{
  "index": "1957-A-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. \\( P(z) \\) is a complex polynomial whose roots (as points in the Argand plane) can be covered by a closed circular disc of radius \\( R \\). Show that the roots of \\( n P(z)-k P^{\\prime}(z) \\) can be covered by a closed circular disc of radius \\( R \\) \\( +|k| \\), where \\( n \\) is the degree of \\( P(z), k \\) is any complex number, and \\( P^{\\prime}(z) \\) is the derivative of \\( P(z) \\).",
  "solution": "Solution. Suppose that the roots of \\( P(z) \\) all lie in the closed disc, \\( D_{1} \\), \\( |z-c| \\leq R \\). We shali prove that the roots of \\( n P(z)-k P^{\\prime}(z) \\) lie in the closed disc, \\( D_{2},|z-c| \\leq R+|k| \\).\n\nIf the roots of \\( P(z) \\) are \\( \\lambda_{1}, \\lambda_{2}, \\ldots, \\lambda_{n} \\), then \\( P(z)=A\\left(z-\\lambda_{1}\\right)\\left(z-\\lambda_{2}\\right) \\) \\( \\cdots\\left(z-\\lambda_{n}\\right) \\) where \\( A \\) is a constant. Taking the logarithmic derivative, we find\n\\[\n\\frac{P^{\\prime}(z)}{P(z)}=\\sum_{i=1}^{n} \\frac{1}{z-\\lambda_{i}} .\n\\]\n\nSuppose \\( u \\notin D_{2} \\). Then\n\\[\n\\left|u-\\lambda_{i}\\right| \\geq|u-c|-\\left|\\lambda_{i}-c\\right|>R+|k|-R=|k|\n\\]\nsince \\( \\lambda_{i} \\in D_{1} \\). Therefore, for \\( i=1,2, \\ldots, n, \\frac{|k|}{\\left|u-\\lambda_{i}\\right|}<1 \\), and\n\\[\n\\begin{aligned}\n\\left|n P(u)-k P^{\\prime}(u)\\right| & =|P(u)| \\cdot\\left|n-k \\sum_{i=1}^{n} \\frac{1}{u-\\lambda_{i}}\\right| \\\\\n& \\geq|P(u)| \\cdot\\left(n-\\sum_{i=1}^{n} \\frac{|k|}{\\left|u-\\lambda_{i}\\right|}\\right)>0\n\\end{aligned}\n\\]\n\nThus \\( u \\) is not a root of \\( n P(z)-k P^{\\prime}(z) \\).\nThis proves that all the roots of \\( n P(z)-k P^{\\prime}(z) \\) lie in \\( D_{2} \\) as claimed.",
  "vars": [
    "z",
    "u",
    "i",
    "\\\\lambda_i"
  ],
  "params": [
    "P",
    "R",
    "n",
    "k",
    "c",
    "A",
    "D_1",
    "D_2"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "complexvar",
        "u": "testpt",
        "i": "indexvar",
        "\\lambda_i": "rootvar",
        "P": "polyfunc",
        "R": "radiusr",
        "n": "polydeg",
        "k": "shiftvar",
        "c": "centerpt",
        "A": "constam",
        "D_1": "discone",
        "D_2": "disctwo"
      },
      "question": "4. \\( polyfunc(complexvar) \\) is a complex polynomial whose roots (as points in the Argand plane) can be covered by a closed circular disc of radius \\( radiusr \\). Show that the roots of \\( polydeg\\, polyfunc(complexvar)-shiftvar\\, polyfunc^{\\prime}(complexvar) \\) can be covered by a closed circular disc of radius \\( radiusr \\) \\( +|shiftvar| \\), where \\( polydeg \\) is the degree of \\( polyfunc(complexvar)\\), \\( shiftvar \\) is any complex number, and \\( polyfunc^{\\prime}(complexvar) \\) is the derivative of \\( polyfunc(complexvar) \\).",
      "solution": "Solution. Suppose that the roots of \\( polyfunc(complexvar) \\) all lie in the closed disc, \\( discone \\), \\( |complexvar-centerpt| \\leq radiusr \\). We shall prove that the roots of \\( polydeg\\, polyfunc(complexvar)-shiftvar\\, polyfunc^{\\prime}(complexvar) \\) lie in the closed disc, \\( disctwo,|complexvar-centerpt| \\leq radiusr+|shiftvar| \\).\n\nIf the roots of \\( polyfunc(complexvar) \\) are \\( rootvar_{1}, rootvar_{2}, \\ldots, rootvar_{polydeg} \\), then \\( polyfunc(complexvar)=constam\\left(complexvar-rootvar_{1}\\right)\\left(complexvar-rootvar_{2}\\right) \\cdots\\left(complexvar-rootvar_{polydeg}\\right) \\) where \\( constam \\) is a constant. Taking the logarithmic derivative, we find\n\\[\n\\frac{polyfunc^{\\prime}(complexvar)}{polyfunc(complexvar)}=\\sum_{indexvar=1}^{polydeg} \\frac{1}{complexvar-rootvar_{indexvar}} .\n\\]\n\nSuppose \\( testpt \\notin disctwo \\). Then\n\\[\n\\left|testpt-rootvar_{indexvar}\\right| \\geq|testpt-centerpt|-\\left|rootvar_{indexvar}-centerpt\\right|>radiusr+|shiftvar|-radiusr=|shiftvar|\n\\]\nsince \\( rootvar_{indexvar} \\in discone \\). Therefore, for \\( indexvar=1,2, \\ldots, polydeg, \\frac{|shiftvar|}{\\left|testpt-rootvar_{indexvar}\\right|}<1 \\), and\n\\[\n\\begin{aligned}\n\\left|polydeg\\, polyfunc(testpt)-shiftvar\\, polyfunc^{\\prime}(testpt)\\right| & =|polyfunc(testpt)| \\cdot\\left|polydeg-shiftvar \\sum_{indexvar=1}^{polydeg} \\frac{1}{testpt-rootvar_{indexvar}}\\right| \\\\\n& \\geq|polyfunc(testpt)| \\cdot\\left(polydeg-\\sum_{indexvar=1}^{polydeg} \\frac{|shiftvar|}{\\left|testpt-rootvar_{indexvar}\\right|}\\right)>0\n\\end{aligned}\n\\]\n\nThus \\( testpt \\) is not a root of \\( polydeg\\, polyfunc(complexvar)-shiftvar\\, polyfunc^{\\prime}(complexvar) \\).\nThis proves that all the roots of \\( polydeg\\, polyfunc(complexvar)-shiftvar\\, polyfunc^{\\prime}(complexvar) \\) lie in \\( disctwo \\) as claimed."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "pineapple",
        "u": "marigold",
        "i": "teaspoon",
        "\\lambda_i": "dragonfly",
        "P": "hemisphere",
        "R": "sandcastle",
        "n": "turnpike",
        "k": "blueberry",
        "c": "chandelier",
        "A": "sailboat",
        "D_1": "blackbird",
        "D_2": "hamilton"
      },
      "question": "4. \\( hemisphere(pineapple) \\) is a complex polynomial whose roots (as points in the Argand plane) can be covered by a closed circular disc of radius \\( sandcastle \\). Show that the roots of \\( turnpike\\, hemisphere(pineapple)-blueberry\\, hemisphere^{\\prime}(pineapple) \\) can be covered by a closed circular disc of radius \\( sandcastle \\) \\( +|blueberry| \\), where \\( turnpike \\) is the degree of \\( hemisphere(pineapple), blueberry \\) is any complex number, and \\( hemisphere^{\\prime}(pineapple) \\) is the derivative of \\( hemisphere(pineapple) \\).",
      "solution": "Solution. Suppose that the roots of \\( hemisphere(pineapple) \\) all lie in the closed disc, \\( blackbird \\), \\( |pineapple-chandelier| \\leq sandcastle \\). We shali prove that the roots of \\( turnpike hemisphere(pineapple)-blueberry hemisphere^{\\prime}(pineapple) \\) lie in the closed disc, \\( hamilton,|pineapple-chandelier| \\leq sandcastle+|blueberry| \\).\n\nIf the roots of \\( hemisphere(pineapple) \\) are \\( dragonfly_{1}, dragonfly_{2}, \\ldots, dragonfly_{turnpike} \\), then \\( hemisphere(pineapple)=sailboat\\left(pineapple-dragonfly_{1}\\right)\\left(pineapple-dragonfly_{2}\\right) \\cdots\\left(pineapple-dragonfly_{turnpike}\\right) \\) where \\( sailboat \\) is a constant. Taking the logarithmic derivative, we find\n\\[\n\\frac{hemisphere^{\\prime}(pineapple)}{hemisphere(pineapple)}=\\sum_{teaspoon=1}^{turnpike} \\frac{1}{pineapple-dragonfly_{teaspoon}} .\n\\]\n\nSuppose \\( marigold \\notin hamilton \\). Then\n\\[\n\\left|marigold-dragonfly_{teaspoon}\\right| \\geq|marigold-chandelier|-\\left|dragonfly_{teaspoon}-chandelier\\right|>sandcastle+|blueberry|-sandcastle=|blueberry|\n\\]\nsince \\( dragonfly_{teaspoon} \\in blackbird \\). Therefore, for \\( teaspoon=1,2, \\ldots, turnpike, \\frac{|blueberry|}{\\left|marigold-dragonfly_{teaspoon}\\right|}<1 \\), and\n\\[\n\\begin{aligned}\n\\left|turnpike\\, hemisphere(marigold)-blueberry\\, hemisphere^{\\prime}(marigold)\\right| & =|hemisphere(marigold)| \\cdot\\left|turnpike-blueberry \\sum_{teaspoon=1}^{turnpike} \\frac{1}{marigold-dragonfly_{teaspoon}}\\right| \\\\\n& \\geq|hemisphere(marigold)| \\cdot\\left(turnpike-\\sum_{teaspoon=1}^{turnpike} \\frac{|blueberry|}{\\left|marigold-dragonfly_{teaspoon}\\right|}\\right)>0\n\\end{aligned}\n\\]\n\nThus \\( marigold \\) is not a root of \\( turnpike hemisphere(pineapple)-blueberry hemisphere^{\\prime}(pineapple) \\).\nThis proves that all the roots of \\( turnpike hemisphere(pineapple)-blueberry hemisphere^{\\prime}(pineapple) \\) lie in \\( hamilton \\) as claimed."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "fixednumber",
        "u": "insidepoint",
        "\\lambda": "leafvalue",
        "P": "randomnoise",
        "R": "diameterlength",
        "n": "flatness",
        "k": "nullfactor",
        "c": "borderpoint",
        "A": "variablecoeff",
        "D_1": "annulusone",
        "D_2": "annulustwo"
      },
      "question": "4. \\( randomnoise(fixednumber) \\) is a complex polynomial whose roots (as points in the Argand plane) can be covered by a closed circular disc of radius \\( diameterlength \\). Show that the roots of \\( flatness\\,randomnoise(fixednumber)-nullfactor\\,randomnoise^{\\prime}(fixednumber) \\) can be covered by a closed circular disc of radius \\( diameterlength \\) \\( +|nullfactor| \\), where \\( flatness \\) is the degree of \\( randomnoise(fixednumber), nullfactor \\) is any complex number, and \\( randomnoise^{\\prime}(fixednumber) \\) is the derivative of \\( randomnoise(fixednumber) \\).",
      "solution": "Solution. Suppose that the roots of \\( randomnoise(fixednumber) \\) all lie in the closed disc, \\( annulusone \\), \\( |fixednumber-borderpoint| \\leq diameterlength \\). We shall prove that the roots of \\( flatness\\,randomnoise(fixednumber)-nullfactor\\,randomnoise^{\\prime}(fixednumber) \\) lie in the closed disc, \\( annulustwo,|fixednumber-borderpoint| \\leq diameterlength+|nullfactor| \\).\n\nIf the roots of \\( randomnoise(fixednumber) \\) are \\( leafvalue_{1}, leafvalue_{2}, \\ldots, leafvalue_{flatness} \\), then\n\\[\nrandomnoise(fixednumber)=variablecoeff\\left(fixednumber-leafvalue_{1}\\right)\\left(fixednumber-leafvalue_{2}\\right) \\cdots\\left(fixednumber-leafvalue_{flatness}\\right)\n\\]\nwhere \\( variablecoeff \\) is a constant. Taking the logarithmic derivative, we find\n\\[\n\\frac{randomnoise^{\\prime}(fixednumber)}{randomnoise(fixednumber)}=\\sum_{i=1}^{flatness} \\frac{1}{fixednumber-leafvalue_{i}} .\n\\]\n\nSuppose \\( insidepoint \\notin annulustwo \\). Then\n\\[\n\\left|insidepoint-leafvalue_{i}\\right| \\geq|insidepoint-borderpoint|-\\left|leafvalue_{i}-borderpoint\\right|>diameterlength+|nullfactor|-diameterlength=|nullfactor|\n\\]\nsince \\( leafvalue_{i} \\in annulusone \\). Therefore, for \\( i=1,2, \\ldots, flatness, \\frac{|nullfactor|}{\\left|insidepoint-leafvalue_{i}\\right|}<1 \\), and\n\\[\n\\begin{aligned}\n\\left|flatness\\,randomnoise(insidepoint)-nullfactor\\,randomnoise^{\\prime}(insidepoint)\\right| & =|randomnoise(insidepoint)| \\cdot\\left|flatness-nullfactor \\sum_{i=1}^{flatness} \\frac{1}{insidepoint-leafvalue_{i}}\\right| \\\\\n& \\geq|randomnoise(insidepoint)| \\cdot\\left(flatness-\\sum_{i=1}^{flatness} \\frac{|nullfactor|}{\\left|insidepoint-leafvalue_{i}\\right|}\\right)>0\n\\end{aligned}\n\\]\n\nThus \\( insidepoint \\) is not a root of \\( flatness\\,randomnoise(fixednumber)-nullfactor\\,randomnoise^{\\prime}(fixednumber) \\).\nThis proves that all the roots of \\( flatness\\,randomnoise(fixednumber)-nullfactor\\,randomnoise^{\\prime}(fixednumber) \\) lie in \\( annulustwo \\) as claimed."
    },
    "garbled_string": {
      "map": {
        "z": "qjfkdlsa",
        "u": "zmvpxqoe",
        "i": "bghtrmle",
        "\\\\lambda_i": "hqsnvagp",
        "P": "wlfkzmsq",
        "R": "pxdvgrla",
        "n": "vrgzoubk",
        "k": "tlasmqwe",
        "c": "jkdashcz",
        "A": "yprnwlse",
        "D_1": "skvjmqer",
        "D_2": "kdfplqva"
      },
      "question": "4. \\( wlfkzmsq(qjfkdlsa) \\) is a complex polynomial whose roots (as points in the Argand plane) can be covered by a closed circular disc of radius \\( pxdvgrla \\). Show that the roots of \\( vrgzoubk wlfkzmsq(qjfkdlsa)-tlasmqwe wlfkzmsq^{\\prime}(qjfkdlsa) \\) can be covered by a closed circular disc of radius \\( pxdvgrla \\) \\( +|tlasmqwe| \\), where \\( vrgzoubk \\) is the degree of \\( wlfkzmsq(qjfkdlsa) \\), tlasmqwe is any complex number, and \\( wlfkzmsq^{\\prime}(qjfkdlsa) \\) is the derivative of \\( wlfkzmsq(qjfkdlsa) \\).",
      "solution": "Solution. Suppose that the roots of \\( wlfkzmsq(qjfkdlsa) \\) all lie in the closed disc, \\( skvjmqer \\), \\( |qjfkdlsa-jkdashcz| \\leq pxdvgrla \\). We shall prove that the roots of \\( vrgzoubk wlfkzmsq(qjfkdlsa)-tlasmqwe wlfkzmsq^{\\prime}(qjfkdlsa) \\) lie in the closed disc, \\( kdfplqva,|qjfkdlsa-jkdashcz| \\leq pxdvgrla+|tlasmqwe| \\).\n\nIf the roots of \\( wlfkzmsq(qjfkdlsa) \\) are \\( hqsnvagp_{1}, hqsnvagp_{2}, \\ldots, hqsnvagp_{vrgzoubk} \\), then \\( wlfkzmsq(qjfkdlsa)=yprnwlse\\left(qjfkdlsa-hqsnvagp_{1}\\right)\\left(qjfkdlsa-hqsnvagp_{2}\\right) \\cdots\\left(qjfkdlsa-hqsnvagp_{vrgzoubk}\\right) \\) where \\( yprnwlse \\) is a constant. Taking the logarithmic derivative, we find\n\\[\n\\frac{wlfkzmsq^{\\prime}(qjfkdlsa)}{wlfkzmsq(qjfkdlsa)}=\\sum_{bghtrmle=1}^{vrgzoubk} \\frac{1}{qjfkdlsa-hqsnvagp_{bghtrmle}} .\n\\]\n\nSuppose \\( zmvpxqoe \\notin kdfplqva \\). Then\n\\[\n\\left|zmvpxqoe-hqsnvagp_{bghtrmle}\\right| \\geq|zmvpxqoe-jkdashcz|-\\left|hqsnvagp_{bghtrmle}-jkdashcz\\right|>pxdvgrla+|tlasmqwe|-pxdvgrla=|tlasmqwe|\n\\]\nsince \\( hqsnvagp_{bghtrmle} \\in skvjmqer \\). Therefore, for \\( bghtrmle=1,2, \\ldots, vrgzoubk, \\frac{|tlasmqwe|}{\\left|zmvpxqoe-hqsnvagp_{bghtrmle}\\right|}<1 \\), and\n\\[\n\\begin{aligned}\n\\left|vrgzoubk wlfkzmsq(zmvpxqoe)-tlasmqwe wlfkzmsq^{\\prime}(zmvpxqoe)\\right| & =|wlfkzmsq(zmvpxqoe)| \\cdot\\left|vrgzoubk-tlasmqwe \\sum_{bghtrmle=1}^{vrgzoubk} \\frac{1}{zmvpxqoe-hqsnvagp_{bghtrmle}}\\right| \\\\\n& \\geq|wlfkzmsq(zmvpxqoe)| \\cdot\\left(vrgzoubk-\\sum_{bghtrmle=1}^{vrgzoubk} \\frac{|tlasmqwe|}{\\left|zmvpxqoe-hqsnvagp_{bghtrmle}\\right|}\\right)>0\n\\end{aligned}\n\\]\n\nThus \\( zmvpxqoe \\) is not a root of \\( vrgzoubk wlfkzmsq(qjfkdlsa)-tlasmqwe wlfkzmsq^{\\prime}(qjfkdlsa) \\).\nThis proves that all the roots of \\( vrgzoubk wlfkzmsq(qjfkdlsa)-tlasmqwe wlfkzmsq^{\\prime}(qjfkdlsa) \\) lie in \\( kdfplqva \\) as claimed."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nP(z)=A\\prod_{i=1}^{n}(z-\\lambda_i),\\qquad A\\in\\mathbb C\\setminus\\{0\\},\\;n\\ge 2,\n\\]  \nbe a complex polynomial whose (possibly repeated) zeros all lie in the closed disc  \n\\[\nD(a,S):=\\{z\\in\\mathbb C:\\,|z-a|\\le S\\},\\qquad a\\in\\mathbb C,\\;S>0 .\n\\]\n\nFix an integer $m$ with $1\\le m\\le n$ and complex parameters $k_{1},\\dots ,k_{m}$.  \nPut $k_{0}:=1$ and, for $0\\le j\\le m$, denote the falling factorial  \n\\[\n(n)_{j}:=n(n-1)\\cdots(n-j+1).\n\\]\nIntroduce the linear differential operator  \n\\[\n\\mathcal L_{(k_{1},\\dots ,k_{m})}[P](z):=\\sum_{j=0}^{m}(-1)^{j}k_{j}(n)_{j}\\,P^{(j)}(z)\n\\]\nand set  \n\\[\nM:=\\max_{1\\le j\\le m}j|k_{j}|\\qquad(\\text{define }M:=0\\text{ if }k_{1}=\\dots =k_{m}=0).\n\\tag{1}\n\\]\n\nFor a finite set $E\\subset\\mathbb C$ we write  \n\\[\n\\operatorname{rad}_{a}(E):=\\max_{z\\in E}|z-a|\n\\]\nfor the \\emph{outer radius of $E$ with respect to $a$}.  \nIn particular,\n\\[\n\\operatorname{rad}_{a}\\bigl(\\operatorname{Zeros}(Q)\\bigr)=\n\\max_{\\zeta:\\,Q(\\zeta)=0}|\\,\\zeta-a|\n\\]\nfor any polynomial $Q$.\n\na)  Prove that every zero $\\zeta$ of $\\mathcal L_{(k_{1},\\dots ,k_{m})}[P]$ satisfies the universal bound  \n\\[\n|\\zeta-a|\\le S+n^{2}\\bigl(M+1\\bigr).\n\\tag{2}\n\\]\n\nb)  Show that the factor $n^{2}$ in front of $M$ is best possible:  \nfor every fixed admissible data $n,m,S$ and for every $\\varepsilon>0$ one can choose coefficients $(k_{j})$ and a polynomial $P$ with all zeros in $D(a,S)$ such that  \n\\[\n\\operatorname{rad}_{a}\\bigl(\\operatorname{Zeros}(\\mathcal L_{(k_{1},\\dots ,k_{m})}[P])\\bigr)\n   \\;\\ge\\; S+\\bigl(1-\\varepsilon\\bigr)n^{2}M.\n\\tag{3}\n\\]\nHence no universal constant strictly smaller than $n^{2}$ can replace $n^{2}$ in inequality (2).\n\n\\vspace{6pt}",
      "solution": "Throughout write  \n\\[\n\\rho:=|z-a|,\\qquad \\delta:=\\rho-S>0\\quad\\bigl(z\\notin D(a,S)\\bigr),\n\\]\nand denote the zero–set of $P$ by $\\Lambda:=\\{\\lambda_{1},\\dots ,\\lambda_{n}\\}$.\n\n\\bigskip\n\\textbf{Step 1 – A uniform higher–derivative estimate.}  \n\nFor every point $z\\in\\mathbb C\\setminus D(a,S)$ define  \n\\[\n\\Delta(z):=\\min_{\\lambda\\in\\Lambda}|z-\\lambda| .\n\\]\nSince $|\\lambda-a|\\le S$ we have $\\Delta(z)\\ge\\rho-S=\\delta$.  \nWe claim that  \n\\[\n\\boxed{\\;\n   |(n)_{j}\\,P^{(j)}(z)|\\le n^{2j}\\,\\Delta(z)^{-j}\\,|P(z)|\n   \\qquad(j=0,1,\\dots ,n).\n\\;}\n\\tag{4}\n\\]\n\n\\emph{Proof of (4).}  \nBecause each factor $z-\\lambda_{i}$ is linear, its second and higher derivatives vanish; consequently every $j$–th partial derivative of $P$ selects $j$ \\emph{distinct} factors to differentiate once.  \nLeibniz’ rule therefore gives the exact formula  \n\\[\nP^{(j)}(z)=j!\\sum_{|I|=j}\\prod_{i\\notin I}(z-\\lambda_{i}),\n\\tag{5}\n\\]\nwhere the sum ranges over all $j$–element subsets $I\\subset\\{1,\\dots ,n\\}$.  \nSince  \n\\[\n\\prod_{i\\notin I}(z-\\lambda_{i})=P(z)\\prod_{i\\in I}\\frac1{z-\\lambda_{i}},\n\\]\nwe obtain from (5)  \n\\[\nP^{(j)}(z)=P(z)\\,j!\\!\\!\\sum_{|I|=j}\\prod_{i\\in I}\\frac1{z-\\lambda_{i}}.\n\\tag{6}\n\\]\nTaking absolute values and using $|z-\\lambda_{i}|\\ge\\Delta(z)$ gives  \n\\[\n|P^{(j)}(z)|\n   \\le |P(z)|\\,j!\\,\\binom{n}{j}\\,\\Delta(z)^{-j}.\n\\tag{7}\n\\]\nMultiplying (7) by $(n)_{j}$ yields  \n\\[\n|(n)_{j}P^{(j)}(z)|\\le(n)_{j}\\,j!\\,\\binom{n}{j}\\,\\Delta(z)^{-j}\\,|P(z)|.\n\\]\nBecause  \n\\[\nj!\\,\\binom{n}{j}=\\frac{n!}{(n-j)!}=(n)_{j},\n\\]\nwe have  \n\\[\n(n)_{j}\\,j!\\,\\binom{n}{j}=(n)_{j}^{2}\\le n^{2j},\n\\]\nand (4) follows immediately since $\\Delta(z)\\ge\\delta$.\n\n\\bigskip\n\\textbf{Step 2 – From \\boldmath{$\\mathcal L[P]/P$} to a geometric series.}  \n\nIntroduce  \n\\[\nx:=\\frac{n^{2}}{\\delta}\\qquad(0<x<1\\;\\Longleftrightarrow\\;\\delta>n^{2}).\n\\tag{8}\n\\]\nUsing (4) in the definition of $\\mathcal L$ we find for every $z$ with $\\delta>0$\n\\[\n\\left|\\frac{\\mathcal L_{(k_{1},\\dots ,k_{m})}[P](z)}{P(z)}-1\\right|\n  \\le\\sum_{j=1}^{m}|k_{j}|\\,n^{2j}\\,\\delta^{-j}\n     =\\sum_{j=1}^{m}|k_{j}|\\,x^{j}.\n\\tag{9}\n\\]\nBecause of (1), $|k_{j}|\\le M/j$, whence  \n\\[\n\\left|\\frac{\\mathcal L[P](z)}{P(z)}-1\\right|\n   \\le M\\sum_{j=1}^{m}\\frac{x^{j}}{j}\n   \\le M\\bigl(-\\ln(1-x)\\bigr)\\qquad(|x|<1).\n\\tag{10}\n\\]\n\n\\bigskip\n\\textbf{Step 3 – Choosing a special radius.}  \n\nIf $M=0$ then $\\mathcal L[P]\\equiv P$ and (2) is trivial.  \nAssume $M>0$ and set  \n\\[\n\\delta_{0}:=n^{2}(M+1),\\qquad\nx_{0}:=\\frac{n^{2}}{\\delta_{0}}=\\frac1{M+1}<1.\n\\tag{11}\n\\]\nFor this $x_{0}$ we have  \n\\[\nM\\bigl(-\\ln(1-x_{0})\\bigr)=M\\ln\\!\\bigl(1+\\tfrac1M\\bigr)<1,\n\\tag{12}\n\\]\nbecause $\\ln(1+u)<u$ for all $u>0$.  \nHence, for every $z$ with $|z-a|=S+\\delta_{0}$,\n\\[\n|\\mathcal L[P](z)-P(z)|<|P(z)|.\n\\tag{13}\n\\]\n\n\\bigskip\n\\textbf{Step 4 – Application of Rouché’s theorem.}  \n\nOn the circle $C:\\,|z-a|=S+\\delta_{0}$ inequality (13) shows that\n$\\mathcal L[P]$ and $P$ have the same number of zeros inside $C$.  \nThe polynomial $P$ possesses exactly $n$ zeros inside $C$ (because all its zeros satisfy $|z-a|\\le S$), hence $\\mathcal L[P]$ also has $n$ zeros inside $C$ and none outside.  \nTherefore every zero $\\zeta$ of $\\mathcal L[P]$ satisfies  \n\\[\n|\\zeta-a|\\le S+\\delta_{0}=S+n^{2}(M+1),\n\\]\nwhich is precisely (2).\n\n\\bigskip\n\\textbf{Step 5 – Sharpness of the factor \\boldmath{$n^{2}$}.}  \n\nFix $n,m,S$ and $\\varepsilon\\in(0,1)$.  \nChoose  \n\\[\nk_{1}:=M>0,\\qquad k_{j}:=0\\quad(j\\ge 2),\\qquad(m\\ge 1),\n\\]\nand the \\emph{monomial} polynomial  \n\\[\nP_{t}(z):=\\bigl(z-(a+S-t)\\bigr)^{n},\\qquad\n0<t<\\min\\{S,\\varepsilon n^{2}M\\}.\n\\tag{14}\n\\]\nAll zeros of $P_{t}$ equal $a+S-t$ and lie in $D(a,S)$.\n\nSince $P_{t}'(z)=n\\bigl(z-(a+S-t)\\bigr)^{n-1}$, we have  \n\\[\n\\begin{aligned}\n\\mathcal L_{(k_{1},0,\\dots ,0)}[P_{t}](z)\n   &=P_{t}(z)-k_{1}(n)_{1}P_{t}'(z) \\\\\n   &=\\bigl(z-(a+S-t)\\bigr)^{n-1}\n      \\Bigl((z-(a+S-t))-n^{2}M\\Bigr).\n\\end{aligned}\n\\tag{15}\n\\]\nThus $\\mathcal L[P_{t}]$ has the simple zero  \n\\[\n\\zeta_{t}:=a+(S-t)+n^{2}M .\n\\tag{16}\n\\]\nConsequently  \n\\[\n|\\zeta_{t}-a|=S-t+n^{2}M\\;\\ge\\;\nS+\\bigl(1-\\varepsilon\\bigr)n^{2}M .\n\\tag{17}\n\\]\nBecause $\\varepsilon>0$ is arbitrary, inequality (3) follows and shows that the constant $n^{2}$ in (2) cannot be lowered.\n\n\\bigskip\nThe proof of parts (a) and (b) is complete. \\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.487635",
        "was_fixed": false,
        "difficulty_analysis": "• Multiple derivatives: the problem involves an arbitrary finite collection of derivatives up to order \\(m\\) (instead of a single first derivative), each weighted by a free complex constant \\(k_j\\).  \n\n• Non-uniform weights: the appearance of the falling factorial \\((n)_j\\) forces the solver to keep careful track of combinatorial multiplicities when estimating the higher derivatives.  \n\n• Sophisticated bounding: one must bound entire families of elementary symmetric sums of reciprocals of linear factors; naive triangle-inequality estimates fail, so a delicate use of series estimates (and sometimes of the logarithmic bound \\(-\\ln(1-x)\\)) is required.  \n\n• Rouché’s theorem in a higher-order setting: one must build a majorant for a whole differential sum, not just for the first-order term, and confirm the inequality on a large circle.  \n\n• Sharpness: the problem is not finished after establishing containment; it also demands the construction of extremal examples showing that the disc radius cannot be diminished even by an arbitrarily small amount.  This requires both ingenuity and a good control of the algebraic form of the operator \\(\\mathcal L_{(k_1,\\dots ,k_m)}\\).  \n\nBecause of these extra layers—handling many derivatives simultaneously, using combinatorial estimates, performing non-trivial analytic bounds, and establishing optimality—the enhanced variant is substantially harder than both the original exercise and the simpler kernel variant supplied earlier."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nP(z)=A\\prod_{i=1}^{n}(z-\\lambda_i),\\qquad A\\in\\mathbb C\\setminus\\{0\\},\\;n\\ge 2,\n\\]  \nbe a complex polynomial whose (possibly repeated) zeros all lie in the closed disc  \n\\[\nD(a,S):=\\{z\\in\\mathbb C:\\,|z-a|\\le S\\},\\qquad a\\in\\mathbb C,\\;S>0 .\n\\]\n\nFix an integer $m$ with $1\\le m\\le n$ and complex parameters $k_{1},\\dots ,k_{m}$.  \nPut $k_{0}:=1$ and, for $0\\le j\\le m$, denote the falling factorial  \n\\[\n(n)_{j}:=n(n-1)\\cdots(n-j+1).\n\\]\nIntroduce the linear differential operator  \n\\[\n\\mathcal L_{(k_{1},\\dots ,k_{m})}[P](z):=\\sum_{j=0}^{m}(-1)^{j}k_{j}(n)_{j}\\,P^{(j)}(z)\n\\]\nand set  \n\\[\nM:=\\max_{1\\le j\\le m}j|k_{j}|\\qquad(\\text{define }M:=0\\text{ if }k_{1}=\\dots =k_{m}=0).\n\\tag{1}\n\\]\n\nFor a finite set $E\\subset\\mathbb C$ we write  \n\\[\n\\operatorname{rad}_{a}(E):=\\max_{z\\in E}|z-a|\n\\]\nfor the \\emph{outer radius of $E$ with respect to $a$}.  \nIn particular,\n\\[\n\\operatorname{rad}_{a}\\bigl(\\operatorname{Zeros}(Q)\\bigr)=\n\\max_{\\zeta:\\,Q(\\zeta)=0}|\\,\\zeta-a|\n\\]\nfor any polynomial $Q$.\n\na)  Prove that every zero $\\zeta$ of $\\mathcal L_{(k_{1},\\dots ,k_{m})}[P]$ satisfies the universal bound  \n\\[\n|\\zeta-a|\\le S+n^{2}\\bigl(M+1\\bigr).\n\\tag{2}\n\\]\n\nb)  Show that the factor $n^{2}$ in front of $M$ is best possible:  \nfor every fixed admissible data $n,m,S$ and for every $\\varepsilon>0$ one can choose coefficients $(k_{j})$ and a polynomial $P$ with all zeros in $D(a,S)$ such that  \n\\[\n\\operatorname{rad}_{a}\\bigl(\\operatorname{Zeros}(\\mathcal L_{(k_{1},\\dots ,k_{m})}[P])\\bigr)\n   \\;\\ge\\; S+\\bigl(1-\\varepsilon\\bigr)n^{2}M.\n\\tag{3}\n\\]\nHence no universal constant strictly smaller than $n^{2}$ can replace $n^{2}$ in inequality (2).\n\n\\vspace{6pt}",
      "solution": "Throughout write  \n\\[\n\\rho:=|z-a|,\\qquad \\delta:=\\rho-S>0\\quad\\bigl(z\\notin D(a,S)\\bigr),\n\\]\nand denote the zero–set of $P$ by $\\Lambda:=\\{\\lambda_{1},\\dots ,\\lambda_{n}\\}$.\n\n\\bigskip\n\\textbf{Step 1 – A uniform higher–derivative estimate.}  \n\nFor every point $z\\in\\mathbb C\\setminus D(a,S)$ define  \n\\[\n\\Delta(z):=\\min_{\\lambda\\in\\Lambda}|z-\\lambda| .\n\\]\nSince $|\\lambda-a|\\le S$ we have $\\Delta(z)\\ge\\rho-S=\\delta$.  \nWe claim that  \n\\[\n\\boxed{\\;\n   |(n)_{j}\\,P^{(j)}(z)|\\le n^{2j}\\,\\Delta(z)^{-j}\\,|P(z)|\n   \\qquad(j=0,1,\\dots ,n).\n\\;}\n\\tag{4}\n\\]\n\n\\emph{Proof of (4).}  \nBecause each factor $z-\\lambda_{i}$ is linear, its second and higher derivatives vanish; consequently every $j$–th partial derivative of $P$ selects $j$ \\emph{distinct} factors to differentiate once.  \nLeibniz’ rule therefore gives the exact formula  \n\\[\nP^{(j)}(z)=j!\\sum_{|I|=j}\\prod_{i\\notin I}(z-\\lambda_{i}),\n\\tag{5}\n\\]\nwhere the sum ranges over all $j$–element subsets $I\\subset\\{1,\\dots ,n\\}$.  \nSince  \n\\[\n\\prod_{i\\notin I}(z-\\lambda_{i})=P(z)\\prod_{i\\in I}\\frac1{z-\\lambda_{i}},\n\\]\nwe obtain from (5)  \n\\[\nP^{(j)}(z)=P(z)\\,j!\\!\\!\\sum_{|I|=j}\\prod_{i\\in I}\\frac1{z-\\lambda_{i}}.\n\\tag{6}\n\\]\nTaking absolute values and using $|z-\\lambda_{i}|\\ge\\Delta(z)$ gives  \n\\[\n|P^{(j)}(z)|\n   \\le |P(z)|\\,j!\\,\\binom{n}{j}\\,\\Delta(z)^{-j}.\n\\tag{7}\n\\]\nMultiplying (7) by $(n)_{j}$ yields  \n\\[\n|(n)_{j}P^{(j)}(z)|\\le(n)_{j}\\,j!\\,\\binom{n}{j}\\,\\Delta(z)^{-j}\\,|P(z)|.\n\\]\nBecause  \n\\[\nj!\\,\\binom{n}{j}=\\frac{n!}{(n-j)!}=(n)_{j},\n\\]\nwe have  \n\\[\n(n)_{j}\\,j!\\,\\binom{n}{j}=(n)_{j}^{2}\\le n^{2j},\n\\]\nand (4) follows immediately since $\\Delta(z)\\ge\\delta$.\n\n\\bigskip\n\\textbf{Step 2 – From \\boldmath{$\\mathcal L[P]/P$} to a geometric series.}  \n\nIntroduce  \n\\[\nx:=\\frac{n^{2}}{\\delta}\\qquad(0<x<1\\;\\Longleftrightarrow\\;\\delta>n^{2}).\n\\tag{8}\n\\]\nUsing (4) in the definition of $\\mathcal L$ we find for every $z$ with $\\delta>0$\n\\[\n\\left|\\frac{\\mathcal L_{(k_{1},\\dots ,k_{m})}[P](z)}{P(z)}-1\\right|\n  \\le\\sum_{j=1}^{m}|k_{j}|\\,n^{2j}\\,\\delta^{-j}\n     =\\sum_{j=1}^{m}|k_{j}|\\,x^{j}.\n\\tag{9}\n\\]\nBecause of (1), $|k_{j}|\\le M/j$, whence  \n\\[\n\\left|\\frac{\\mathcal L[P](z)}{P(z)}-1\\right|\n   \\le M\\sum_{j=1}^{m}\\frac{x^{j}}{j}\n   \\le M\\bigl(-\\ln(1-x)\\bigr)\\qquad(|x|<1).\n\\tag{10}\n\\]\n\n\\bigskip\n\\textbf{Step 3 – Choosing a special radius.}  \n\nIf $M=0$ then $\\mathcal L[P]\\equiv P$ and (2) is trivial.  \nAssume $M>0$ and set  \n\\[\n\\delta_{0}:=n^{2}(M+1),\\qquad\nx_{0}:=\\frac{n^{2}}{\\delta_{0}}=\\frac1{M+1}<1.\n\\tag{11}\n\\]\nFor this $x_{0}$ we have  \n\\[\nM\\bigl(-\\ln(1-x_{0})\\bigr)=M\\ln\\!\\bigl(1+\\tfrac1M\\bigr)<1,\n\\tag{12}\n\\]\nbecause $\\ln(1+u)<u$ for all $u>0$.  \nHence, for every $z$ with $|z-a|=S+\\delta_{0}$,\n\\[\n|\\mathcal L[P](z)-P(z)|<|P(z)|.\n\\tag{13}\n\\]\n\n\\bigskip\n\\textbf{Step 4 – Application of Rouché’s theorem.}  \n\nOn the circle $C:\\,|z-a|=S+\\delta_{0}$ inequality (13) shows that\n$\\mathcal L[P]$ and $P$ have the same number of zeros inside $C$.  \nThe polynomial $P$ possesses exactly $n$ zeros inside $C$ (because all its zeros satisfy $|z-a|\\le S$), hence $\\mathcal L[P]$ also has $n$ zeros inside $C$ and none outside.  \nTherefore every zero $\\zeta$ of $\\mathcal L[P]$ satisfies  \n\\[\n|\\zeta-a|\\le S+\\delta_{0}=S+n^{2}(M+1),\n\\]\nwhich is precisely (2).\n\n\\bigskip\n\\textbf{Step 5 – Sharpness of the factor \\boldmath{$n^{2}$}.}  \n\nFix $n,m,S$ and $\\varepsilon\\in(0,1)$.  \nChoose  \n\\[\nk_{1}:=M>0,\\qquad k_{j}:=0\\quad(j\\ge 2),\\qquad(m\\ge 1),\n\\]\nand the \\emph{monomial} polynomial  \n\\[\nP_{t}(z):=\\bigl(z-(a+S-t)\\bigr)^{n},\\qquad\n0<t<\\min\\{S,\\varepsilon n^{2}M\\}.\n\\tag{14}\n\\]\nAll zeros of $P_{t}$ equal $a+S-t$ and lie in $D(a,S)$.\n\nSince $P_{t}'(z)=n\\bigl(z-(a+S-t)\\bigr)^{n-1}$, we have  \n\\[\n\\begin{aligned}\n\\mathcal L_{(k_{1},0,\\dots ,0)}[P_{t}](z)\n   &=P_{t}(z)-k_{1}(n)_{1}P_{t}'(z) \\\\\n   &=\\bigl(z-(a+S-t)\\bigr)^{n-1}\n      \\Bigl((z-(a+S-t))-n^{2}M\\Bigr).\n\\end{aligned}\n\\tag{15}\n\\]\nThus $\\mathcal L[P_{t}]$ has the simple zero  \n\\[\n\\zeta_{t}:=a+(S-t)+n^{2}M .\n\\tag{16}\n\\]\nConsequently  \n\\[\n|\\zeta_{t}-a|=S-t+n^{2}M\\;\\ge\\;\nS+\\bigl(1-\\varepsilon\\bigr)n^{2}M .\n\\tag{17}\n\\]\nBecause $\\varepsilon>0$ is arbitrary, inequality (3) follows and shows that the constant $n^{2}$ in (2) cannot be lowered.\n\n\\bigskip\nThe proof of parts (a) and (b) is complete. \\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.408002",
        "was_fixed": false,
        "difficulty_analysis": "• Multiple derivatives: the problem involves an arbitrary finite collection of derivatives up to order \\(m\\) (instead of a single first derivative), each weighted by a free complex constant \\(k_j\\).  \n\n• Non-uniform weights: the appearance of the falling factorial \\((n)_j\\) forces the solver to keep careful track of combinatorial multiplicities when estimating the higher derivatives.  \n\n• Sophisticated bounding: one must bound entire families of elementary symmetric sums of reciprocals of linear factors; naive triangle-inequality estimates fail, so a delicate use of series estimates (and sometimes of the logarithmic bound \\(-\\ln(1-x)\\)) is required.  \n\n• Rouché’s theorem in a higher-order setting: one must build a majorant for a whole differential sum, not just for the first-order term, and confirm the inequality on a large circle.  \n\n• Sharpness: the problem is not finished after establishing containment; it also demands the construction of extremal examples showing that the disc radius cannot be diminished even by an arbitrarily small amount.  This requires both ingenuity and a good control of the algebraic form of the operator \\(\\mathcal L_{(k_1,\\dots ,k_m)}\\).  \n\nBecause of these extra layers—handling many derivatives simultaneously, using combinatorial estimates, performing non-trivial analytic bounds, and establishing optimality—the enhanced variant is substantially harder than both the original exercise and the simpler kernel variant supplied earlier."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}