Complex polynomials… need help












-1














I am very confused on how to start this question, I know that x=0 when i=1/5 and i=1 but not sure where to go from here, any help would be really appreciated.



"Find a polynomial
P
of a complex variable
x
that has zeros at exactly the following points:
x=5i−1,x=3i−3.



All these zeros should be of first order. Be sure to expand all products in your answer; also expand products of the form
(a+bi)⋅x
if they occur."






linear-algebra polynomials complex-numbers







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  • You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
    – A.Γ.
    2 days ago










  • $i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
    – Andrei
    2 days ago










  • $i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
    – Henrik
    2 days ago










  • There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
    – José Carlos Santos
    2 days ago
















-1














I am very confused on how to start this question, I know that x=0 when i=1/5 and i=1 but not sure where to go from here, any help would be really appreciated.



"Find a polynomial
P
of a complex variable
x
that has zeros at exactly the following points:
x=5i−1,x=3i−3.



All these zeros should be of first order. Be sure to expand all products in your answer; also expand products of the form
(a+bi)⋅x
if they occur."






linear-algebra polynomials complex-numbers







share|cite|improve this question







New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
    – A.Γ.
    2 days ago










  • $i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
    – Andrei
    2 days ago










  • $i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
    – Henrik
    2 days ago










  • There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
    – José Carlos Santos
    2 days ago














-1












-1








-1







I am very confused on how to start this question, I know that x=0 when i=1/5 and i=1 but not sure where to go from here, any help would be really appreciated.



"Find a polynomial
P
of a complex variable
x
that has zeros at exactly the following points:
x=5i−1,x=3i−3.



All these zeros should be of first order. Be sure to expand all products in your answer; also expand products of the form
(a+bi)⋅x
if they occur."






linear-algebra polynomials complex-numbers







share|cite|improve this question







New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am very confused on how to start this question, I know that x=0 when i=1/5 and i=1 but not sure where to go from here, any help would be really appreciated.



"Find a polynomial
P
of a complex variable
x
that has zeros at exactly the following points:
x=5i−1,x=3i−3.



All these zeros should be of first order. Be sure to expand all products in your answer; also expand products of the form
(a+bi)⋅x
if they occur."






linear-algebra polynomials complex-numbers




linear-algebra polynomials complex-numbers






share|cite|improve this question







New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Kiryne

135




135




New contributor




Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kiryne is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
    – A.Γ.
    2 days ago










  • $i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
    – Andrei
    2 days ago










  • $i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
    – Henrik
    2 days ago










  • There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
    – José Carlos Santos
    2 days ago


















  • You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
    – A.Γ.
    2 days ago










  • $i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
    – Andrei
    2 days ago










  • $i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
    – Henrik
    2 days ago










  • There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
    – José Carlos Santos
    2 days ago
















You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
– A.Γ.
2 days ago




You seem to misunderstand what $i$ means. It is not a variable, it is a complex constant such that $i^2=-1$.
– A.Γ.
2 days ago












$i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
– Andrei
2 days ago




$i=sqrt{-1}$ since we are talking about complex polynomials, so $ine 1/5$. $i$ is not a variable, it's a constant.
– Andrei
2 days ago












$i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
– Henrik
2 days ago




$i$ is the complex unit, not a variable so it's never $frac15$ nor $1$.
– Henrik
2 days ago












There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
– José Carlos Santos
2 days ago




There's a lot of confusion here. The complex (non-real) number $i$ cannot possibly be the real number $frac15$ or the real number $1$.
– José Carlos Santos
2 days ago










1 Answer
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Polynomials can be written in a form that shows its zeros directly, that is, if $x_1, x_2, ldots, x_n$ are zeros of a polynomial $p$ and $a$ is the leading coefficient, $p(x) = a(x - x_1)(x - x_2)ldots(x - x_n)$.

To get the the standard form you only have to multiply it all together.



So in your case, $x_1 = 5i - 1$ and $x_2 = 3i - 3$.
You do not have any conditions on what the leading polynomial coefficient should be, so let it be $1$. That gives you
$$p(x) = (x - (5i - 1))(x - (3i -3))= \
x^2 - x(3i -3) - (5i - 1)x + (5i - 1)(3i - 3) $$

You are instructed to expand those produsts...
$$ p(x) = x^2 - 3ix + 3x - 5ix + x + 15i^2 - 15i - 3i + 3= \
x^2 - 8ix + 4x -18i -12$$.






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    1 Answer
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    Polynomials can be written in a form that shows its zeros directly, that is, if $x_1, x_2, ldots, x_n$ are zeros of a polynomial $p$ and $a$ is the leading coefficient, $p(x) = a(x - x_1)(x - x_2)ldots(x - x_n)$.

    To get the the standard form you only have to multiply it all together.



    So in your case, $x_1 = 5i - 1$ and $x_2 = 3i - 3$.
    You do not have any conditions on what the leading polynomial coefficient should be, so let it be $1$. That gives you
    $$p(x) = (x - (5i - 1))(x - (3i -3))= \
    x^2 - x(3i -3) - (5i - 1)x + (5i - 1)(3i - 3) $$

    You are instructed to expand those produsts...
    $$ p(x) = x^2 - 3ix + 3x - 5ix + x + 15i^2 - 15i - 3i + 3= \
    x^2 - 8ix + 4x -18i -12$$    .






    share|cite|improve this answer


























      0














      Polynomials can be written in a form that shows its zeros directly, that is, if $x_1, x_2, ldots, x_n$ are zeros of a polynomial $p$ and $a$ is the leading coefficient, $p(x) = a(x - x_1)(x - x_2)ldots(x - x_n)$.

      To get the the standard form you only have to multiply it all together.



      So in your case, $x_1 = 5i - 1$ and $x_2 = 3i - 3$.
      You do not have any conditions on what the leading polynomial coefficient should be, so let it be $1$. That gives you
      $$p(x) = (x - (5i - 1))(x - (3i -3))= \
      x^2 - x(3i -3) - (5i - 1)x + (5i - 1)(3i - 3) $$

      You are instructed to expand those produsts...
      $$ p(x) = x^2 - 3ix + 3x - 5ix + x + 15i^2 - 15i - 3i + 3= \
      x^2 - 8ix + 4x -18i -12$$      .






      share|cite|improve this answer
























        0












        0








        0






        Polynomials can be written in a form that shows its zeros directly, that is, if $x_1, x_2, ldots, x_n$ are zeros of a polynomial $p$ and $a$ is the leading coefficient, $p(x) = a(x - x_1)(x - x_2)ldots(x - x_n)$.

        To get the the standard form you only have to multiply it all together.



        So in your case, $x_1 = 5i - 1$ and $x_2 = 3i - 3$.
        You do not have any conditions on what the leading polynomial coefficient should be, so let it be $1$. That gives you
        $$p(x) = (x - (5i - 1))(x - (3i -3))= \
        x^2 - x(3i -3) - (5i - 1)x + (5i - 1)(3i - 3) $$

        You are instructed to expand those produsts...
        $$ p(x) = x^2 - 3ix + 3x - 5ix + x + 15i^2 - 15i - 3i + 3= \
        x^2 - 8ix + 4x -18i -12$$        .






        share|cite|improve this answer












        Polynomials can be written in a form that shows its zeros directly, that is, if $x_1, x_2, ldots, x_n$ are zeros of a polynomial $p$ and $a$ is the leading coefficient, $p(x) = a(x - x_1)(x - x_2)ldots(x - x_n)$.

        To get the the standard form you only have to multiply it all together.



        So in your case, $x_1 = 5i - 1$ and $x_2 = 3i - 3$.
        You do not have any conditions on what the leading polynomial coefficient should be, so let it be $1$. That gives you
        $$p(x) = (x - (5i - 1))(x - (3i -3))= \
        x^2 - x(3i -3) - (5i - 1)x + (5i - 1)(3i - 3) $$

        You are instructed to expand those produsts...
        $$ p(x) = x^2 - 3ix + 3x - 5ix + x + 15i^2 - 15i - 3i + 3= \
        x^2 - 8ix + 4x -18i -12$$        .







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










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