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Given two complex numbers, find by inspection the one that is a root of a given quartic real polynomial and hence find the other roots.

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Given two complex numbers, find by inspection the one that is a root of a given quartic real polynomial $f(z)$ and hence find the other roots.

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a) We use the method given in Show steps for part a).

Note that $|\\var{z1}|^2=\\var{mz1}$ divides the constant term $\\var{mz1*mz2}$,

but that $|\\var{z3}|^2=\\var{mz3}$ does not divides the constant term $\\var{mz1*mz2}$.

Hence $\\var{z1}$ is the root we are looking for.

b) A quadratic factor of $f(z)$.

Since $f(z)$ is a polynomial with real coefficients then if $z=z_1$ is a root we have that the conjugate $z=\\overline{z_1}$ is also a root.

Hence the complex number $z_2=\\overline{\\var{z1}}=\\var{conj(z1)}$ is a root.

Hence $q_1(z) = (z-(\\var{z1}))(z-(\\var{conj(z1)}))=\\simplify[std]{z^2-{2*a1}*z+{abs(z1)^2}}$ is a factor of $f(z)$.

c)The other quadratic factor and the other roots.

We have that $f(z)=q_1(z)q_2(z)$, where $q_1(z)$ is as above and we have to find the quadratic $q_2(z)=z^2+az+b$ with real coefficients $a$ and $b$.

\\[\\begin{eqnarray*}f(z) &=& \\simplify[std]{z ^ 4+ {( -2) * r12}*z ^ 3+ {mz1+mz2+4*re(z1)*re(z2)} * z^2 -{2*(re(z2)*mz1+re(z1)*mz2)}z+{mz1*mz2}}\\\\&=&q_1(z)q_2(z)\\\\&=&(\\simplify[std]{z^2-{2*a1}*z+{mz1}})(z^2+az+b)\\\\&=&\\simplify[std]{z^4+(a-{2*a1})z^3+(b-{2*a1}*a+{mz1})*z^2+({mz1}a-{2*a1}b)*z+{mz1}*b}\\end{eqnarray*}\\]

Identifying the constant terms and the coefficients of $z^3$ on both sides of this equation gives:

$a=\\var{-2*a2},\\;\\;b=\\var{mz2}$

Hence $q_2(z)=\\simplify[std]{z^2-{2*a2}*z+{mz2}}$

You can then find the roots of this quadratic, giving the other roots of $f(z)$:

$z_3=\\simplify[std]{{a2}-{b2}*i}$ (negative imaginary part)

$z_4=\\simplify[std]{{a2}+{b2}*i}$ (positive imaginary part)

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Given $\\displaystyle f(z) = \\simplify[std]{z ^ 4+ {( -2) * r12}*z ^ 3+ {mz1+mz2+4*re(z1)*re(z2)} * z^2 -{2*(re(z2)*mz1+re(z1)*mz2)}z+{mz1*mz2}}$, one of the following complex numbers is a root $z_1$ of the equation $f(z)=0$.

Choose the correct value for $z_1$:[[0]]

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Since you are given that $f(z)$ has a complex root $z_1$ and since $f(z)$ is a polynomial with real coefficients then the complex conjugate $\\overline{z_1}$ must also be a root.

Since $(z-z_1)(z-\\overline{z_1})=(z^2-2\\operatorname{Re}(z)+|z_1|^2)$ we have that:\\[f(z)=(z^2-2\\operatorname{Re}(z)+|z_1|^2)(z^2+az+b)=\\simplify{z ^ 4+ {( -2) * r12}*z ^ 3+ {mz1+mz2+4*re(z1)*re(z2)} * z^2 -{2*(re(z2)*mz1+re(z1)*mz2)}z+{mz1*mz2}}\\] where $a$ and $b$ are real.

Looking at the constant term we see that :

\\[|z_1|^2b = \\var{mz1*mz2}\\]

Hence $|z_1|^2$ divides $ \\var{mz1*mz2}$.

An easy test to see if one of the complex numbers given is not a root is to see if its modulus squared does not divide $ \\var{mz1*mz2}$. If it does not divide then the other must be the root.

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$\\simplify[std]{{a1}+{b1}i}$

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$\\simplify[std]{{z3}}$

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Write down the quadratic factor with real coefficients, $q_1(z)$, of $f(z)$ which has $z_1$ as a root:

$q_1(z)=\\;$[[0]]

Apart from $z_1$, $q_1(z)$ has another root $z_2$, which is also a root of $f(z)$.

$z_2=\\;$[[1]]

If $z_1$ is a root then its conjugate $z_2$= $\\overline{z_1}$ is also a root.

Since $q_1(z)$ is a factor of $f(z)$ the other roots are given by finding the other quadratic factor $q_2(z)$ of $f(z)=q_1(z)q_2(z)$

$q_2(z)\\;=$[[0]]

Find the roots of $q_2(z)$ and hence the remaining two roots $z_3,\\;z_4$ of $f(z)$

$z_3=\\;$[[1]] (imaginary part negative)

$z_4=\\;$[[2]] (imaginary part positive).

$q_1(z)q_2(z)=f(z)$.

Once you have found $q_1(z)$ then the easiest way to find $q_2(z)$ is to compare the terms in $z^3$ and the constant terms.

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