{
  "index": "2005-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $p(z)$ be a polynomial of degree $n$ all of whose zeros have absolute\nvalue 1 in the complex  plane. Put $g(z) = p(z)/z^{n/2}$. Show that all zeros\nof $g'(z) = 0$ have absolute value 1.",
  "solution": "Note that it is implicit in the problem that $p$ is nonconstant,\none may take any branch of the square root, and that $z=0$ should be\nignored.\n\n\\textbf{First solution:}\nWrite $p(z) = c \\prod_{j=1}^n (z-r_j)$, so that\n\\[\n\\frac{g'(z)}{g(z)} = \\frac{1}{2z} \\sum_{j=1}^n \\frac{z+r_j}{z-r_j}.\n\\]\nNow if $z\\neq r_j$ for all $j$,then\n\\[\n\\frac{z+r_j}{z-r_j}\n= \\frac{(z+r_j)(\\overline{z}-\\overline{r_j})}{|z-r_j|^2}\n= \\frac{|z|^2-1+2 \\mathrm{Im}(\\overline{z} r_j)}{|z-r_j|^2},\n\\]\nand so\n\\[\n\\mathrm{Re}\\, \\frac{z g'(z)}{g(z)} = \\frac{|z|^2-1}{2} \\left(\n\\sum_j \\frac{1}{|z-r_j|^2} \\right).\n\\]\nSince the quantity in parentheses\nis positive, $g'(z)/g(z)$ can be $0$ only if $|z|=1$. If on the other\nhand $z = r_j$ for some $j$, then $|z|=1$ anyway.\n\n\\textbf{Second solution:}\nWrite $p(z) = c\\prod_{j=1}^n (z-r_j)$, so that\n\\[\n\\frac{g'(z)}{g(z)}\n= \\sum_{j=1}^n \\left( \\frac{1}{z-r_j} - \\frac{1}{2z} \\right).\n\\]\nWe first check that $g'(z) \\neq 0$ whenever $z$ is real and $z>1$.\nIn this case, for $r_j = e^{i \\theta_j}$, we have $z - r_j =\n(z - \\cos (\\theta_j)) + \\sin(\\theta_j) i$, so the real part of\n$\\frac{1}{z-r_j} - \\frac{1}{2z}$ is\n\\[\n\\frac{z - \\cos(\\theta_j)}{z^2 - 2z \\cos(\\theta_j) + 1} - \\frac{1}{2z}\n= \\frac{z^2-1}{2z(z^2 - 2z \\cos(\\theta_j) + 1)} > 0.\n\\]\nHence $g'(z)/g(z)$ has positive real part, so $g'(z)/g(z)$ and hence\n$g(z)$ are nonzero.\n\nApplying the same argument after replacing $p(z)$ by $p(e^{i \\theta} z)$,\nwe deduce that $g'$ cannot have any roots outside the unit circle.\nApplying the same argument after replacing $p(z)$ by $z^n p(1/z)$, we\nalso deduce that $g'$ cannot have any roots inside the unit circle.\nHence all roots of $g'$ have absolute value 1, as desired.\n\n\\textbf{Third solution:}\nWrite $p(z) = c \\prod_{j=1}^n (z - r_j)$ and\nput $r_j = e^{2 i \\theta_j}$. Note that $g(e^{2 i \\theta})$\nis equal to a nonzero constant times\n\\begin{align*}\nh(\\theta) &= \\prod_{j=1}^n \\frac{e^{i (\\theta + \\theta_j)}\n- e^{-i(\\theta + \\theta_j)}}{2i} = \\prod_{j=1}^n \\sin(\\theta +\\theta_j).\n\\end{align*}\nSince $h$ has at least $2n$ roots (counting multiplicity)\nin the interval $[0, 2\\pi)$, $h'$ does also by repeated application of\nRolle's theorem. Since $g'(e^{2 i \\theta}) = 2i e^{2i \\theta} h'(\\theta)$,\n$g'(z^2)$ has at least $2n$ roots on the unit circle. Since $g'(z^2)$ is equal to\n$z^{-n-1}$ times a polynomial of degree $2n$, $g'(z^2)$ has all roots on the\nunit circle, as then does $g'(z)$.\n\n\\textbf{Remarks:}\nThe second solution imitates the proof of the Gauss-Lucas theorem:\nthe roots of the derivative of a complex polynomial lie in the convex hull of\nthe roots of the original polynomial.\nThe second solution is close to problem B3 from the 2000 Putnam.\nA hybrid between the first and third solutions is to check that\non the unit circle, $\\mathrm{Re}(zg'(z)/g(z)) = 0$ while between any\ntwo roots of $p$, $\\mathrm{Im}(zg'(z)/g(z))$ runs from $+\\infty$ to $-\\infty$\nand so must have a zero crossing. (This only works when $p$ has distinct roots,\nbut the general case follows by the continuity of the roots of a polynomial\nas functions of the coefficients.)\nOne can also construct a  solution using Rouch\\'e's theorem.",
  "vars": [
    "z",
    "g",
    "p",
    "h"
  ],
  "params": [
    "c",
    "n",
    "r_j",
    "j",
    "\\\\theta",
    "\\\\theta_j"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "complexvar",
        "g": "quotientfun",
        "p": "polynomial",
        "h": "sineprod",
        "c": "leadingconst",
        "n": "degreenum",
        "r_j": "unitroot",
        "j": "indexvar",
        "\\theta": "anglevar",
        "\\theta_j": "angleindex"
      },
      "question": "Let $polynomial(complexvar)$ be a polynomial of degree $degreenum$ all of whose zeros have absolute\nvalue 1 in the complex  plane. Put $quotientfun(complexvar) = polynomial(complexvar)/complexvar^{degreenum/2}$. Show that all zeros\nof $quotientfun'(complexvar) = 0$ have absolute value 1.",
      "solution": "Note that it is implicit in the problem that $polynomial$ is nonconstant,\none may take any branch of the square root, and that $complexvar=0$ should be\nignored.\n\n\\textbf{First solution:}\nWrite $polynomial(complexvar) = leadingconst \\prod_{indexvar=1}^{degreenum} (complexvar-unitroot)$, so that\n\\[\n\\frac{quotientfun'(complexvar)}{quotientfun(complexvar)} = \\frac{1}{2complexvar} \\sum_{indexvar=1}^{degreenum} \\frac{complexvar+unitroot}{complexvar-unitroot}.\n\\]\nNow if $complexvar\\neq unitroot$ for all $indexvar$,then\n\\[\n\\frac{complexvar+unitroot}{complexvar-unitroot}\n= \\frac{(complexvar+unitroot)(\\overline{complexvar}-\\overline{unitroot})}{|complexvar-unitroot|^2}\n= \\frac{|complexvar|^2-1+2 \\mathrm{Im}(\\overline{complexvar} unitroot)}{|complexvar-unitroot|^2},\n\\]\nand so\n\\[\n\\mathrm{Re}\\, \\frac{complexvar \\!\\, quotientfun'(complexvar)}{quotientfun(complexvar)} = \\frac{|complexvar|^2-1}{2} \\left(\n\\sum_{indexvar} \\frac{1}{|complexvar-unitroot|^2} \\right).\n\\]\nSince the quantity in parentheses\nis positive, $quotientfun'(complexvar)/quotientfun(complexvar)$ can be $0$ only if $|complexvar|=1$. If on the other\nhand $complexvar = unitroot$ for some $indexvar$, then $|complexvar|=1$ anyway.\n\n\\textbf{Second solution:}\nWrite $polynomial(complexvar) = leadingconst\\prod_{indexvar=1}^{degreenum} (complexvar-unitroot)$, so that\n\\[\n\\frac{quotientfun'(complexvar)}{quotientfun(complexvar)}\n= \\sum_{indexvar=1}^{degreenum} \\left( \\frac{1}{complexvar-unitroot} - \\frac{1}{2complexvar} \\right).\n\\]\nWe first check that $quotientfun'(complexvar) \\neq 0$ whenever $complexvar$ is real and $complexvar>1$.\nIn this case, for $unitroot = e^{i angleindex}$, we have $complexvar - unitroot =\n(complexvar - \\cos (angleindex)) + \\sin(angleindex) i$, so the real part of\n$\\frac{1}{complexvar-unitroot} - \\frac{1}{2complexvar}$ is\n\\[\n\\frac{complexvar - \\cos(angleindex)}{complexvar^2 - 2complexvar \\cos(angleindex) + 1} - \\frac{1}{2complexvar}\n= \\frac{complexvar^2-1}{2complexvar(complexvar^2 - 2complexvar \\cos(angleindex) + 1)} > 0.\n\\]\nHence $quotientfun'(complexvar)/quotientfun(complexvar)$ has positive real part, so $quotientfun'(complexvar)/quotientfun(complexvar)$ and hence\n$quotientfun(complexvar)$ are nonzero.\n\nApplying the same argument after replacing $polynomial(complexvar)$ by $polynomial(e^{i anglevar} complexvar)$,\nwe deduce that $quotientfun'$ cannot have any roots outside the unit circle.\nApplying the same argument after replacing $polynomial(complexvar)$ by $complexvar^{degreenum} polynomial(1/complexvar)$, we\nalso deduce that $quotientfun'$ cannot have any roots inside the unit circle.\nHence all roots of $quotientfun'$ have absolute value 1, as desired.\n\n\\textbf{Third solution:}\nWrite $polynomial(complexvar) = leadingconst \\prod_{indexvar=1}^{degreenum} (complexvar - unitroot)$ and\nput $unitroot = e^{2 i angleindex}$. Note that $quotientfun(e^{2 i anglevar})$\nis equal to a nonzero constant times\n\\begin{align*}\nsineprod(anglevar) &= \\prod_{indexvar=1}^{degreenum} \\frac{e^{i (anglevar + angleindex)}\n- e^{-i(anglevar + angleindex)}}{2i} = \\prod_{indexvar=1}^{degreenum} \\sin(anglevar +angleindex).\n\\end{align*}\nSince $sineprod$ has at least $2degreenum$ roots (counting multiplicity)\nin the interval $[0, 2\\pi)$, $sineprod'$ does also by repeated application of\nRolle's theorem. Since $quotientfun'(e^{2 i anglevar}) = 2i e^{2i anglevar} sineprod'(anglevar)$,\n$quotientfun'(complexvar^2)$ has at least $2degreenum$ roots on the unit circle. Since $quotientfun'(complexvar^2)$ is equal to\n$complexvar^{-degreenum-1}$ times a polynomial of degree $2degreenum$, $quotientfun'(complexvar^2)$ has all roots on the\nunit circle, as then does $quotientfun'(complexvar)$.\n\n\\textbf{Remarks:}\nThe second solution imitates the proof of the Gauss-Lucas theorem:\nthe roots of the derivative of a complex polynomial lie in the convex hull of\nthe roots of the original polynomial.\nThe second solution is close to problem B3 from the 2000 Putnam.\nA hybrid between the first and third solutions is to check that\non the unit circle, $\\mathrm{Re}(complexvar \\, quotientfun'(complexvar)/quotientfun(complexvar)) = 0$ while between any\ntwo roots of $polynomial$, $\\mathrm{Im}(complexvar \\, quotientfun'(complexvar)/quotientfun(complexvar))$ runs from $+\\infty$ to $-\\infty$\nand so must have a zero crossing. (This only works when $polynomial$ has distinct roots,\nbut the general case follows by the continuity of the roots of a polynomial\nas functions of the coefficients.)\nOne can also construct a  solution using Rouch\\'e's theorem."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "skyladder",
        "g": "riverstone",
        "p": "cloudmirror",
        "h": "embertrail",
        "c": "lanternwood",
        "n": "foxglovese",
        "r_j": "whisperpath",
        "j": "moonshaper",
        "\\\\theta": "starquartz",
        "\\\\theta_j": "starquartzway"
      },
      "question": "Let $cloudmirror(skyladder)$ be a polynomial of degree $foxglovese$ all of whose zeros have absolute\nvalue 1 in the complex  plane. Put $riverstone(skyladder) = cloudmirror(skyladder)/skyladder^{foxglovese/2}$. Show that all zeros\nof $riverstone'(skyladder) = 0$ have absolute value 1.",
      "solution": "Note that it is implicit in the problem that $cloudmirror$ is nonconstant,\none may take any branch of the square root, and that $skyladder=0$ should be\nignored.\n\n\\textbf{First solution:}\nWrite $cloudmirror(skyladder) = lanternwood \\prod_{moonshaper=1}^{foxglovese} (skyladder-whisperpath)$, so that\n\\[\n\\frac{riverstone'(skyladder)}{riverstone(skyladder)} = \\frac{1}{2skyladder} \\sum_{moonshaper=1}^{foxglovese} \\frac{skyladder+whisperpath}{skyladder-whisperpath}.\n\\]\nNow if $skyladder\\neq whisperpath$ for all $moonshaper$,then\n\\[\n\\frac{skyladder+whisperpath}{skyladder-whisperpath}\n= \\frac{(skyladder+whisperpath)(\\overline{skyladder}-\\overline{whisperpath})}{|skyladder-whisperpath|^2}\n= \\frac{|skyladder|^2-1+2 \\mathrm{Im}(\\overline{skyladder} whisperpath)}{|skyladder-whisperpath|^2},\n\\]\nand so\n\\[\n\\mathrm{Re}\\, \\frac{skyladder riverstone'(skyladder)}{riverstone(skyladder)} = \\frac{|skyladder|^2-1}{2} \\left(\n\\sum_{moonshaper} \\frac{1}{|skyladder-whisperpath|^2} \\right).\n\\]\nSince the quantity in parentheses\nis positive, $riverstone'(skyladder)/riverstone(skyladder)$ can be $0$ only if $|skyladder|=1$. If on the other\nhand $skyladder = whisperpath$ for some $moonshaper$, then $|skyladder|=1$ anyway.\n\n\\textbf{Second solution:}\nWrite $cloudmirror(skyladder) = lanternwood\\prod_{moonshaper=1}^{foxglovese} (skyladder-whisperpath)$, so that\n\\[\n\\frac{riverstone'(skyladder)}{riverstone(skyladder)}\n= \\sum_{moonshaper=1}^{foxglovese} \\left( \\frac{1}{skyladder-whisperpath} - \\frac{1}{2skyladder} \\right).\n\\]\nWe first check that $riverstone'(skyladder) \\neq 0$ whenever $skyladder$ is real and $skyladder>1$.\nIn this case, for $whisperpath = e^{i starquartzway}$, we have $skyladder - whisperpath =\n(skyladder - \\cos (starquartzway)) + \\sin(starquartzway) i$, so the real part of\n$\\frac{1}{skyladder-whisperpath} - \\frac{1}{2skyladder}$ is\n\\[\n\\frac{skyladder - \\cos(starquartzway)}{skyladder^2 - 2skyladder \\cos(starquartzway) + 1} - \\frac{1}{2skyladder}\n= \\frac{skyladder^2-1}{2skyladder(skyladder^2 - 2skyladder \\cos(starquartzway) + 1)} > 0.\n\\]\nHence $riverstone'(skyladder)/riverstone(skyladder)$ has positive real part, so $riverstone'(skyladder)/riverstone(skyladder)$ and hence\n$riverstone(skyladder)$ are nonzero.\n\nApplying the same argument after replacing $cloudmirror(skyladder)$ by $cloudmirror(e^{i starquartz} skyladder)$,\nwe deduce that $riverstone'$ cannot have any roots outside the unit circle.\nApplying the same argument after replacing $cloudmirror(skyladder)$ by $skyladder^{foxglovese} cloudmirror(1/skyladder)$, we\nalso deduce that $riverstone'$ cannot have any roots inside the unit circle.\nHence all roots of $riverstone'$ have absolute value 1, as desired.\n\n\\textbf{Third solution:}\nWrite $cloudmirror(skyladder) = lanternwood \\prod_{moonshaper=1}^{foxglovese} (skyladder - whisperpath)$ and\nput $whisperpath = e^{2 i starquartzway}$. Note that $riverstone(e^{2 i starquartz})$\nis equal to a nonzero constant times\n\\begin{align*}\nembertrail(starquartz) &= \\prod_{moonshaper=1}^{foxglovese} \\frac{e^{i (starquartz + starquartzway)}\n- e^{-i(starquartz + starquartzway)}}{2i} = \\prod_{moonshaper=1}^{foxglovese} \\sin(starquartz +starquartzway).\n\\end{align*}\nSince $embertrail$ has at least $2foxglovese$ roots (counting multiplicity)\nin the interval $[0, 2\\pi)$, $embertrail'$ does also by repeated application of\nRolle's theorem. Since $riverstone'(e^{2 i starquartz}) = 2i e^{2i starquartz} embertrail'(starquartz)$,\n$riverstone'(skyladder^2)$ has at least $2foxglovese$ roots on the unit circle. Since $riverstone'(skyladder^2)$ is equal to\n$skyladder^{-foxglovese-1}$ times a polynomial of degree $2foxglovese$, $riverstone'(skyladder^2)$ has all roots on the\nunit circle, as then does $riverstone'(skyladder)$.\n\n\\textbf{Remarks:}\nThe second solution imitates the proof of the Gauss-Lucas theorem:\nthe roots of the derivative of a complex polynomial lie in the convex hull of\nthe roots of the original polynomial.\nThe second solution is close to problem B3 from the 2000 Putnam.\nA hybrid between the first and third solutions is to check that\non the unit circle, $\\mathrm{Re}(skyladder riverstone'(skyladder)/riverstone(skyladder)) = 0$ while between any\ntwo roots of $cloudmirror$, $\\mathrm{Im}(skyladder riverstone'(skyladder)/riverstone(skyladder))$ runs from $+\\infty$ to $-\\infty$\nand so must have a zero crossing. (This only works when $cloudmirror$ has distinct roots,\nbut the general case follows by the continuity of the roots of a polynomial\nas functions of the coefficients.)\nOne can also construct a  solution using Rouch\\'e's theorem."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "constantval",
        "g": "staticfunc",
        "p": "transcend",
        "h": "monotone",
        "c": "variable",
        "n": "infinite",
        "r_j": "summitval",
        "j": "aggregate",
        "\\theta": "\\counterangle",
        "\\theta_j": "\\counteranglej"
      },
      "question": "Let $transcend(constantval)$ be a polynomial of degree $infinite$ all of whose zeros have absolute\nvalue 1 in the complex  plane. Put $staticfunc(constantval) = transcend(constantval)/constantval^{infinite/2}$. Show that all zeros\nof $staticfunc'(constantval) = 0$ have absolute value 1.",
      "solution": "Note that it is implicit in the problem that $transcend$ is nonconstant,\none may take any branch of the square root, and that $constantval=0$ should be\nignored.\n\n\\textbf{First solution:}\nWrite $transcend(constantval) = variable \\prod_{aggregate=1}^{infinite} (constantval-summitval)$, so that\n\\[\n\\frac{staticfunc'(constantval)}{staticfunc(constantval)} = \\frac{1}{2 constantval} \\sum_{aggregate=1}^{infinite} \\frac{constantval+ summitval}{constantval- summitval}.\n\\]\nNow if $constantval\\neq summitval$ for all $aggregate$,then\n\\[\n\\frac{constantval+ summitval}{constantval- summitval}\n= \\frac{(constantval+ summitval)(\\overline{constantval}-\\overline{summitval})}{|constantval-summitval|^2}\n= \\frac{|constantval|^2-1+2 \\\\mathrm{Im}(\\overline{constantval} summitval)}{|constantval-summitval|^2},\n\\]\nand so\n\\[\n\\\\mathrm{Re}\\, \\frac{constantval\\, staticfunc'(constantval)}{staticfunc(constantval)} = \\frac{|constantval|^2-1}{2} \\left(\n\\sum_{aggregate} \\frac{1}{|constantval-summitval|^2} \\right).\n\\]\nSince the quantity in parentheses\nis positive, $staticfunc'(constantval)/staticfunc(constantval)$ can be $0$ only if $|constantval|=1$. If on the other\nhand $constantval = summitval$ for some $aggregate$, then $|constantval|=1$ anyway.\n\n\\textbf{Second solution:}\nWrite $transcend(constantval) = variable\\prod_{aggregate=1}^{infinite} (constantval-summitval)$, so that\n\\[\n\\frac{staticfunc'(constantval)}{staticfunc(constantval)}\n= \\sum_{aggregate=1}^{infinite} \\left( \\frac{1}{constantval-summitval} - \\frac{1}{2 constantval} \\right).\n\\]\nWe first check that $staticfunc'(constantval) \\neq 0$ whenever $constantval$ is real and $constantval>1$.\nIn this case, for $summitval = e^{i \\counteranglej}$, we have $constantval - summitval =\n(constantval - \\cos (\\counteranglej)) + \\sin(\\counteranglej) i$, so the real part of\n$\\frac{1}{constantval-summitval} - \\frac{1}{2 constantval}$ is\n\\[\n\\frac{constantval - \\cos(\\counteranglej)}{constantval^2 - 2 constantval \\cos(\\counteranglej) + 1} - \\frac{1}{2 constantval}\n= \\frac{constantval^2-1}{2 constantval(constantval^2 - 2 constantval \\cos(\\counteranglej) + 1)} > 0.\n\\]\nHence $staticfunc'(constantval)/staticfunc(constantval)$ has positive real part, so $staticfunc'(constantval)/staticfunc(constantval)$ and hence\n$staticfunc(constantval)$ are nonzero.\n\nApplying the same argument after replacing $transcend(constantval)$ by $transcend(e^{i \\counterangle} constantval)$,\nwe deduce that $staticfunc'$ cannot have any roots outside the unit circle.\nApplying the same argument after replacing $transcend(constantval)$ by $constantval^{infinite} transcend(1/constantval)$, we\nalso deduce that $staticfunc'$ cannot have any roots inside the unit circle.\nHence all roots of $staticfunc'$ have absolute value $1$, as desired.\n\n\\textbf{Third solution:}\nWrite $transcend(constantval) = variable \\prod_{aggregate=1}^{infinite} (constantval - summitval)$ and\nput $summitval = e^{2 i \\counteranglej}$. Note that $staticfunc(e^{2 i \\counterangle})$\nis equal to a nonzero constant times\n\\begin{align*}\nmonotone(\\counterangle) &= \\prod_{aggregate=1}^{infinite} \\frac{e^{i (\\counterangle + \\counteranglej)}\n- e^{-i(\\counterangle + \\counteranglej)}}{2i} = \\prod_{aggregate=1}^{infinite} \\sin(\\counterangle +\\counteranglej).\n\\end{align*}\nSince $monotone$ has at least $2infinite$ roots (counting multiplicity)\nin the interval $[0, 2\\pi)$, $monotone'$ does also by repeated application of\nRolle's theorem. Since $staticfunc'(e^{2 i \\counterangle}) = 2i e^{2i \\counterangle} monotone'(\\counterangle)$,\n$staticfunc'(constantval^2)$ has at least $2infinite$ roots on the unit circle. Since $staticfunc'(constantval^2)$ is equal to\n$constantval^{-infinite-1}$ times a polynomial of degree $2infinite$, $staticfunc'(constantval^2)$ has all roots on the\nunit circle, as then does $staticfunc'(constantval)$.\n\n\\textbf{Remarks:}\nThe second solution imitates the proof of the Gauss-Lucas theorem:\nthe roots of the derivative of a complex polynomial lie in the convex hull of\nthe roots of the original polynomial.\nThe second solution is close to problem B3 from the 2000 Putnam.\nA hybrid between the first and third solutions is to check that\non the unit circle, $\\mathrm{Re}(constantval staticfunc'(constantval)/staticfunc(constantval)) = 0$ while between any\ntwo roots of $transcend$, $\\mathrm{Im}(constantval staticfunc'(constantval)/staticfunc(constantval))$ runs from $+\\infty$ to $-\\infty$\nand so must have a zero crossing. (This only works when $transcend$ has distinct roots,\nbut the general case follows by the continuity of the roots of a polynomial\nas functions of the coefficients.)\nOne can also construct a  solution using Rouch\\'e's theorem."
    },
    "garbled_string": {
      "map": {
        "z": "qzxwvtnp",
        "g": "hjgrksla",
        "p": "mnbvcxqe",
        "h": "trsdifgh",
        "c": "lkjhgfdsa",
        "n": "poiuytrew",
        "r_j": "qazplmok",
        "j": "wsxedcrfv",
        "\\theta": "\\klmnbvcz",
        "\\theta_j": "\\xswedcvfr"
      },
      "question": "Let $mnbvcxqe(qzxwvtnp)$ be a polynomial of degree $poiuytrew$ all of whose zeros have absolute\nvalue 1 in the complex  plane. Put $hjgrksla(qzxwvtnp) = mnbvcxqe(qzxwvtnp)/qzxwvtnp^{poiuytrew/2}$. Show that all zeros\nof $hjgrksla'(qzxwvtnp) = 0$ have absolute value 1.",
      "solution": "Note that it is implicit in the problem that $mnbvcxqe$ is nonconstant,\none may take any branch of the square root, and that $qzxwvtnp=0$ should be\nignored.\n\n\\textbf{First solution:}\nWrite $mnbvcxqe(qzxwvtnp) = lkjhgfdsa \\prod_{wsxedcrfv=1}^{poiuytrew} (qzxwvtnp-qazplmok)$, so that\n\\[\n\\frac{hjgrksla'(qzxwvtnp)}{hjgrksla(qzxwvtnp)} = \\frac{1}{2qzxwvtnp} \\sum_{wsxedcrfv=1}^{poiuytrew} \\frac{qzxwvtnp+qazplmok}{qzxwvtnp-qazplmok}.\n\\]\nNow if $qzxwvtnp\\neq qazplmok$ for all $wsxedcrfv$, then\n\\[\n\\frac{qzxwvtnp+qazplmok}{qzxwvtnp-qazplmok}\n= \\frac{(qzxwvtnp+qazplmok)(\\overline{qzxwvtnp}-\\overline{qazplmok})}{|qzxwvtnp-qazplmok|^2}\n= \\frac{|qzxwvtnp|^2-1+2 \\\\mathrm{Im}(\\overline{qzxwvtnp} \\, qazplmok)}{|qzxwvtnp-qazplmok|^2},\n\\]\nand so\n\\[\n\\\\mathrm{Re}\\, \\frac{qzxwvtnp \\, hjgrksla'(qzxwvtnp)}{hjgrksla(qzxwvtnp)} = \\frac{|qzxwvtnp|^2-1}{2} \\left(\n\\sum_{wsxedcrfv} \\frac{1}{|qzxwvtnp-qazplmok|^2} \\right).\n\\]\nSince the quantity in parentheses\nis positive, $hjgrksla'(qzxwvtnp)/hjgrksla(qzxwvtnp)$ can be $0$ only if $|qzxwvtnp|=1$. If on the other\nhand $qzxwvtnp = qazplmok$ for some $wsxedcrfv$, then $|qzxwvtnp|=1$ anyway.\n\n\\textbf{Second solution:}\nWrite $mnbvcxqe(qzxwvtnp) = lkjhgfdsa\\prod_{wsxedcrfv=1}^{poiuytrew} (qzxwvtnp-qazplmok)$, so that\n\\[\n\\frac{hjgrksla'(qzxwvtnp)}{hjgrksla(qzxwvtnp)}\n= \\sum_{wsxedcrfv=1}^{poiuytrew} \\left( \\frac{1}{qzxwvtnp-qazplmok} - \\frac{1}{2qzxwvtnp} \\right).\n\\]\nWe first check that $hjgrksla'(qzxwvtnp) \\neq 0$ whenever $qzxwvtnp$ is real and $qzxwvtnp>1$.\nIn this case, for $qazplmok = e^{i \\xswedcvfr}$, we have $qzxwvtnp - qazplmok =\n(qzxwvtnp - \\cos (\\xswedcvfr)) + \\sin(\\xswedcvfr) i$, so the real part of\n$\\frac{1}{qzxwvtnp-qazplmok} - \\frac{1}{2qzxwvtnp}$ is\n\\[\n\\frac{qzxwvtnp - \\cos(\\xswedcvfr)}{qzxwvtnp^2 - 2qzxwvtnp \\cos(\\xswedcvfr) + 1} - \\frac{1}{2qzxwvtnp}\n= \\frac{qzxwvtnp^2-1}{2qzxwvtnp(qzxwvtnp^2 - 2qzxwvtnp \\cos(\\xswedcvfr) + 1)} > 0.\n\\]\nHence $hjgrksla'(qzxwvtnp)/hjgrksla(qzxwvtnp)$ has positive real part, so $hjgrksla'(qzxwvtnp)/hjgrksla(qzxwvtnp)$ and hence\n$hjgrksla(qzxwvtnp)$ are nonzero.\n\nApplying the same argument after replacing $mnbvcxqe(qzxwvtnp)$ by $mnbvcxqe(e^{i \\klmnbvcz} qzxwvtnp)$,\nwe deduce that $hjgrksla'$ cannot have any roots outside the unit circle.\nApplying the same argument after replacing $mnbvcxqe(qzxwvtnp)$ by $qzxwvtnp^{poiuytrew} mnbvcxqe(1/qzxwvtnp)$, we\nalso deduce that $hjgrksla'$ cannot have any roots inside the unit circle.\nHence all roots of $hjgrksla'$ have absolute value 1, as desired.\n\n\\textbf{Third solution:}\nWrite $mnbvcxqe(qzxwvtnp) = lkjhgfdsa \\prod_{wsxedcrfv=1}^{poiuytrew} (qzxwvtnp - qazplmok)$ and\nput $qazplmok = e^{2 i \\xswedcvfr}$. Note that $hjgrksla(e^{2 i \\klmnbvcz})$\nis equal to a nonzero constant times\n\\begin{align*}\ntrsdifgh(\\klmnbvcz) &= \\prod_{wsxedcrfv=1}^{poiuytrew} \\frac{e^{i (\\klmnbvcz + \\xswedcvfr)}\n- e^{-i(\\klmnbvcz + \\xswedcvfr)}}{2i} = \\prod_{wsxedcrfv=1}^{poiuytrew} \\sin(\\klmnbvcz +\\xswedcvfr).\n\\end{align*}\nSince $trsdifgh$ has at least $2poiuytrew$ roots (counting multiplicity)\nin the interval $[0, 2\\pi)$, $trsdifgh'$ does also by repeated application of\nRolle's theorem. Since $hjgrksla'(e^{2 i \\klmnbvcz}) = 2i e^{2i \\klmnbvcz} trsdifgh'(\\klmnbvcz)$,\n$hjgrksla'(qzxwvtnp^2)$ has at least $2poiuytrew$ roots on the unit circle. Since $hjgrksla'(qzxwvtnp^2)$ is equal to\n$qzxwvtnp^{-poiuytrew-1}$ times a polynomial of degree $2poiuytrew$, $hjgrksla'(qzxwvtnp^2)$ has all roots on the\nunit circle, as then does $hjgrksla'(qzxwvtnp)$.\n\n\\textbf{Remarks:}\nThe second solution imitates the proof of the Gauss-Lucas theorem:\nthe roots of the derivative of a complex polynomial lie in the convex hull of\nthe roots of the original polynomial.\nThe second solution is close to problem B3 from the 2000 Putnam.\nA hybrid between the first and third solutions is to check that\non the unit circle, $\\\\mathrm{Re}(qzxwvtnp hjgrksla'(qzxwvtnp)/hjgrksla(qzxwvtnp)) = 0$ while between any\ntwo roots of $mnbvcxqe$, $\\\\mathrm{Im}(qzxwvtnp hjgrksla'(qzxwvtnp)/hjgrksla(qzxwvtnp))$ runs from $+\\\\infty$ to $-\\\\infty$\nand so must have a zero crossing. (This only works when $mnbvcxqe$ has distinct roots,\nbut the general case follows by the continuity of the roots of a polynomial\nas functions of the coefficients.)\nOne can also construct a  solution using Rouch\\'e's theorem."
    },
    "kernel_variant": {
      "question": "Fix the radius R = 3.  Let f(z) be a non-constant polynomial of degree k ≥ 1 whose zeros s₁,…,s_k all satisfy |s_j| = R.  Choose a ray L emanating from the origin – for definiteness the negative real axis (−∞,0] – and delete it from the plane:\n  Ω := ℂ \\ L.\nOn the simply–connected domain Ω fix the principal branch Log z of the logarithm and, for every z ∈ Ω, put\n  z^{k/2} := exp((k/2)·Log z).\n(When k is odd, this amounts to fixing a single-valued branch of the square root.)  Define the holomorphic function\n  h(z) := f(z) / z^{k/2}\t (z ∈ Ω, z ≠ 0).\nProve that every zero of the derivative h′(z) lies on the circle |z| = R.",
      "solution": "Factor f :\n  f(z) = a ∏_{j = 1}^{k} (z − s_j),  |s_j| = R = 3, a ∈ ℂ\\{0}.\nBecause h(z) = f(z) / z^{k/2} on Ω, logarithmic differentiation gives\n  h′(z)/h(z) = f′(z)/f(z) − k/(2z)\n        = Σ_{j = 1}^{k} ( 1/(z − s_j) − 1/(2z) )\n        = (1/2z) Σ_{j = 1}^{k} (z + s_j)/(z − s_j).\nHence\n(1)\t z·h′(z)/h(z) = (1/2) Σ_{j = 1}^{k} (z + s_j)/(z − s_j).\n\nA geometric identity.\nFor any fixed zero s with |s| = R and any z ∈ Ω \\ {s}\n  Re [ (z + s)/(z − s) ]\n  = Re [ (z + s)(\\overline z − \\overline s) / |z − s|² ]\n  = (|z|² − |s|²) / |z − s|²\n  = (|z|² − R²) / |z − s|².\n\nTaking real parts in (1) therefore yields\n  Re [ z·h′(z)/h(z) ]\n  = (|z|² − R²)/2 · Σ_{j = 1}^{k} 1/|z − s_j|².   (2)\n\nThe sum in (2) is strictly positive whenever z ≠ s_j for every j.  Consequently\n  Re [ z·h′(z)/h(z) ] = 0 ⇒ |z| = R.                               (3)\n\nLocating the zeros of h′.\nLet z₀ ∈ Ω satisfy h′(z₀) = 0.\n\n• If z₀ ≠ s_j for all j, then h′(z₀) = 0 implies h′(z₀)/h(z₀) = 0, hence z₀·h′(z₀)/h(z₀) = 0.  By (3) we obtain |z₀| = R.\n\n• If z₀ = s_m for some m, then by the hypothesis on the zeros of f we again have |z₀| = R.\n\nThus every zero of h′ lies on the circle |z| = R, completing the proof.\n\nBranch considerations.\nAll operations (division by z, logarithmic differentiation, taking real parts) are carried out in the domain Ω on which the chosen branch z^{k/2} is single-valued and holomorphic.  The argument above therefore applies equally whether k is even or odd.",
      "_meta": {
        "core_steps": [
          "Express p(z)=c·∏(z−r_j) with |r_j|=1 and write the logarithmic derivative: g'(z)/g(z)= (1/2z)∑(z+r_j)/(z−r_j).",
          "Multiply by z and take real parts: Re[z·g'(z)/g(z)] = (|z|^2−1)/2 · ∑ 1/|z−r_j|^2.",
          "Note the summation factor is positive unless z coincides with a root.",
          "Hence Re[z·g'(z)/g(z)] vanishes only when |z|=1 (or when z=r_j, which already has |z|=1).",
          "Conclude that every zero of g'(z) lies on the unit circle."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Radius of the circle on which all roots r_j lie (scaling of the problem)",
            "original": "1"
          },
          "slot2": {
            "description": "Symbol n denoting the (positive) degree/number of roots of p",
            "original": "n"
          },
          "slot3": {
            "description": "Leading constant c in the factorization p(z)=c∏(z−r_j)",
            "original": "c"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}