{
  "index": "1949-A-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. How many roots of the equation \\( z^{6}+6 z+10=0 \\) lie in each quadrant of the complex plane?",
  "solution": "First Solution. Let \\( P(z)=z^{6}+6 z+10 \\). The minimum value of \\( P(z) \\) for real \\( z \\) is \\( P(-1)=5 \\). Hence the equation has no real roots. There can be no purely imaginary roots since \\( \\operatorname{Im}(P(i y))=6 y \\neq 0 \\) unless \\( y=0 \\), and \\( P(0) \\neq 0 \\).\n\nThe roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since \\( P \\) has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:\n(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or\n(2) there are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.\n\nWe can decide between these two possibilities through the argument principle of complex variable theory. This states:\n\nIf a function \\( f \\) is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of \\( f \\) in the interior of the curve (counting multiplicities) is \\( 1 / 2 \\pi \\) times the variation of \\( \\arg f(z) \\) as \\( z \\) describes the curve once in the positive direction.\n\nIn this problem take \\( f=P \\) and the simple closed curve formed by the real axis from \\( O \\) to \\( R \\), the arc of the circle \\( |z|=R \\) from \\( R \\) to \\( i R \\), and the imaginary axis from \\( i R \\) to \\( O \\), where \\( R \\) is a large positive number.\n\nThe argument variation of \\( P \\) along \\( [O, R] \\) is zero since \\( P \\) remains real and positive. If \\( R \\) is large, then the argument variation of \\( P(z) \\) along the circular arc is approximately the same as that of \\( z^{6} \\) along that same arc, since \\( \\left|\\left(P(z) / z^{6}\\right)-1\\right| \\) is small for all \\( z \\) on this arc. Therefore the argument variation of \\( P(z) \\) along the circular arc is about \\( 6 \\cdot \\pi / 2=3 \\pi \\). Finally, as \\( z \\) goes from \\( i R \\) to \\( O \\) along the imaginary axis, \\( P(z) \\) goes back to the value 10 , keeping always in the upper half-plane; hence the argument variation along this part of the path is about \\( -\\pi \\).\n\nBut the total argument variation, seen to be about \\( (3 \\pi-\\pi)=2 \\pi \\), must be an integral multiple of \\( 2 \\pi \\), so it must be exactly \\( 2 \\pi \\) (assuming that \\( R \\) is large enough). Hence the total number of zeros enclosed by the path is one for any large \\( R \\). Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1).\n\nSecond Solution. Another way to decide between possibilities (1) and (2) is to study carefully a root in the first quadrant.\n\nSuppose \\( z_{1}=r e^{i \\theta}, 0<\\theta<\\pi / 2 \\), is a root of \\( z^{6}+6 z+10=0 \\) in the first quadrant. Then \\( 6 z_{1}+10 \\) also lies in the first quadrant, and \\( \\arg \\left(6 z_{1}\\right. \\) \\( +10)<\\arg z_{1}=\\theta \\). Since \\( z_{1}{ }^{6}=r^{6} e^{6 i \\theta}=-\\left(6 z_{1}+10\\right), z_{1}{ }^{6} \\) lies in the third quadrant. Since also \\( 0<6 \\theta<3 \\pi \\), we see that \\( \\pi<6 \\theta<\\frac{3}{2} \\pi \\), and hence \\( 6 \\theta-\\pi \\) is that argument of \\( -z_{1}{ }^{6} \\) which lies between 0 and \\( \\pi / 2 \\). Therefore \\( 6 \\theta-\\pi=\\arg \\left(6 z_{1}+10\\right)<\\theta \\), giving \\( \\theta<\\pi / 5 \\). Thus any root in the first quadrant has argument in ( \\( 0, \\pi / 5 \\) ).\n\nSuppose that \\( z_{1} \\) and \\( z_{2} \\) are different roots in the first quadrant. Then\n\\[\n\\begin{array}{l} \n0=\\frac{P\\left(z_{1}\\right)-P\\left(z_{2}\\right)}{z_{1}-z_{2}} \\\\\n=z_{1}{ }^{5}+z_{1}{ }^{4} z_{2}+z_{1}{ }^{3} z_{2}^{2}+z_{1}{ }^{2} z_{2}^{3}+z_{1} z_{2}{ }^{4}+z_{2}^{5}+6 .\n\\end{array}\n\\]\n\nSince both \\( z_{1} \\) and \\( z_{2} \\) have arguments in \\( (0, \\pi / 5) \\) every term in the right member of (3) has argument in \\( [0, \\pi) \\), so the sum of these terms cannot be zero, a contradiction. Therefore there cannot be two distinct roots of the equation in the first quadrant, and we conclude that (1) holds.\n\nThird Solution. The zeros of \\( z^{6}+6 z+t \\) are continuous functions of the real parameter \\( t \\). For no positive \\( t \\) does this polynomial have a purely imaginary root. Hence for all positive \\( t \\) the number \\( N \\) of roots in the right half-plane remains fixed. As we saw above, \\( N \\) is even for large positive \\( t \\), so \\( N \\) is even for all positive \\( t \\). For \\( t=0 \\), there are three roots in the left half-plane, two in the right half-plane, and one at zero. For small positive \\( t, N \\) must be either two or three by continuity. So \\( N=2 \\) for small, and hence all, positive \\( t \\). Therefore (1) holds.",
  "vars": [
    "z",
    "y",
    "r",
    "z_1",
    "z_2",
    "\\\\theta"
  ],
  "params": [
    "P",
    "f",
    "O",
    "R",
    "t",
    "N"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "complexvar",
        "y": "imagcoord",
        "r": "radiusvar",
        "z_1": "firstroot",
        "z_2": "secondroot",
        "\\theta": "anglevar",
        "P": "polynom",
        "f": "funcsymbol",
        "O": "originpt",
        "R": "bigradius",
        "t": "parameter",
        "N": "numtotal"
      },
      "question": "5. How many roots of the equation \\( complexvar^{6}+6 complexvar+10=0 \\) lie in each quadrant of the complex plane?",
      "solution": "First Solution. Let \\( polynom(complexvar)=complexvar^{6}+6 complexvar+10 \\). The minimum value of \\( polynom(complexvar) \\) for real \\( complexvar \\) is \\( polynom(-1)=5 \\). Hence the equation has no real roots. There can be no purely imaginary roots since \\( \\operatorname{Im}(polynom(i imagcoord))=6 imagcoord \\neq 0 \\) unless \\( imagcoord=0 \\), and \\( polynom(0) \\neq 0 \\).\n\nThe roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since \\( polynom \\) has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:\n(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or\n(2) there are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.\n\nWe can decide between these two possibilities through the argument principle of complex variable theory. This states:\n\nIf a function \\( funcsymbol \\) is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of \\( funcsymbol \\) in the interior of the curve (counting multiplicities) is \\( 1 / 2 \\pi \\) times the variation of \\( \\arg funcsymbol(complexvar) \\) as \\( complexvar \\) describes the curve once in the positive direction.\n\nIn this problem take \\( funcsymbol=polynom \\) and the simple closed curve formed by the real axis from \\( originpt \\) to \\( bigradius \\), the arc of the circle \\( |complexvar|=bigradius \\) from \\( bigradius \\) to \\( i bigradius \\), and the imaginary axis from \\( i bigradius \\) to \\( originpt \\), where \\( bigradius \\) is a large positive number.\n\nThe argument variation of \\( polynom \\) along \\( [originpt, bigradius] \\) is zero since \\( polynom \\) remains real and positive. If \\( bigradius \\) is large, then the argument variation of \\( polynom(complexvar) \\) along the circular arc is approximately the same as that of \\( complexvar^{6} \\) along that same arc, since \\( \\left|\\left(polynom(complexvar) / complexvar^{6}\\right)-1\\right| \\) is small for all \\( complexvar \\) on this arc. Therefore the argument variation of \\( polynom(complexvar) \\) along the circular arc is about \\( 6 \\cdot \\pi / 2=3 \\pi \\). Finally, as \\( complexvar \\) goes from \\( i bigradius \\) to \\( originpt \\) along the imaginary axis, \\( polynom(complexvar) \\) goes back to the value 10 , keeping always in the upper half-plane; hence the argument variation along this part of the path is about \\( -\\pi \\).\n\nBut the total argument variation, seen to be about \\( (3 \\pi-\\pi)=2 \\pi \\), must be an integral multiple of \\( 2 \\pi \\), so it must be exactly \\( 2 \\pi \\) (assuming that \\( bigradius \\) is large enough). Hence the total number of zeros enclosed by the path is one for any large \\( bigradius \\). Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1).\n\nSecond Solution. Another way to decide between possibilities (1) and (2) is to study carefully a root in the first quadrant.\n\nSuppose \\( firstroot=radiusvar e^{i anglevar}, 0<anglevar<\\pi / 2 \\), is a root of \\( complexvar^{6}+6 complexvar+10=0 \\) in the first quadrant. Then \\( 6 firstroot+10 \\) also lies in the first quadrant, and \\( \\arg \\left(6 firstroot+10\\right)<\\arg firstroot=anglevar \\). Since \\( firstroot^{6}=radiusvar^{6} e^{6 i anglevar}=-\\left(6 firstroot+10\\right), firstroot^{6} \\) lies in the third quadrant. Since also \\( 0<6 anglevar<3 \\pi \\), we see that \\( \\pi<6 anglevar<\\frac{3}{2} \\pi \\), and hence \\( 6 anglevar-\\pi \\) is that argument of \\( -firstroot^{6} \\) which lies between 0 and \\( \\pi / 2 \\). Therefore \\( 6 anglevar-\\pi=\\arg \\left(6 firstroot+10\\right)<anglevar \\), giving \\( anglevar<\\pi / 5 \\). Thus any root in the first quadrant has argument in ( \\( 0, \\pi / 5 \\) ).\n\nSuppose that \\( firstroot \\) and \\( secondroot \\) are different roots in the first quadrant. Then\n\\[\n\\begin{array}{l} \n0=\\frac{polynom\\left(firstroot\\right)-polynom\\left(secondroot\\right)}{firstroot-secondroot} \\\\\n=firstroot^{5}+firstroot^{4} secondroot+firstroot^{3} secondroot^{2}+firstroot^{2} secondroot^{3}+firstroot secondroot^{4}+secondroot^{5}+6 .\n\\end{array}\n\\]\n\nSince both \\( firstroot \\) and \\( secondroot \\) have arguments in \\( (0, \\pi / 5) \\) every term in the right member of (3) has argument in \\( [0, \\pi) \\), so the sum of these terms cannot be zero, a contradiction. Therefore there cannot be two distinct roots of the equation in the first quadrant, and we conclude that (1) holds.\n\nThird Solution. The zeros of \\( complexvar^{6}+6 complexvar+parameter \\) are continuous functions of the real parameter \\( parameter \\). For no positive \\( parameter \\) does this polynomial have a purely imaginary root. Hence for all positive \\( parameter \\) the number \\( numtotal \\) of roots in the right half-plane remains fixed. As we saw above, \\( numtotal \\) is even for large positive \\( parameter \\), so \\( numtotal \\) is even for all positive \\( parameter \\). For \\( parameter=0 \\), there are three roots in the left half-plane, two in the right half-plane, and one at zero. For small positive \\( parameter, numtotal \\) must be either two or three by continuity. So \\( numtotal=2 \\) for small, and hence all, positive \\( parameter \\). Therefore (1) holds."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "sunflower",
        "y": "bicycle",
        "r": "lantern",
        "z_1": "raincloud",
        "z_2": "snowflake",
        "\\\\theta": "marshland",
        "P": "compass",
        "f": "skylight",
        "O": "emberstone",
        "R": "ravenwood",
        "t": "parchment",
        "N": "stargazer"
      },
      "question": "5. How many roots of the equation \\( sunflower^{6}+6 sunflower+10=0 \\) lie in each quadrant of the complex plane?",
      "solution": "First Solution. Let \\( compass(sunflower)=sunflower^{6}+6 sunflower+10 \\). The minimum value of \\( compass(sunflower) \\) for real \\( sunflower \\) is \\( compass(-1)=5 \\). Hence the equation has no real roots. There can be no purely imaginary roots since \\( \\operatorname{Im}(compass(i bicycle))=6 bicycle \\neq 0 \\) unless \\( bicycle=0 \\), and \\( compass(0) \\neq 0 \\).\n\nThe roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since compass has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:\n(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or\n(2) there are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.\n\nWe can decide between these two possibilities through the argument principle of complex variable theory. This states:\n\nIf a function \\( skylight \\) is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of \\( skylight \\) in the interior of the curve (counting multiplicities) is \\( 1 / 2 \\pi \\) times the variation of \\( \\arg skylight(sunflower) \\) as \\( sunflower \\) describes the curve once in the positive direction.\n\nIn this problem take \\( skylight = compass \\) and the simple closed curve formed by the real axis from \\( emberstone \\) to \\( ravenwood \\), the arc of the circle \\( |sunflower|=ravenwood \\) from \\( ravenwood \\) to \\( i\\,ravenwood \\), and the imaginary axis from \\( i\\,ravenwood \\) to \\( emberstone \\), where \\( ravenwood \\) is a large positive number.\n\nThe argument variation of compass along \\( [emberstone, ravenwood] \\) is zero since compass remains real and positive. If \\( ravenwood \\) is large, then the argument variation of \\( compass(sunflower) \\) along the circular arc is approximately the same as that of \\( sunflower^{6} \\) along that same arc, since \\( \\left|\\left(compass(sunflower) / sunflower^{6}\\right)-1\\right| \\) is small for all \\( sunflower \\) on this arc. Therefore the argument variation of \\( compass(sunflower) \\) along the circular arc is about \\( 6 \\cdot \\pi / 2=3 \\pi \\). Finally, as \\( sunflower \\) goes from \\( i\\,ravenwood \\) to \\( emberstone \\) along the imaginary axis, \\( compass(sunflower) \\) goes back to the value 10, keeping always in the upper half-plane; hence the argument variation along this part of the path is about \\( -\\pi \\).\n\nBut the total argument variation, seen to be about \\( (3 \\pi-\\pi)=2 \\pi \\), must be an integral multiple of \\( 2 \\pi \\), so it must be exactly \\( 2 \\pi \\) (assuming that \\( ravenwood \\) is large enough). Hence the total number of zeros enclosed by the path is one for any large \\( ravenwood \\). Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1).\n\nSecond Solution. Another way to decide between possibilities (1) and (2) is to study carefully a root in the first quadrant.\n\nSuppose \\( raincloud=lantern e^{i\\,marshland},\\;0<marshland<\\pi / 2 \\), is a root of \\( sunflower^{6}+6 sunflower+10=0 \\) in the first quadrant. Then \\( 6\\,raincloud+10 \\) also lies in the first quadrant, and \\( \\arg \\left(6\\,raincloud+10\\right)<\\arg raincloud=marshland \\). Since \\( raincloud^{6}=lantern^{6} e^{6 i\\,marshland}=-\\left(6\\,raincloud+10\\right),\\;raincloud^{6} \\) lies in the third quadrant. Since also \\( 0<6\\,marshland<3 \\pi \\), we see that \\( \\pi<6\\,marshland<\\frac{3}{2} \\pi \\), and hence \\( 6\\,marshland-\\pi \\) is that argument of \\( -raincloud^{6} \\) which lies between 0 and \\( \\pi / 2 \\). Therefore \\( 6\\,marshland-\\pi=\\arg \\left(6\\,raincloud+10\\right)<marshland \\), giving \\( marshland<\\pi / 5 \\). Thus any root in the first quadrant has argument in (\\( 0, \\pi / 5 \\)).\n\nSuppose that \\( raincloud \\) and \\( snowflake \\) are different roots in the first quadrant. Then\n\\[\n\\begin{array}{l} \n0=\\frac{compass\\left(raincloud\\right)-compass\\left(snowflake\\right)}{raincloud-snowflake} \\\\\n=raincloud^{5}+raincloud^{4}\\,snowflake+raincloud^{3}\\,snowflake^{2}+raincloud^{2}\\,snowflake^{3}+raincloud\\,snowflake^{4}+snowflake^{5}+6 .\n\\end{array}\n\\]\n\nSince both \\( raincloud \\) and \\( snowflake \\) have arguments in \\( (0, \\pi / 5) \\) every term in the right member of (3) has argument in \\( [0, \\pi) \\), so the sum of these terms cannot be zero, a contradiction. Therefore there cannot be two distinct roots of the equation in the first quadrant, and we conclude that (1) holds.\n\nThird Solution. The zeros of \\( sunflower^{6}+6 sunflower+parchment \\) are continuous functions of the real parameter \\( parchment \\). For no positive \\( parchment \\) does this polynomial have a purely imaginary root. Hence for all positive \\( parchment \\) the number \\( stargazer \\) of roots in the right half-plane remains fixed. As we saw above, \\( stargazer \\) is even for large positive \\( parchment \\), so \\( stargazer \\) is even for all positive \\( parchment \\). For \\( parchment=0 \\), there are three roots in the left half-plane, two in the right half-plane, and one at zero. For small positive \\( parchment \\), \\( stargazer \\) must be either two or three by continuity. So \\( stargazer=2 \\) for small, and hence all, positive \\( parchment \\). Therefore (1) holds."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "realaxisval",
        "y": "constantval",
        "r": "zeroextent",
        "z_1": "lastbranch",
        "z_2": "leafpoint",
        "\\\\theta": "lengthunit",
        "P": "linearfun",
        "f": "constantfun",
        "O": "infinitypoint",
        "R": "smallvalue",
        "t": "fixpoint",
        "N": "emptiness"
      },
      "question": "5. How many roots of the equation \\( realaxisval^{6}+6 realaxisval+10=0 \\) lie in each quadrant of the complex plane?",
      "solution": "First Solution. Let \\( linearfun(realaxisval)=realaxisval^{6}+6 realaxisval+10 \\). The minimum value of \\( linearfun(realaxisval) \\) for real \\( realaxisval \\) is \\( linearfun(-1)=5 \\). Hence the equation has no real roots. There can be no purely imaginary roots since \\( \\operatorname{Im}(linearfun(i constantval))=6 constantval \\neq 0 \\) unless \\( constantval=0 \\), and \\( linearfun(0) \\neq 0 \\).\n\nThe roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since \\( linearfun \\) has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:\n(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or\n(2) there are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.\n\nWe can decide between these two possibilities through the argument principle of complex variable theory. This states:\n\nIf a function \\( constantfun \\) is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of \\( constantfun \\) in the interior of the curve (counting multiplicities) is \\( 1 / 2 \\pi \\) times the variation of \\( \\arg constantfun(realaxisval) \\) as \\( realaxisval \\) describes the curve once in the positive direction.\n\nIn this problem take \\( constantfun=linearfun \\) and the simple closed curve formed by the real axis from \\( infinitypoint \\) to \\( smallvalue \\), the arc of the circle \\( |realaxisval|=smallvalue \\) from \\( smallvalue \\) to \\( i smallvalue \\), and the imaginary axis from \\( i smallvalue \\) to \\( infinitypoint \\), where \\( smallvalue \\) is a large positive number.\n\nThe argument variation of \\( linearfun \\) along \\( [infinitypoint, smallvalue] \\) is zero since \\( linearfun \\) remains real and positive. If \\( smallvalue \\) is large, then the argument variation of \\( linearfun(realaxisval) \\) along the circular arc is approximately the same as that of \\( realaxisval^{6} \\) along that same arc, since \\( \\left|\\left(linearfun(realaxisval) / realaxisval^{6}\\right)-1\\right| \\) is small for all \\( realaxisval \\) on this arc. Therefore the argument variation of \\( linearfun(realaxisval) \\) along the circular arc is about \\( 6 \\cdot \\pi / 2=3 \\pi \\). Finally, as \\( realaxisval \\) goes from \\( i smallvalue \\) to \\( infinitypoint \\) along the imaginary axis, \\( linearfun(realaxisval) \\) goes back to the value 10 , keeping always in the upper half-plane; hence the argument variation along this part of the path is about \\( -\\pi \\).\n\nBut the total argument variation, seen to be about \\( (3 \\pi-\\pi)=2 \\pi \\), must be an integral multiple of \\( 2 \\pi \\), so it must be exactly \\( 2 \\pi \\) (assuming that \\( smallvalue \\) is large enough). Hence the total number of zeros enclosed by the path is one for any large \\( smallvalue \\). Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1).\n\nSecond Solution. Another way to decide between possibilities (1) and (2) is to study carefully a root in the first quadrant.\n\nSuppose \\( lastbranch=zeroextent e^{i lengthunit}, 0<lengthunit<\\pi / 2 \\), is a root of \\( realaxisval^{6}+6 realaxisval+10=0 \\) in the first quadrant. Then \\( 6 lastbranch+10 \\) also lies in the first quadrant, and \\( \\arg \\left(6 lastbranch +10\\right)<\\arg lastbranch=lengthunit \\). Since \\( lastbranch^{6}=zeroextent^{6} e^{6 i lengthunit}=-\\left(6 lastbranch+10\\right), lastbranch^{6} \\) lies in the third quadrant. Since also \\( 0<6 lengthunit<3 \\pi \\), we see that \\( \\pi<6 lengthunit<\\frac{3}{2} \\pi \\), and hence \\( 6 lengthunit-\\pi \\) is that argument of \\( -lastbranch^{6} \\) which lies between 0 and \\( \\pi / 2 \\). Therefore \\( 6 lengthunit-\\pi=\\arg \\left(6 lastbranch+10\\right)<lengthunit \\), giving \\( lengthunit<\\pi / 5 \\). Thus any root in the first quadrant has argument in ( \\( 0, \\pi / 5 \\) ).\n\nSuppose that \\( lastbranch \\) and \\( leafpoint \\) are different roots in the first quadrant. Then\n\\[\n\\begin{array}{l} \n0=\\frac{linearfun\\left(lastbranch\\right)-linearfun\\left(leafpoint\\right)}{lastbranch-leafpoint} \\\n=lastbranch^{5}+lastbranch^{4} leafpoint+lastbranch^{3} leafpoint^{2}+lastbranch^{2} leafpoint^{3}+lastbranch leafpoint^{4}+leafpoint^{5}+6 .\n\\end{array}\n\\]\n\nSince both \\( lastbranch \\) and \\( leafpoint \\) have arguments in \\( (0, \\pi / 5) \\) every term in the right member of (3) has argument in \\( [0, \\pi) \\), so the sum of these terms cannot be zero, a contradiction. Therefore there cannot be two distinct roots of the equation in the first quadrant, and we conclude that (1) holds.\n\nThird Solution. The zeros of \\( realaxisval^{6}+6 realaxisval+fixpoint \\) are continuous functions of the real parameter \\( fixpoint \\). For no positive \\( fixpoint \\) does this polynomial have a purely imaginary root. Hence for all positive \\( fixpoint \\) the number \\( emptiness \\) of roots in the right half-plane remains fixed. As we saw above, \\( emptiness \\) is even for large positive \\( fixpoint \\), so \\( emptiness \\) is even for all positive \\( fixpoint \\). For \\( fixpoint=0 \\), there are three roots in the left half-plane, two in the right half-plane, and one at zero. For small positive \\( fixpoint, emptiness \\) must be either two or three by continuity. So \\( emptiness=2 \\) for small, and hence all, positive \\( fixpoint \\). Therefore (1) holds."
    },
    "garbled_string": {
      "map": {
        "z": "vwjdlskk",
        "y": "kntxqzbm",
        "r": "szfpldve",
        "z_1": "lqkzmnpa",
        "z_2": "cfwtdhoy",
        "\\theta": "mpgrxsvu",
        "P": "qzjdpwkf",
        "f": "spclirga",
        "O": "uhsambov",
        "R": "enxtvzla",
        "t": "dakljhqp",
        "N": "vqzlpest"
      },
      "question": "5. How many roots of the equation \\( vwjdlskk^{6}+6 vwjdlskk+10=0 \\) lie in each quadrant of the complex plane?",
      "solution": "First Solution. Let \\( qzjdpwkf(vwjdlskk)=vwjdlskk^{6}+6 vwjdlskk+10 \\). The minimum value of \\( qzjdpwkf(vwjdlskk) \\) for real \\( vwjdlskk \\) is \\( qzjdpwkf(-1)=5 \\). Hence the equation has no real roots. There can be no purely imaginary roots since \\( \\operatorname{Im}(qzjdpwkf(i kntxqzbm))=6 kntxqzbm \\neq 0 \\) unless \\( kntxqzbm=0 \\), and \\( qzjdpwkf(0) \\neq 0 \\).\n\nThe roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since \\( qzjdpwkf \\) has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:\n(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or\n(2) there are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.\n\nWe can decide between these two possibilities through the argument principle of complex variable theory. This states:\n\nIf a function \\( spclirga \\) is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of \\( spclirga \\) in the interior of the curve (counting multiplicities) is \\( 1 / 2 \\pi \\) times the variation of \\( \\arg spclirga(vwjdlskk) \\) as \\( vwjdlskk \\) describes the curve once in the positive direction.\n\nIn this problem take \\( spclirga=qzjdpwkf \\) and the simple closed curve formed by the real axis from \\( uhsambov \\) to \\( enxtvzla \\), the arc of the circle \\( |vwjdlskk|=enxtvzla \\) from \\( enxtvzla \\) to \\( i\\,enxtvzla \\), and the imaginary axis from \\( i\\,enxtvzla \\) to \\( uhsambov \\), where \\( enxtvzla \\) is a large positive number.\n\nThe argument variation of \\( qzjdpwkf \\) along \\( [uhsambov, enxtvzla] \\) is zero since \\( qzjdpwkf \\) remains real and positive. If \\( enxtvzla \\) is large, then the argument variation of \\( qzjdpwkf(vwjdlskk) \\) along the circular arc is approximately the same as that of \\( vwjdlskk^{6} \\) along that same arc, since \\( \\left|\\left(qzjdpwkf(vwjdlskk) / vwjdlskk^{6}\\right)-1\\right| \\) is small for all \\( vwjdlskk \\) on this arc. Therefore the argument variation of \\( qzjdpwkf(vwjdlskk) \\) along the circular arc is about \\( 6 \\cdot \\pi / 2=3 \\pi \\). Finally, as \\( vwjdlskk \\) goes from \\( i\\,enxtvzla \\) to \\( uhsambov \\) along the imaginary axis, \\( qzjdpwkf(vwjdlskk) \\) goes back to the value 10, keeping always in the upper half-plane; hence the argument variation along this part of the path is about \\( -\\pi \\).\n\nBut the total argument variation, seen to be about \\( (3 \\pi-\\pi)=2 \\pi \\), must be an integral multiple of \\( 2 \\pi \\), so it must be exactly \\( 2 \\pi \\) (assuming that \\( enxtvzla \\) is large enough). Hence the total number of zeros enclosed by the path is one for any large \\( enxtvzla \\). Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1).\n\nSecond Solution. Another way to decide between possibilities (1) and (2) is to study carefully a root in the first quadrant.\n\nSuppose \\( lqkzmnpa=szfpldve e^{i mpgrxsvu}, 0<mpgrxsvu<\\pi / 2 \\), is a root of \\( vwjdlskk^{6}+6 vwjdlskk+10=0 \\) in the first quadrant. Then \\( 6 lqkzmnpa+10 \\) also lies in the first quadrant, and \\( \\arg \\left(6 lqkzmnpa+10\\right)<\\arg lqkzmnpa=mpgrxsvu \\). Since \\( lqkzmnpa^{6}=szfpldve^{6} e^{6 i mpgrxsvu}=-\\left(6 lqkzmnpa+10\\right), lqkzmnpa^{6} \\) lies in the third quadrant. Since also \\( 0<6 mpgrxsvu<3 \\pi \\), we see that \\( \\pi<6 mpgrxsvu<\\frac{3}{2} \\pi \\), and hence \\( 6 mpgrxsvu-\\pi \\) is that argument of \\( -lqkzmnpa^{6} \\) which lies between 0 and \\( \\pi / 2 \\). Therefore \\( 6 mpgrxsvu-\\pi=\\arg \\left(6 lqkzmnpa+10\\right)<mpgrxsvu \\), giving \\( mpgrxsvu<\\pi / 5 \\). Thus any root in the first quadrant has argument in \\( (0, \\pi / 5) \\).\n\nSuppose that \\( lqkzmnpa \\) and \\( cfwtdhoy \\) are different roots in the first quadrant. Then\n\\[\n\\begin{array}{l} \n0=\\frac{qzjdpwkf\\left(lqkzmnpa\\right)-qzjdpwkf\\left(cfwtdhoy\\right)}{lqkzmnpa-cfwtdhoy} \\\\ \n=lqkzmnpa^{5}+lqkzmnpa^{4} cfwtdhoy+lqkzmnpa^{3} cfwtdhoy^{2}+lqkzmnpa^{2} cfwtdhoy^{3}+lqkzmnpa cfwtdhoy^{4}+cfwtdhoy^{5}+6 .\n\\end{array}\n\\]\n\nSince both \\( lqkzmnpa \\) and \\( cfwtdhoy \\) have arguments in \\( (0, \\pi / 5) \\) every term in the right member of (3) has argument in \\( [0, \\pi) \\), so the sum of these terms cannot be zero, a contradiction. Therefore there cannot be two distinct roots of the equation in the first quadrant, and we conclude that (1) holds.\n\nThird Solution. The zeros of \\( vwjdlskk^{6}+6 vwjdlskk+dakljhqp \\) are continuous functions of the real parameter \\( dakljhqp \\). For no positive \\( dakljhqp \\) does this polynomial have a purely imaginary root. Hence for all positive \\( dakljhqp \\) the number \\( vqzlpest \\) of roots in the right half-plane remains fixed. As we saw above, \\( vqzlpest \\) is even for large positive \\( dakljhqp \\), so \\( vqzlpest \\) is even for all positive \\( dakljhqp \\). For \\( dakljhqp=0 \\), there are three roots in the left half-plane, two in the right half-plane, and one at zero. For small positive \\( dakljhqp, vqzlpest \\) must be either two or three by continuity. So \\( vqzlpest=2 \\) for small, and hence all, positive \\( dakljhqp \\). Therefore (1) holds."
    },
    "kernel_variant": {
      "question": "Let\n\\[\nP(z)=z^{10}+2z+5.\n\\]\nDetermine how many zeros of the equation\n\\[\nP(z)=0\n\\]\nlie in each of the four open quadrants of the complex plane.",
      "solution": "We wish to find the number of zeros of P(z)=z¹⁰+2z+5 in each open quadrant Q₁, Q₂, Q₃, Q₄.  We proceed in three steps.\n\n1.  No zeros on the real or imaginary axes.\n(a)  If x∈ℝ then P(x)=x¹⁰+2x+5.  For x≥0 clearly P(x)>0.  For x<0 write x=−t, t>0, and set f(t)=t¹⁰−2t+5.  Then f′(t)=10t⁹−2, which has the single positive root t₀=(2/10)^{1/9}=(1/5)^{1/9}.  At t₀ we have f(t₀)=t₀^{10}−2t₀+5=(t₀/5)−2t₀+5=5−(9/5)t₀>5−9/5=16/5>0, and f(t)→+∞ as t→∞, so f(t)>0 for all t>0.  Hence P(x)>0 for every real x: no real zeros.\n(b)  If z=iy (y real) then P(iy)=−y¹⁰+5+2iy.  Its imaginary part is 2y, which vanishes only at y=0, but P(0)=5≠0.  Hence no purely imaginary zeros.\n\n2.  Conjugation symmetry and reduction to two unknowns.\nSince P has real coefficients, nonreal zeros occur in conjugate pairs.  Conjugation sends Q₁↔Q₄ and Q₂↔Q₃, so let a=# {zeros in Q₁} = # in Q₄, and b=# in Q₂ =# in Q₃.  Then the total is 2a+2b=10 ⇒ a+b=5.  Also the sum of all zeros (by the vanishing z⁹‐coefficient) is 0, so they cannot all lie in Re z>0 nor all in Re z<0, which merely excludes a=5 or b=5.  Hence 1≤a,b≤4 and a+b=5, but (a,b) could still be (1,4),(2,3),(3,2),(4,1).  To decide among these we must compute a directly.\n\n3.  Use the argument principle on the first quadrant.\nLet Γ_R be the boundary of the quarter‐disk D_R={z:|z|≤R,0≤Arg z≤π/2}, traversed positively:\n  (i) along [0,R] on the real axis, P(x)>0 ⇒ arg P=0 ⇒ Δ₁=0.\n  (ii) along the circular arc z=Re^{iθ}, θ from 0 to π/2.  For large R, P(z)=z¹⁰(1+o(1)), so arg P increases by ≃10·(π/2)=5π: call this Δ₂=5π.\n  (iii) along the imaginary segment from iR down to 0: z=iy, y from R→0.  Then P(iy)=−y¹⁰+5+2iy has Im P=2y>0, so P(iy) stays in the upper half‐plane.  At y=R large, Re P≈−R¹⁰<0 so arg P≈π⁻; at y=0, P(0)=5>0 so arg P=0.  Thus arg P decreases continuously from ≈π to 0, giving Δ₃≈−π.\nTotal change Δarg P=Δ₁+Δ₂+Δ₃=0+5π−π=4π.  By the argument principle the number of zeros of P inside D_R is Δ/(2π)=2.  Since D_R exhausts Q₁ as R→∞, we conclude a=2.\n\n4.  Conclusion.\nHence # in Q₁=2, # in Q₄=2, and b=5−a=3 ⇒ # in Q₂=3=# in Q₃.  Answer:\nThere are 2 zeros in Q₁, 3 in Q₂, 3 in Q₃, and 2 in Q₄.\n∎",
      "_meta": {
        "core_steps": [
          "Exclude real and purely imaginary zeros by direct evaluation on the axes.",
          "Use real-coefficient symmetry: roots occur in conjugate pairs and (because the z^{n-1} term is missing) their sum is 0, giving only two quadrant-count patterns to check.",
          "Apply the argument principle on the quarter-circle contour (real axis → big arc → imaginary axis) to find the change of arg P(z) and hence the number of zeros in Q1.",
          "Compare that count with the two admissible patterns and pick the one that matches."
        ],
        "mutable_slots": {
          "slot_degree": {
            "description": "The even degree n = 4k+2 of the leading term z^n (keeps arg-variation n·π/2 so that the net change along the contour is an integer multiple of 2π).",
            "original": 6
          },
          "slot_linear_coeff": {
            "description": "Any non-zero real coefficient of the linear term; guarantees Im P(iy)=c·y ≠ 0 for y≠0, so no pure-imaginary roots.",
            "original": 6
          },
          "slot_constant": {
            "description": "Any positive real constant term; ensures P(0) ≠ 0 and that along the imaginary axis the image stays in the upper half-plane, giving an arg change of −π.",
            "original": 10
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}