{
  "index": "1959-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. Find all complex-valued functions \\( f \\) of a complex variable such that \\( f(z)+z f(1-z)=1+z \\) for all \\( z \\).",
  "solution": "Solution. We have\n\\[\nf(z)+z f(1-z)=1+z\n\\]\nfor all \\( z \\). Replace \\( z \\) by \\( 1-z \\) in this equation to get\n\\[\nf(1-z)+(1-z) f(z)=2-z\n\\]\n\nEliminating \\( f(1-z) \\) from (1) and (2), we get\n\\[\n\\left(1-z+z^{2}\\right) f(z)=1-z+z^{2}\n\\]\n\nHence \\( f(z)=1 \\) for all \\( z \\) except possibly the two values \\( e^{ \\pm \\pi i / 3} \\) for which \\( 1-z+z^{2}=0 \\).\n\nLet \\( \\alpha=e^{i \\pi / 3}, \\vec{\\alpha}=e^{-i \\pi / 3} \\). Then \\( \\alpha+\\bar{\\alpha}=1, \\alpha \\bar{\\alpha}=1 \\). Let \\( f(\\alpha)=1+\\beta \\), \\( f(\\bar{\\alpha})=1+\\gamma \\). Then equation (1) with \\( z=\\alpha \\) becomes\n\\[\n\\beta+\\alpha \\gamma=0,\n\\]\nso we choose \\( \\beta \\) arbitrarily and set \\( \\gamma=-\\bar{\\alpha} \\beta \\). With these values (1) is readily checked for \\( z=\\alpha, \\bar{\\alpha} \\). Hence a function \\( f \\) satisfies the given equation if and only if it has the form\n\\[\n\\begin{array}{l}\nf(\\alpha)=1+\\beta \\\\\nf(\\bar{\\alpha})=1-\\bar{\\alpha} \\beta\n\\end{array}\n\\]\n\\[\nf(z)=1 \\text { for all other values of } z,\n\\]\nwhere \\( \\beta \\) is any complex number.",
  "vars": [
    "z",
    "f"
  ],
  "params": [
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "complexvar",
        "f": "complexfunc",
        "\\alpha": "rootalpha",
        "\\beta": "coeffbeta",
        "\\gamma": "coeffgamma"
      },
      "question": "3. Find all complex-valued functions \\( complexfunc \\) of a complex variable such that \\( complexfunc(complexvar)+complexvar complexfunc(1-complexvar)=1+complexvar \\) for all \\( complexvar \\).",
      "solution": "Solution. We have\n\\[\ncomplexfunc(complexvar)+complexvar complexfunc(1-complexvar)=1+complexvar\n\\]\nfor all \\( complexvar \\). Replace \\( complexvar \\) by \\( 1-complexvar \\) in this equation to get\n\\[\ncomplexfunc(1-complexvar)+(1-complexvar) complexfunc(complexvar)=2-complexvar\n\\]\n\nEliminating complexfunc(1-complexvar) from (1) and (2), we get\n\\[\n\\left(1-complexvar+complexvar^{2}\\right) complexfunc(complexvar)=1-complexvar+complexvar^{2}\n\\]\n\nHence \\( complexfunc(complexvar)=1 \\) for all \\( complexvar \\) except possibly the two values \\( e^{ \\pm \\pi i / 3} \\) for which \\( 1-complexvar+complexvar^{2}=0 \\).\n\nLet \\( rootalpha=e^{i \\pi / 3}, \\vec{rootalpha}=e^{-i \\pi / 3} \\). Then \\( rootalpha+\\bar{rootalpha}=1, rootalpha \\bar{rootalpha}=1 \\). Let \\( complexfunc(rootalpha)=1+coeffbeta \\), \\( complexfunc(\\bar{rootalpha})=1+coeffgamma \\). Then equation (1) with \\( complexvar=rootalpha \\) becomes\n\\[\ncoeffbeta+rootalpha coeffgamma=0,\n\\]\nso we choose \\( coeffbeta \\) arbitrarily and set \\( coeffgamma=-\\bar{rootalpha} coeffbeta \\). With these values (1) is readily checked for \\( complexvar=rootalpha, \\bar{rootalpha} \\). Hence a function \\( complexfunc \\) satisfies the given equation if and only if it has the form\n\\[\n\\begin{array}{l}\ncomplexfunc(rootalpha)=1+coeffbeta \\\\\ncomplexfunc(\\bar{rootalpha})=1-\\bar{rootalpha} coeffbeta\n\\end{array}\n\\]\n\\[\ncomplexfunc(complexvar)=1 \\text { for all other values of } complexvar,\n\\]\nwhere \\( coeffbeta \\) is any complex number."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "pineapple",
        "f": "horizonline",
        "\\alpha": "starlight",
        "\\beta": "moonshadow",
        "\\gamma": "cloudburst"
      },
      "question": "3. Find all complex-valued functions \\( horizonline \\) of a complex variable such that \\( horizonline(pineapple)+pineapple horizonline(1-pineapple)=1+pineapple \\) for all \\( pineapple \\).",
      "solution": "Solution. We have\n\\[\nhorizonline(pineapple)+pineapple horizonline(1-pineapple)=1+pineapple\n\\]\nfor all \\( pineapple \\). Replace \\( pineapple \\) by \\( 1-pineapple \\) in this equation to get\n\\[\nhorizonline(1-pineapple)+(1-pineapple) horizonline(pineapple)=2-pineapple\n\\]\n\nEliminating \\( horizonline(1-pineapple) \\) from (1) and (2), we get\n\\[\n\\left(1-pineapple+pineapple^{2}\\right) horizonline(pineapple)=1-pineapple+pineapple^{2}\n\\]\n\nHence \\( horizonline(pineapple)=1 \\) for all \\( pineapple \\) except possibly the two values \\( e^{ \\pm \\pi i / 3} \\) for which \\( 1-pineapple+pineapple^{2}=0 \\).\n\nLet \\( starlight=e^{i \\pi / 3}, \\vec{starlight}=e^{-i \\pi / 3} \\). Then \\( starlight+\\bar{starlight}=1, starlight \\bar{starlight}=1 \\). Let \\( horizonline(starlight)=1+moonshadow \\), \\( horizonline(\\bar{starlight})=1+cloudburst \\). Then equation (1) with \\( pineapple=starlight \\) becomes\n\\[\nmoonshadow+starlight cloudburst=0,\n\\]\nso we choose \\( moonshadow \\) arbitrarily and set \\( cloudburst=-\\bar{starlight} moonshadow \\). With these values (1) is readily checked for \\( pineapple=starlight, \\bar{starlight} \\). Hence a function \\( horizonline \\) satisfies the given equation if and only if it has the form\n\\[\n\\begin{array}{l}\nhorizonline(starlight)=1+moonshadow \\\\\nhorizonline(\\bar{starlight})=1-\\bar{starlight} moonshadow\n\\end{array}\n\\]\n\\[\nhorizonline(pineapple)=1 \\text { for all other values of } pineapple,\n\\]\nwhere \\( moonshadow \\) is any complex number."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "realconstant",
        "f": "nonfunction",
        "\\alpha": "lastletter",
        "\\beta": "startletter",
        "\\gamma": "middleletter"
      },
      "question": "3. Find all complex-valued functions \\( nonfunction \\) of a complex variable such that \\( nonfunction(realconstant)+realconstant\\,nonfunction(1-realconstant)=1+realconstant \\) for all \\( realconstant \\).",
      "solution": "Solution. We have\n\\[\nnonfunction(realconstant)+realconstant\\,nonfunction(1-realconstant)=1+realconstant\n\\]\nfor all \\( realconstant \\). Replace \\( realconstant \\) by \\( 1-realconstant \\) in this equation to get\n\\[\nnonfunction(1-realconstant)+(1-realconstant)\\,nonfunction(realconstant)=2-realconstant\n\\]\n\nEliminating \\( nonfunction(1-realconstant) \\) from (1) and (2), we get\n\\[\n\\left(1-realconstant+realconstant^{2}\\right) nonfunction(realconstant)=1-realconstant+realconstant^{2}\n\\]\n\nHence \\( nonfunction(realconstant)=1 \\) for all \\( realconstant \\) except possibly the two values \\( e^{ \\pm \\pi i / 3} \\) for which \\( 1-realconstant+realconstant^{2}=0 \\).\n\nLet \\( lastletter=e^{i \\pi / 3}, \\vec{lastletter}=e^{-i \\pi / 3} \\). Then \\( lastletter+\\bar{lastletter}=1, lastletter\\,\\bar{lastletter}=1 \\). Let \\( nonfunction(lastletter)=1+startletter \\), \\( nonfunction(\\bar{lastletter})=1+middleletter \\). Then equation (1) with \\( realconstant=lastletter \\) becomes\n\\[\nstartletter+lastletter\\,middleletter=0,\n\\]\nso we choose \\( startletter \\) arbitrarily and set \\( middleletter=-\\bar{lastletter}\\,startletter \\). With these values (1) is readily checked for \\( realconstant=lastletter, \\bar{lastletter} \\). Hence a function \\( nonfunction \\) satisfies the given equation if and only if it has the form\n\\[\n\\begin{array}{l}\nnonfunction(lastletter)=1+startletter \\\\\nnonfunction(\\bar{lastletter})=1-\\bar{lastletter}\\,startletter\n\\end{array}\n\\]\n\\[\nnonfunction(realconstant)=1 \\text { for all other values of } realconstant,\n\\]\nwhere \\( startletter \\) is any complex number."
    },
    "garbled_string": {
      "map": {
        "z": "qzxwvtnp",
        "f": "hjgrksla",
        "\\alpha": "plmsdfgh",
        "\\beta": "rtyuiojk",
        "\\gamma": "cvbnmqwe"
      },
      "question": "Find all complex-valued functions \\( hjgrksla \\) of a complex variable such that \\( hjgrksla(qzxwvtnp)+qzxwvtnp hjgrksla(1-qzxwvtnp)=1+qzxwvtnp \\) for all \\( qzxwvtnp \\).",
      "solution": "Solution. We have\n\\[\nhjgrksla(qzxwvtnp)+qzxwvtnp hjgrksla(1-qzxwvtnp)=1+qzxwvtnp\n\\]\nfor all \\( qzxwvtnp \\). Replace \\( qzxwvtnp \\) by \\( 1-qzxwvtnp \\) in this equation to get\n\\[\nhjgrksla(1-qzxwvtnp)+(1-qzxwvtnp) hjgrksla(qzxwvtnp)=2-qzxwvtnp\n\\]\n\nEliminating \\( hjgrksla(1-qzxwvtnp) \\) from (1) and (2), we get\n\\[\n\\left(1-qzxwvtnp+qzxwvtnp^{2}\\right) hjgrksla(qzxwvtnp)=1-qzxwvtnp+qzxwvtnp^{2}\n\\]\n\nHence \\( hjgrksla(qzxwvtnp)=1 \\) for all \\( qzxwvtnp \\) except possibly the two values \\( e^{ \\pm \\pi i / 3} \\) for which \\( 1-qzxwvtnp+qzxwvtnp^{2}=0 \\).\n\nLet \\( plmsdfgh=e^{i \\pi / 3}, \\vec{plmsdfgh}=e^{-i \\pi / 3} \\). Then \\( plmsdfgh+\\bar{plmsdfgh}=1, plmsdfgh \\bar{plmsdfgh}=1 \\). Let \\( hjgrksla(plmsdfgh)=1+rtyuiojk \\), \\( hjgrksla(\\bar{plmsdfgh})=1+cvbnmqwe \\). Then equation (1) with \\( qzxwvtnp=plmsdfgh \\) becomes\n\\[\nrtyuiojk+plmsdfgh cvbnmqwe=0,\n\\]\nso we choose \\( rtyuiojk \\) arbitrarily and set \\( cvbnmqwe=-\\bar{plmsdfgh} rtyuiojk \\). With these values (1) is readily checked for \\( qzxwvtnp=plmsdfgh, \\bar{plmsdfgh} \\). Hence a function \\( hjgrksla \\) satisfies the given equation if and only if it has the form\n\\[\n\\begin{array}{l}\nhjgrksla(plmsdfgh)=1+rtyuiojk \\\\\nhjgrksla(\\bar{plmsdfgh})=1-\\bar{plmsdfgh} rtyuiojk\n\\end{array}\n\\]\n\\[\nhjgrksla(qzxwvtnp)=1 \\text { for all other values of } qzxwvtnp,\n\\]\nwhere \\( rtyuiojk \\) is any complex number."
    },
    "kernel_variant": {
      "question": "Let  \n\n  Δ(z)=1−z²+z³ = ∏_{j=1}^{3}(z−ρ_j)  (ρ₁,ρ₂,ρ₃ pairwise distinct)  \n\nand set S={ρ₁,ρ₂,ρ₃}.  \n\nFor an entire function f:ℂ→ℂ consider the coupled system  \n\n(I) f(z)+z·f(1−z)=1+z  \n(II) f(1−z)+(1−z)·f(−z)=2−z  \n(III) f(−z)−z·f(z)=1−z          (∗)\n\nassumed to hold for every z∈ℂ.\n\n1.  Prove that, for entire f, the system (∗) is equivalent to the single algebraic identity  \n    Δ(z)·[f(z)−1]=0   for all z∈ℂ.  \n\n2.  Deduce that the unique entire (indeed meromorphic) solution of the full system (∗) is the constant function f≡1.\n\n(Harder refinement – selective rigidity)\n\n3.  Let g be an arbitrary entire function and suppose that g satisfies exactly one of the three relations (I)–(III).  \n (a)  Show that if g satisfies (I) then g≡1.  \n (b)  Show that if g satisfies (III) then g≡1.  \n (c)  Show that equation (II) admits non-constant entire solutions. Construct an explicit infinite–dimensional family of such solutions and prove that every function of this family indeed satisfies (II).  \n (d)  Conclude that the relations (I) and (III) are rigid (they force an entire solution to be the constant 1 and automatically imply the other two), whereas relation (II) is flexible: it has an infinite-dimensional space of entire solutions and, by itself, does not entail either (I) or (III).\n\n4.  (Very hard)  Let  \n U := {z∈ℂ : {z,1−z}∩S = ∅} = ℂ\\bigl(S∪(1-S)\\bigr)  \nand let h be a meromorphic function on U that satisfies (I) there.  \n (a) Prove that h is holomorphic at the two points  \n   ζ₊=e^{iπ/3}, ζ₋=e^{−iπ/3}  (the zeros of D(z):=1−z+z²).  \n (b) Show that h extends meromorphically through every ρ∈S and that the extension coincides with the constant 1; in particular h≡1 on ℂ.\n\nDetermine all functions that satisfy the system (∗) and provide complete, rigorous proofs of every assertion.\n\n──────────────────────────────────────────────────────────────",
      "solution": "Throughout we use the matrices  \n\n M(z)=⎛1  z  0⎞   b(z)=⎛1+z⎞,   X(z)=⎛f(z)  ⎞,  \n    ⎜0  1  1−z⎟     ⎜2−z⎟       ⎜f(1−z)⎟,  \n    ⎝−z 0  1 ⎠    ⎝1−z⎠       ⎝f(−z) ⎠ ,\n\nso that (∗) is equivalent to M(z)X(z)=b(z).\n\n————————————————————————————————————————\n1.  Equivalence (∗) ⇔ Δ(z)[f(z)−1]=0    \n\nStep 1. det M(z).  \nA cofactor expansion gives  \n\n  det M(z)=1−z²+z³=:Δ(z).  (1)\n\nStep 2. Cramer’s rule for f(z).  \nReplace the first column of M(z) by b(z) and call the new matrix M₁(z).  \nAgain by a cofactor expansion  \n\n  det M₁(z)=1−z²+z³=Δ(z).  \n\nHence  \n\n  Δ(z)·f(z)=det M₁(z)=Δ(z) ⇒ Δ(z)[f(z)−1]=0  for all z∈ℂ. (2)\n\nStep 3. Converse.  \nIf an entire f satisfies (2) then f(z)=1 on ℂ\\S, an open set possessing an\naccumulation point; by the Identity Theorem f≡1.  Substituting f≡1\ninto (∗) confirms the three relations, whence (∗) and (2) are equivalent.\n\n————————————————————————————————————————\n2.  Uniqueness of the entire solution  \n\nPart 1 already enforces f≡1; consequently the full coupled system (I)–(III)\nadmits exactly one entire (indeed meromorphic) solution:\n\n    f(z)=1  (∀z∈ℂ).\n\n————————————————————————————————————————\n3.  Selective rigidity\n\nPut h(z):=g(z)−1.  The three relations become homogeneous equations for h.\n\n3(a) Equation (I).  \n(I) gives h(z)+z h(1−z)=0.  Replacing z by 1−z and eliminating h(1−z)\nyields  \n\n  (1−z+z²) h(z)=0=:D(z) h(z).  \n\nBecause h is entire, D(z) h(z)≡0 implies h vanishes identically: outside\nits two isolated zeros ζ₊,ζ₋ of D the factor in front of h is non-zero,\nand the Identity Theorem forces h≡0.  Hence g≡1.\n\n3(b) Equation (III).  \n(III) becomes  \n\n  h(−z)−z h(z)=0.  (3)\n\nReplacing z by −z and substituting the first relation into the second one\ngives  \n\n  (1+z²) h(z)=0.  \n\nAgain h is entire, so it must vanish identically; whence g≡1.\n\n3(c) Equation (II) – an infinite–dimensional solution space.  \n\nThe inhomogeneous equation  \n\n  h(1−z)+(1−z) h(−z)=0   (4)\n\ndoes not force h to vanish.  We construct an explicit family of\nnon-trivial entire solutions.  Let Φ be any entire function satisfying  \n\n  Φ(1−z)=Φ(−z)  for every z∈ℂ.  (5)\n\n(An abundance of such functions exists; for instance every exponential\nΦ_n(z)=e^{2π i n z} (n∈ℤ) has this property because\ne^{2π i n(1−z)}=e^{2π i n}e^{-2π i n z}=e^{-2π i n z}=Φ_n(−z).)\n\nDefine  \n\n  h_Φ(z):=Γ(z+1)\\sin(π z) Φ(z),    (6)\n\nand put  \n\n  g_Φ(z):=1+h_Φ(z).        (7)\n\nEntirety of h_Φ.\nΓ(z+1) has simple poles at the negative integers, while sin πz has the\nsame zeros; hence their product is an entire function.  Multiplying by\nthe entire Φ preserves entire-ness.\n\nVerification of (4).  \nWe use the two standard formulas  \n\n  Γ(2−z)=(1−z) Γ(1−z),    sin(π(1−z))=sin πz.  (8)\n\nFrom (6) and (8) we obtain  \n\nh_Φ(1−z)=(1−z) Γ(1−z) sin(π z) Φ(1−z),  \nh_Φ(−z)=−Γ(1−z) sin(π z) Φ(−z).\n\nBecause Φ satisfies (5), Φ(1−z)=Φ(−z), whence  \n\nh_Φ(1−z)+(1−z) h_Φ(−z)  \n  = (1−z) Γ(1−z) sin(π z) Φ(−z)  \n    −(1−z) Γ(1−z) sin(π z) Φ(−z)  = 0.\n\nThus every g_Φ defined in (7) satisfies equation (II).  Varying the\nentire function Φ subject to (5) produces an infinite-dimensional family\nof solutions; for Φ≡1 we recover the elementary example\n\n  g(z)=1+Γ(z+1)\\sin(π z)\n\nmentioned in the review.\n\nTherefore equation (II) alone is flexible: it has infinitely many\nentire solutions, none of which (except the constant one) satisfies\neither (I) or (III).\n\n3(d) Consequences.  \n• Relations (I) and (III) are rigid: every entire solution equals 1;  \n this solution necessarily fulfils the other two relations, so the whole\n system (∗) holds.  \n• Relation (II) is flexible: it possesses an infinite-dimensional space\n of entire solutions and does not imply (I) or (III).\n\n————————————————————————————————————————\n4.  Meromorphic solutions defined on U  \n\nPut k(z)=h(z)−1.  Throughout we only use relation (I); hence again the\npolynomial  \n\n  D(z)=1−z+z²  (9)\n\nreappears.\n\nEquation (I) reads on U  \n\n  k(z)+z k(1−z)=0, (10)  \n\nand replacing z by 1−z gives  \n\n  k(1−z)+(1−z) k(z)=0. (11)\n\nEliminating k(1−z) from (10)–(11) gives  \n\n  D(z) k(z)=0.   (12)\n\n4(a) The zeros ζ₊, ζ₋ of D.  \nOutside the points ζ± the factor D(z) is non-zero, so (12) forces\nk≡0.  Near ζ± the same argument as in the original solution shows that h\nis holomorphic and assumes the value 1 at each of ζ±.\n\n4(b) Extension across the roots of Δ.  \nLet ρ∈S.  Because ρ∉U, h is a priori undefined at ρ, but U contains a\npunctured neighbourhood of ρ.  On U we already know k≡0, so\nlim_{z→ρ,z∈U}k(z)=0.  If h had a pole at ρ the same limit would be\ninfinite – contradiction.  Hence ρ is a removable singularity of h with\nvalue 1.  Doing this for every root of Δ yields an entire extension that\nequals 1 on U and therefore everywhere.\n\n————————————————————————————————————————\nAnswer.  The coupled system (I)–(III) possesses exactly one entire—and in\nfact one meromorphic—solution:\n\n      f(z)=1  for every z∈ℂ. ∎\n\n──────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.511080",
        "was_fixed": false,
        "difficulty_analysis": "•  Multiple equations  –   The original problem has a single\nfunctional equation; here one must satisfy {\\em three} coupled relations\nsimultaneously.\n\n•  Three interacting arguments  –   \nThe unknown values \\(f(z),\\,f(1-z),\\,f(-z)\\) appear together, forcing a\n\\(3\\times3\\) linear system instead of the \\(2\\times2\\) system in the\nkernel variant.  This raises both algebraic complexity and bookkeeping.\n\n•  Non-trivial determinant  –  \nSolving demands computation of a cubic determinant\n\\(\\Delta(z)=1-z^{2}+z^{3}\\) whose vanishing introduces a {\\em singular\nlocus} of three complex points.  One must treat the regular and\nsingular cases separately, recognise the rank drop, and manage the\nresulting free parameters.\n\n•  Rational–function core  –   \nOutside the singular set the solution is a genuinely {\\em rational}\nfunction of degree three over degree three, rather than the constant\nsolution of the original problem.  Obtaining it requires systematic\nelimination, not mere substitution.\n\n•  Parametrisation on the singular set  –   \nAt each root of \\(\\Delta\\) the system loses rank, so compatibility must be\nchecked and the local freedom carefully parametrised; this demands a\nfull analysis of linear dependence, something entirely absent in the\noriginal exercise.\n\nIn summary, the problem keeps the original idea\n“relate \\(f\\) at two (now three) symmetric points and isolate a\nmultiplicative polynomial”, but it escalates the dimension, the number\nof constraints, the algebraic labour, and the subtlety of the singular\ncase, making it significantly harder than both the original and the\ncurrent kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  Δ(z)=1−z²+z³ = ∏_{j=1}^{3}(z−ρ_j)  (ρ₁,ρ₂,ρ₃ pairwise distinct)  \n\nand set S={ρ₁,ρ₂,ρ₃}.  \n\nFor an entire function f:ℂ→ℂ consider the coupled system  \n\n(I) f(z)+z·f(1−z)=1+z  \n(II) f(1−z)+(1−z)·f(−z)=2−z  \n(III) f(−z)−z·f(z)=1−z          (∗)\n\nassumed to hold for every z∈ℂ.\n\n1.  Prove that, for entire f, the system (∗) is equivalent to the single algebraic identity  \n    Δ(z)·[f(z)−1]=0   for all z∈ℂ.  \n\n2.  Deduce that the unique entire (indeed meromorphic) solution of the full system (∗) is the constant function f≡1.\n\n(Harder refinement – selective rigidity)\n\n3.  Let g be an arbitrary entire function and suppose that g satisfies exactly one of the three relations (I)–(III).  \n (a)  Show that if g satisfies (I) then g≡1.  \n (b)  Show that if g satisfies (III) then g≡1.  \n (c)  Show that equation (II) admits non-constant entire solutions. Construct an explicit infinite–dimensional family of such solutions and prove that every function of this family indeed satisfies (II).  \n (d)  Conclude that the relations (I) and (III) are rigid (they force an entire solution to be the constant 1 and automatically imply the other two), whereas relation (II) is flexible: it has an infinite-dimensional space of entire solutions and, by itself, does not entail either (I) or (III).\n\n4.  (Very hard)  Let  \n U := {z∈ℂ : {z,1−z}∩S = ∅} = ℂ\\bigl(S∪(1-S)\\bigr)  \nand let h be a meromorphic function on U that satisfies (I) there.  \n (a) Prove that h is holomorphic at the two points  \n   ζ₊=e^{iπ/3}, ζ₋=e^{−iπ/3}  (the zeros of D(z):=1−z+z²).  \n (b) Show that h extends meromorphically through every ρ∈S and that the extension coincides with the constant 1; in particular h≡1 on ℂ.\n\nDetermine all functions that satisfy the system (∗) and provide complete, rigorous proofs of every assertion.\n\n──────────────────────────────────────────────────────────────",
      "solution": "Throughout we use the matrices  \n\n M(z)=⎛1  z  0⎞   b(z)=⎛1+z⎞,   X(z)=⎛f(z)  ⎞,  \n    ⎜0  1  1−z⎟     ⎜2−z⎟       ⎜f(1−z)⎟,  \n    ⎝−z 0  1 ⎠    ⎝1−z⎠       ⎝f(−z) ⎠ ,\n\nso that (∗) is equivalent to M(z)X(z)=b(z).\n\n————————————————————————————————————————\n1.  Equivalence (∗) ⇔ Δ(z)[f(z)−1]=0    \n\nStep 1. det M(z).  \nA cofactor expansion gives  \n\n  det M(z)=1−z²+z³=:Δ(z).  (1)\n\nStep 2. Cramer’s rule for f(z).  \nReplace the first column of M(z) by b(z) and call the new matrix M₁(z).  \nAgain by a cofactor expansion  \n\n  det M₁(z)=1−z²+z³=Δ(z).  \n\nHence  \n\n  Δ(z)·f(z)=det M₁(z)=Δ(z) ⇒ Δ(z)[f(z)−1]=0  for all z∈ℂ. (2)\n\nStep 3. Converse.  \nIf an entire f satisfies (2) then f(z)=1 on ℂ\\S, an open set possessing an\naccumulation point; by the Identity Theorem f≡1.  Substituting f≡1\ninto (∗) confirms the three relations, whence (∗) and (2) are equivalent.\n\n————————————————————————————————————————\n2.  Uniqueness of the entire solution  \n\nPart 1 already enforces f≡1; consequently the full coupled system (I)–(III)\nadmits exactly one entire (indeed meromorphic) solution:\n\n    f(z)=1  (∀z∈ℂ).\n\n————————————————————————————————————————\n3.  Selective rigidity\n\nPut h(z):=g(z)−1.  The three relations become homogeneous equations for h.\n\n3(a) Equation (I).  \n(I) gives h(z)+z h(1−z)=0.  Replacing z by 1−z and eliminating h(1−z)\nyields  \n\n  (1−z+z²) h(z)=0=:D(z) h(z).  \n\nBecause h is entire, D(z) h(z)≡0 implies h vanishes identically: outside\nits two isolated zeros ζ₊,ζ₋ of D the factor in front of h is non-zero,\nand the Identity Theorem forces h≡0.  Hence g≡1.\n\n3(b) Equation (III).  \n(III) becomes  \n\n  h(−z)−z h(z)=0.  (3)\n\nReplacing z by −z and substituting the first relation into the second one\ngives  \n\n  (1+z²) h(z)=0.  \n\nAgain h is entire, so it must vanish identically; whence g≡1.\n\n3(c) Equation (II) – an infinite–dimensional solution space.  \n\nThe inhomogeneous equation  \n\n  h(1−z)+(1−z) h(−z)=0   (4)\n\ndoes not force h to vanish.  We construct an explicit family of\nnon-trivial entire solutions.  Let Φ be any entire function satisfying  \n\n  Φ(1−z)=Φ(−z)  for every z∈ℂ.  (5)\n\n(An abundance of such functions exists; for instance every exponential\nΦ_n(z)=e^{2π i n z} (n∈ℤ) has this property because\ne^{2π i n(1−z)}=e^{2π i n}e^{-2π i n z}=e^{-2π i n z}=Φ_n(−z).)\n\nDefine  \n\n  h_Φ(z):=Γ(z+1)\\sin(π z) Φ(z),    (6)\n\nand put  \n\n  g_Φ(z):=1+h_Φ(z).        (7)\n\nEntirety of h_Φ.\nΓ(z+1) has simple poles at the negative integers, while sin πz has the\nsame zeros; hence their product is an entire function.  Multiplying by\nthe entire Φ preserves entire-ness.\n\nVerification of (4).  \nWe use the two standard formulas  \n\n  Γ(2−z)=(1−z) Γ(1−z),    sin(π(1−z))=sin πz.  (8)\n\nFrom (6) and (8) we obtain  \n\nh_Φ(1−z)=(1−z) Γ(1−z) sin(π z) Φ(1−z),  \nh_Φ(−z)=−Γ(1−z) sin(π z) Φ(−z).\n\nBecause Φ satisfies (5), Φ(1−z)=Φ(−z), whence  \n\nh_Φ(1−z)+(1−z) h_Φ(−z)  \n  = (1−z) Γ(1−z) sin(π z) Φ(−z)  \n    −(1−z) Γ(1−z) sin(π z) Φ(−z)  = 0.\n\nThus every g_Φ defined in (7) satisfies equation (II).  Varying the\nentire function Φ subject to (5) produces an infinite-dimensional family\nof solutions; for Φ≡1 we recover the elementary example\n\n  g(z)=1+Γ(z+1)\\sin(π z)\n\nmentioned in the review.\n\nTherefore equation (II) alone is flexible: it has infinitely many\nentire solutions, none of which (except the constant one) satisfies\neither (I) or (III).\n\n3(d) Consequences.  \n• Relations (I) and (III) are rigid: every entire solution equals 1;  \n this solution necessarily fulfils the other two relations, so the whole\n system (∗) holds.  \n• Relation (II) is flexible: it possesses an infinite-dimensional space\n of entire solutions and does not imply (I) or (III).\n\n————————————————————————————————————————\n4.  Meromorphic solutions defined on U  \n\nPut k(z)=h(z)−1.  Throughout we only use relation (I); hence again the\npolynomial  \n\n  D(z)=1−z+z²  (9)\n\nreappears.\n\nEquation (I) reads on U  \n\n  k(z)+z k(1−z)=0, (10)  \n\nand replacing z by 1−z gives  \n\n  k(1−z)+(1−z) k(z)=0. (11)\n\nEliminating k(1−z) from (10)–(11) gives  \n\n  D(z) k(z)=0.   (12)\n\n4(a) The zeros ζ₊, ζ₋ of D.  \nOutside the points ζ± the factor D(z) is non-zero, so (12) forces\nk≡0.  Near ζ± the same argument as in the original solution shows that h\nis holomorphic and assumes the value 1 at each of ζ±.\n\n4(b) Extension across the roots of Δ.  \nLet ρ∈S.  Because ρ∉U, h is a priori undefined at ρ, but U contains a\npunctured neighbourhood of ρ.  On U we already know k≡0, so\nlim_{z→ρ,z∈U}k(z)=0.  If h had a pole at ρ the same limit would be\ninfinite – contradiction.  Hence ρ is a removable singularity of h with\nvalue 1.  Doing this for every root of Δ yields an entire extension that\nequals 1 on U and therefore everywhere.\n\n————————————————————————————————————————\nAnswer.  The coupled system (I)–(III) possesses exactly one entire—and in\nfact one meromorphic—solution:\n\n      f(z)=1  for every z∈ℂ. ∎\n\n──────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.427497",
        "was_fixed": false,
        "difficulty_analysis": "•  Multiple equations  –   The original problem has a single\nfunctional equation; here one must satisfy {\\em three} coupled relations\nsimultaneously.\n\n•  Three interacting arguments  –   \nThe unknown values \\(f(z),\\,f(1-z),\\,f(-z)\\) appear together, forcing a\n\\(3\\times3\\) linear system instead of the \\(2\\times2\\) system in the\nkernel variant.  This raises both algebraic complexity and bookkeeping.\n\n•  Non-trivial determinant  –  \nSolving demands computation of a cubic determinant\n\\(\\Delta(z)=1-z^{2}+z^{3}\\) whose vanishing introduces a {\\em singular\nlocus} of three complex points.  One must treat the regular and\nsingular cases separately, recognise the rank drop, and manage the\nresulting free parameters.\n\n•  Rational–function core  –   \nOutside the singular set the solution is a genuinely {\\em rational}\nfunction of degree three over degree three, rather than the constant\nsolution of the original problem.  Obtaining it requires systematic\nelimination, not mere substitution.\n\n•  Parametrisation on the singular set  –   \nAt each root of \\(\\Delta\\) the system loses rank, so compatibility must be\nchecked and the local freedom carefully parametrised; this demands a\nfull analysis of linear dependence, something entirely absent in the\noriginal exercise.\n\nIn summary, the problem keeps the original idea\n“relate \\(f\\) at two (now three) symmetric points and isolate a\nmultiplicative polynomial”, but it escalates the dimension, the number\nof constraints, the algebraic labour, and the subtlety of the singular\ncase, making it significantly harder than both the original and the\ncurrent kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}