\n\n2. Technical depth: proving the converse requires comparing two polynomials whose coefficients match in modulus but not necessarily in argument, and showing that they must possess the same multiset of zeros.  This employs\n   • reversed–conjugate (reciprocal) polynomials,  \n   • identification of root multisets via coefficient data, and  \n   • a multiset argument to force individual modulus equality.\n\n3. Interacting concepts: the solution blends classical symmetric-function manipulations with complex conjugation, polynomial reciprocity, and root-set arguments.  The final equivalence in part (c) brings in elementary potential theory (location of zeros on a circle) tying algebraic constraints on coefficients to geometric constraints on roots.\n\n4. Increased workload: besides re-proving the original statement, the solver must create and analyse a new polynomial (the reciprocal), handle unimodular ambiguities, and carry out a non-trivial multiset comparison—all steps absent from the original exercise.\n\nThese additions push the problem well beyond routine symmetric-sum manipulation, making it substantially harder than both the original statement and its immediate kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $m\\ge 2$ be an integer and let $b_{1},\\dots ,b_{m}$ be non–zero complex numbers.  \nFor $0\\le k\\le m$ put  \n\\[\n\\sigma_{k}\\;=\\;\\sum_{1\\le i_{1}<\\dots <i_{k}\\le m} b_{i_{1}}\\dots b_{i_{k}},\n\\qquad \n\\sigma_{0}:=1 .\n\\]\n\nFix a real number $\\rho>0$ and introduce the anti–linear involution  \n\\[\n\\varphi_{\\rho}: \\Bbb C\\setminus\\{0\\}\\longrightarrow\\Bbb C\\setminus\\{0\\},\n\\qquad \n\\varphi_{\\rho}(z)=\\dfrac{\\rho^{2}}{\\overline z}.\n\\tag{1}\n\\]\n\nDefine the monic polynomial  \n\\[\nP(z)=\\sum_{k=0}^{m} (-1)^{k}\\sigma_{k}\\,z^{\\,m-k}\n        =\\prod_{j=1}^{m}(z-b_{j}),\n\\tag{2}\n\\]\nand its $\\rho$–conjugate–reciprocal  \n\\[\nP^{\\sharp}(z):=z^{\\,m}\\,\\overline{P\\!\\left(\\dfrac{\\rho^{2}}{\\overline z}\\right)} .\n\\tag{3}\n\\]\n\nThroughout we shall meet two different coefficient relations\n\\[\n(R_{\\rho})\\;:\\; \n\\sigma_{m-k}\\neq 0\\;\\Longrightarrow\\;\n\\left|\\frac{\\sigma_{k}}{\\sigma_{m-k}}\\right|\n          =\\rho^{\\,2k-m}\n\\quad (0\\le k\\le m),\n\\tag{4}\n\\]\n\n\\[\n(A_{\\rho})\\;:\\;\\;\n\\text{there exists } \\lambda\\in\\Bbb C,\\;|\\lambda|=1\n\\text{ such that }\\;\n\\overline{\\sigma_{k}}\n      =\\lambda\\,(-1)^{m}\\rho^{\\,2k-m}\\,\\sigma_{m-k}\n\\quad (0\\le k\\le m).\n\\tag{5}\n\\]\n(Thus $(A_{\\rho})$ refines $(R_{\\rho})$ by prescribing the phases.)\n\nQuestions  \n\na)  Prove that $(A_{\\rho})$ is equivalent to  \n\\[\nP^{\\sharp}(z)=\\lambda\\,\\rho^{\\,m}\\,P(z)\\qquad(\\forall z\\in\\Bbb C);\n\\tag{6}\n\\]\npolynomials satisfying (6) are called $\\rho$–self–reciprocal.\n\nb)  Show that $\\rho$–self–reciprocity is equivalent to invariance of the\nroot multiset under $\\varphi_{\\rho}$; namely,\n\\[\n\\varphi_{\\rho}(B)=B,\n\\qquad B=\\{b_{1},\\dots ,b_{m}\\}.\n\\tag{7}\n\\]\nHence the roots can be reordered as  \n\\[\nB=\\{w_{1},\\varphi_{\\rho}(w_{1}),\\dots ,w_{s},\\varphi_{\\rho}(w_{s}),\n      u_{1},\\dots ,u_{t}\\},\n\\tag{8}\n\\]\nwith $|w_{j}|\\neq\\rho$ and $|u_{i}|=\\rho$, $2s+t=m$.  \nConversely every multiset of the form (8) yields a $\\rho$–self–reciprocal\npolynomial.\n\nc)  Suppose $P$ is $\\rho$–self–reciprocal.  \nProve that the following are equivalent:\n\n(i)  $|b_{j}|=\\rho$ for every $j$;\n\n(ii) $B$ contains no distinct $\\varphi_{\\rho}$–pair, that is,\n$\\varphi_{\\rho}(b_{j})=b_{j}$ for all $j$.\n\nIn particular, under $(A_{\\rho})$ the condition “no distinct\n$\\varphi_{\\rho}$–pair’’ is equivalent to $|b_{j}|=\\rho\\;(\\forall j)$.\n\nd)  Show that $(R_{\\rho})$ by itself does \\emph{not} imply\n$\\rho$–self–reciprocity.  More precisely, let $m=2$.\n\ni)   Prove that $(R_{\\rho})$ is equivalent to $|b_{1}b_{2}|=\\rho^{2}$.\n\nii)  Prove that $(A_{\\rho})$ additionally holds iff, with the unimodular\nconstant \n\\[\n\\lambda=\\frac{\\overline{b_{1}b_{2}}}{|b_{1}b_{2}|},\n\\]\none has  \n\\[\n\\overline{b_{1}+b_{2}}=\\lambda\\,(b_{1}+b_{2}).\n\\tag{9}\n\\]\n\niii) Exhibit concrete numbers\n\\[\nb_{1}=2i,\\qquad b_{2}=1,\\qquad \\rho=\\sqrt 2,\n\\]\nfor which $(R_{\\rho})$ is valid while (9) fails, thereby giving a\nquadratic example with $(R_{\\rho})$ but \\emph{not} $(A_{\\rho})$.\n\n\n\n----------------------------------------------------------------------",
      "solution": "We keep the notation of the statement throughout.\n\n0.  Coefficients of $P^{\\sharp}$.  \nExpanding (3) with (2) gives  \n\\[\n\\begin{aligned}\nP^{\\sharp}(z)\n      &=z^{\\,m}\\sum_{k=0}^{m}(-1)^{k}\\,\n           \\overline{\\sigma_{k}}\\,\n           \\left(\\dfrac{\\rho^{2}}{\\overline z}\\right)^{m-k}\n      =\\sum_{k=0}^{m}(-1)^{k}\\rho^{\\,2(m-k)}\n           \\overline{\\sigma_{k}}\\,\n           z^{\\,k}.\n\\end{aligned}\n\\tag{10}\n\\]\n\nHence, for $0\\le k\\le m$,  \n\\[\np_{k}=(-1)^{\\,m-k}\\sigma_{m-k}= \\operatorname{coeff}_{z^{\\,k}}P,\n\\qquad\nq_{k}=(-1)^{k}\\rho^{\\,2(m-k)}\n      \\overline{\\sigma_{k}}\n      =\\operatorname{coeff}_{z^{\\,k}}P^{\\sharp}.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nPart (a) $(A_{\\rho})\\Longleftrightarrow$ (6)\n--------------------------------------------------------------------\n\n(i)  $(A_{\\rho})\\Rightarrow$ (6).  \nInsert (5) into $q_{k}$:\n\\[\n\\begin{aligned}\nq_{k}\n   &=(-1)^{k}\\rho^{\\,2(m-k)}\n       \\lambda\\,(-1)^{m}\\rho^{\\,2k-m}\\,\\sigma_{m-k}  \\\\\n   &=\\lambda\\,(-1)^{m-k}\\rho^{\\,m}\\sigma_{m-k}\n   =\\lambda\\,\\rho^{\\,m}\\,p_{k}\\qquad(0\\le k\\le m).\n\\end{aligned}\n\\]\nThus $P^{\\sharp}=\\lambda\\rho^{m}P$.\n\n(ii)  (6)$\\Rightarrow(A_{\\rho})$.  \nAssume $P^{\\sharp}=cP$ with $c\\in\\Bbb C^{\\times}$.  \nComparing the coefficients of $z^{m}$ in $P^{\\sharp}$ and $P$ (see (2),\n(10)) gives  \n\\[\nc=(-1)^{m}\\overline{\\sigma_{m}},\\qquad |c|=|\\sigma_{m}|.\n\\tag{12}\n\\]\nComparing the constant terms yields  \n\\[\nc=\\frac{\\rho^{2m}}{(-1)^{m}\\sigma_{m}},\\qquad\n|c|=\\frac{\\rho^{2m}}{|\\sigma_{m}|}.\n\\tag{13}\n\\]\nEquating the moduli in (12) and (13) gives $|\\sigma_{m}|^{2}=\\rho^{2m}$,\nhence $|\\sigma_{m}|=\\rho^{m}$ and $|c|=\\rho^{m}$.\nWrite $c=\\lambda\\rho^{m}$ with $|\\lambda|=1$.  \nBecause $P^{\\sharp}=cP$ we have $q_{k}=c\\,p_{k}$ for \\emph{every}\n$k=0,\\dots ,m$.  Substituting the explicit forms (11) gives\n\\[\n(-1)^{k}\\rho^{\\,2(m-k)}\\overline{\\sigma_{k}}\n      =\\lambda\\rho^{m}(-1)^{\\,m-k}\\sigma_{m-k},\n\\]\nwhich rearranges exactly to (5) for all $k$.  \nThus (6) implies $(A_{\\rho})$.\n\n--------------------------------------------------------------------\nPart (b) $\\rho$–self–reciprocity $\\Longleftrightarrow$\n          $\\varphi_{\\rho}$–invariance\n--------------------------------------------------------------------\n\nSuppose $P$ satisfies (6).  If $b\\in B$ then\n\\[\n0=P(b)=\\lambda^{-1}\\rho^{-m}P^{\\sharp}(b)\n      =\\lambda^{-1}\\rho^{-m}b^{\\,m}\\,\n        \\overline{P\\!\\bigl(\\varphi_{\\rho}(b)\\bigr)} ,\n\\]\nso $P\\!\\bigl(\\varphi_{\\rho}(b)\\bigr)=0$ and $\\varphi_{\\rho}(b)\\in B$.\nCounting multiplicities proves (7).\n\nConversely, assume (7).  \nThen $P$ and $P^{\\sharp}$ share the same zeros, hence\n$P^{\\sharp}=cP$ for some $c\\neq 0$.\nUsing the argument of part (a) we already know $|c|=\\rho^{m}$, so\n$c=\\lambda\\rho^{m}$ with $|\\lambda|=1$ and (6) holds.\n\nRoot decomposition.  \nBecause $\\varphi_{\\rho}$ is an involution, each orbit in $B$ is either a\ntwo–cycle $\\{w,\\varphi_{\\rho}(w)\\}$ with $|w|\\neq\\rho$ or a fixed point\n$u$ with $|u|=\\rho$.  Re–ordering the roots yields (8).  \nConversely any multiset of the form (8) is $\\varphi_{\\rho}$–stable, so\nits monic polynomial is $\\rho$–self–reciprocal by the first part of (b).\n\n--------------------------------------------------------------------\nPart (c) All roots on the circle\n--------------------------------------------------------------------\n\nAssume $P$ is $\\rho$–self–reciprocal.  \n\n(i)$\\Rightarrow$(ii).  \nIf $|b_{j}|=\\rho$ then\n$\\varphi_{\\rho}(b_{j})=\\rho^{2}/\\overline{b_{j}}=b_{j}$, so $B$ contains\nno distinct $\\varphi_{\\rho}$–pair.\n\n(ii)$\\Rightarrow$(i).  \nConversely, if every $\\varphi_{\\rho}$–orbit is trivial then (8) has\n$s=0$, hence every root is fixed by $\\varphi_{\\rho}$ and consequently\nhas modulus $\\rho$.  \n\nSince $\\rho$–self–reciprocity is equivalent to $(A_{\\rho})$ by (a), the\nlast sentence follows immediately.\n\n--------------------------------------------------------------------\nPart (d) $(R_{\\rho})\\nRightarrow(A_{\\rho})$\n          in the quadratic case\n--------------------------------------------------------------------\n\nLet $m=2$.  Then $\\sigma_{1}=b_{1}+b_{2}$ and $\\sigma_{2}=b_{1}b_{2}$.\n\n(i)  From $k=0$ or $k=2$ in (4) we obtain  \n\\[\n|\\sigma_{2}|=\\rho^{2}\\quad\\Longleftrightarrow\\quad\n|b_{1}b_{2}|=\\rho^{2},\n\\]\nwhile the $k=1$ condition is automatically satisfied.\nThus $(R_{\\rho})\\iff |b_{1}b_{2}|=\\rho^{2}$.\n\n(ii)  Assume $(R_{\\rho})$.  \nSet \n\\[\n\\lambda=\\frac{\\overline{\\sigma_{2}}}{|\\sigma_{2}|}\\quad(|\\lambda|=1).\n\\]\nCondition (5) for $k=2$ is then automatically fulfilled, and for $k=0$\nis trivial.  The remaining case $k=1$ yields precisely (9).  \nHence $(A_{\\rho})$ holds iff (9) is satisfied.\n\n(iii)  Take  \n\\[\nb_{1}=2i,\\qquad b_{2}=1,\\qquad \\rho=\\sqrt 2 .\n\\]\nThen $\\sigma_{2}=2i$ so $|\\sigma_{2}|=2=\\rho^{2}$, i.e.\\ $(R_{\\rho})$\nholds.  With \n\\[\n\\lambda=\\frac{\\overline{\\sigma_{2}}}{|\\sigma_{2}|}\n        =\\frac{-2i}{2}=-i\n\\]\nwe have  \n\\[\n\\overline{b_{1}+b_{2}}=1-2i,\n\\qquad\n\\lambda\\,(b_{1}+b_{2})=-i(1+2i)=2-i,\n\\]\nwhich are different; hence (9) fails and $(A_{\\rho})$ is not satisfied.\nThis exhibits a quadratic polynomial for which $(R_{\\rho})$ holds\nwhereas $(A_{\\rho})$ (and therefore $\\rho$–self–reciprocity) fails.\n\n\\hfill$\\square$\n\n\n----------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.421970",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra directions: the original problem only asks for the forward implication “equal moduli ⇒ ratio of symmetric sums”.  The enhanced variant additionally demands  \n   • the converse (deducing equal moduli from the ratios),  \n   • a functional identity between a polynomial and its reciprocal, and  \n   • a spectral characterisation of the location of its zeros.  \n   Each new claim introduces a fresh layer of algebraic and analytic reasoning.\n\n2. Technical depth: proving the converse requires comparing two polynomials whose coefficients match in modulus but not necessarily in argument, and showing that they must possess the same multiset of zeros.  This employs\n   • reversed–conjugate (reciprocal) polynomials,  \n   • identification of root multisets via coefficient data, and  \n   • a multiset argument to force individual modulus equality.\n\n3. Interacting concepts: the solution blends classical symmetric-function manipulations with complex conjugation, polynomial reciprocity, and root-set arguments.  The final equivalence in part (c) brings in elementary potential theory (location of zeros on a circle) tying algebraic constraints on coefficients to geometric constraints on roots.\n\n4. Increased workload: besides re-proving the original statement, the solver must create and analyse a new polynomial (the reciprocal), handle unimodular ambiguities, and carry out a non-trivial multiset comparison—all steps absent from the original exercise.\n\nThese additions push the problem well beyond routine symmetric-sum manipulation, making it substantially harder than both the original statement and its immediate kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}