Using a given list of four complex numbers, find by inspection the one that is a root of a given cubic real polynomial and hence find the other roots.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "From the list of complex numbers, choose the one which is a root of the given equation $f(z)=0$ , and hence find all roots.
", "advice": "a) Finding a root.
\nIn order to find which one of the four choices is a root you have to evaluate $f(z)$ for each choice.If you find for a choice of $z$ that $f(z)=0$ then that choice of $z$ is a root of the equation.
\nNote that\\[\\begin{eqnarray*} \\simplify{f({z1})} &=&\\simplify[std]{{z1}^3+{-2*a1 -c1}*{z1} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\ &=&\\simplify[std]{{z1^3}+{-2*a1 -c1}{z1 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\ &=&\\simplify[std]{{z1^3}+{(( -2) * a1 -c1)*z1^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z1}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\ &=&0 \\end{eqnarray*}. \\]So of the list of choices $z_1=\\var{z1}$ is a root.
\nb) The other roots
\nNow that you have found a complex root it is very simple to find another complex root.
\nSince $f(z)$ is a polynomial with real coefficients then if $z=z_0$ is a root we have that the conjugate $z=\\overline{z_0}$ is also a root.
\nHence the complex number $z_2=\\overline{\\var{z1}}=\\var{conj(z1)}$ is a root.
\nTo find the real root $z_3=c$ we note that the constant term of\\[f(z) =(z-z_1)(z-z_2)(z-c)\\]is $-z_1z_2c = -(\\var{z1})(\\var{conj(z1)})c=\\var{-z1*conj(z1)}c$.
\nBut we know that the constant term of $f(z)$ is $\\simplify{-{c1 * (a1 ^ 2 + b1 ^ 2)}} $.
\nHence \\[\\begin{eqnarray*} \\var{-z1*conj(z1)}c &=&\\simplify{-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\ \\Rightarrow c &=& \\simplify[]{{c1 * (a1 ^ 2 + b1 ^ 2)}/{abs(z1^2)}}\\\\ &=&\\var{c1} \\end{eqnarray*} \\]
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\nChoose the correct value for $z_1$:[[0]]
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\n \n \n \n\\[\\begin{eqnarray*}\n \n \\simplify{f({z1})}\t&=&\\simplify[std]{{z1}^3+{-2*a1 -c1}*{z1} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z1^3}+{-2*a1 -c1}{z1 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z1^3}+{(( -2) * a1 -c1)*z1^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z1}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\t\n \n &=&0\t\t\t\t\t\t\t\t\n \n \\end{eqnarray*}\n \n \\] Hence is a root.
\n \n ", "\n \n \nNot the correct choice as :\\[\\begin{eqnarray*}\n \n \\simplify{f({z2})}\t&=&\\simplify[std]{{z2}^3+{-2*a1 -c1}*{z2} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z2} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z2^3}+{-2*a1 -c1}{z2 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z2} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z2^3}+{(( -2) * a1 -c1)*z2^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z2}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\t\n \n &=&\\simplify[std]{{z2^3+(( -2) * a1 -c1)*z2^2+ (2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z2-c1 * (a1 ^ 2 + b1 ^ 2)}}\\neq 0\t\t\t\t\t\t\t\t\n \n \\end{eqnarray*}\\]
Hence not a root
Not the correct choice as :\\[\\begin{eqnarray*}\n \n \\simplify{f({z3})}\t&=&\\simplify[std]{{z3}^3+{-2*a1 -c1}*{z3} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z3} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z3^3}+{-2*a1 -c1}{z3 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z3} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z3^3}+{(( -2) * a1 -c1)*z3^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z3}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\t\n \n &=&\\simplify[std]{{z3^3+(( -2) * a1 -c1)*z3^2+ (2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z3-c1 * (a1 ^ 2 + b1 ^ 2)}}\\neq 0\t\t\t\t\t\t\t\t\n \n \\end{eqnarray*}\\]
Hence not a root
Not the correct choice as :\\[\\begin{eqnarray*}\n \n \\simplify{f({z2})}\t&=&\\simplify[std]{{z4}^3+{-2*a1 -c1}*{z4} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z4} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z4^3}+{-2*a1 -c1}{z4 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z4} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\n \n &=&\\simplify[std]{{z4^3}+{(( -2) * a1 -c1)*z4^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z4}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\\\\t\t\n \n &=&\\simplify[std]{{z4^3+(( -2) * a1 -c1)*z4^2+ (2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z4-c1 * (a1 ^ 2 + b1 ^ 2)}}\\neq 0\t\t\t\t\t\t\t\t\n \n \\end{eqnarray*}\\]
Hence not a root
The remaining roots of $f(z)$ are:
\n$z_2=\\;\\;$[[0]] (enter the complex root here)
\n$z_3=\\;\\;$[[1]] (enter the real root here)
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