If $z=4+i sqrt{7}$ then find the value of $z^3 -4z^2 -9z + 91$.












2












$begingroup$


So i was learning complex numbers and i came across this problem. In the solution they have made $z-4=isqrt{7}$ and then they squared the above equation resulting in $z^2 -8z+16=-7$ then they proceeded by sending $-7$ to the LHS resulting in $z^2 -8z+23=0$.



They then divided the original cubic term by the resulting equation from the above steps to get answer as $-1$.
I did not understand why did they do it and how they got $-1$.
The final equation looked like this:
$z^3 -4z^2 -9z + 91=(z^2 -8z+23)(z+4)-1=-1$.



I personally think they used the $x=nq+r$ where $x$ is divided by $n$ resulting in $q$ as the quotient and $r$ as the remainder.
If so then can all the problems like this be solved in the same manner?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to Maths SX! What don't you understand? The result of the division?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:57










  • $begingroup$
    I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:02












  • $begingroup$
    The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:05












  • $begingroup$
    Thankyou sir. Also can i solve similar problems in this way?
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:10










  • $begingroup$
    See the answer below, it explains it well.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:11
















2












$begingroup$


So i was learning complex numbers and i came across this problem. In the solution they have made $z-4=isqrt{7}$ and then they squared the above equation resulting in $z^2 -8z+16=-7$ then they proceeded by sending $-7$ to the LHS resulting in $z^2 -8z+23=0$.



They then divided the original cubic term by the resulting equation from the above steps to get answer as $-1$.
I did not understand why did they do it and how they got $-1$.
The final equation looked like this:
$z^3 -4z^2 -9z + 91=(z^2 -8z+23)(z+4)-1=-1$.



I personally think they used the $x=nq+r$ where $x$ is divided by $n$ resulting in $q$ as the quotient and $r$ as the remainder.
If so then can all the problems like this be solved in the same manner?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to Maths SX! What don't you understand? The result of the division?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:57










  • $begingroup$
    I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:02












  • $begingroup$
    The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:05












  • $begingroup$
    Thankyou sir. Also can i solve similar problems in this way?
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:10










  • $begingroup$
    See the answer below, it explains it well.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:11














2












2








2





$begingroup$


So i was learning complex numbers and i came across this problem. In the solution they have made $z-4=isqrt{7}$ and then they squared the above equation resulting in $z^2 -8z+16=-7$ then they proceeded by sending $-7$ to the LHS resulting in $z^2 -8z+23=0$.



They then divided the original cubic term by the resulting equation from the above steps to get answer as $-1$.
I did not understand why did they do it and how they got $-1$.
The final equation looked like this:
$z^3 -4z^2 -9z + 91=(z^2 -8z+23)(z+4)-1=-1$.



I personally think they used the $x=nq+r$ where $x$ is divided by $n$ resulting in $q$ as the quotient and $r$ as the remainder.
If so then can all the problems like this be solved in the same manner?






complex-numbers







share|cite|improve this question











$endgroup$




So i was learning complex numbers and i came across this problem. In the solution they have made $z-4=isqrt{7}$ and then they squared the above equation resulting in $z^2 -8z+16=-7$ then they proceeded by sending $-7$ to the LHS resulting in $z^2 -8z+23=0$.



They then divided the original cubic term by the resulting equation from the above steps to get answer as $-1$.
I did not understand why did they do it and how they got $-1$.
The final equation looked like this:
$z^3 -4z^2 -9z + 91=(z^2 -8z+23)(z+4)-1=-1$.



I personally think they used the $x=nq+r$ where $x$ is divided by $n$ resulting in $q$ as the quotient and $r$ as the remainder.
If so then can all the problems like this be solved in the same manner?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 17:06







HOME WORK AND EXERCISES

















asked Dec 9 '18 at 16:49









HOME WORK AND EXERCISESHOME WORK AND EXERCISES

195




195








  • 1




    $begingroup$
    Welcome to Maths SX! What don't you understand? The result of the division?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:57










  • $begingroup$
    I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:02












  • $begingroup$
    The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:05












  • $begingroup$
    Thankyou sir. Also can i solve similar problems in this way?
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:10










  • $begingroup$
    See the answer below, it explains it well.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:11














  • 1




    $begingroup$
    Welcome to Maths SX! What don't you understand? The result of the division?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:57










  • $begingroup$
    I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:02












  • $begingroup$
    The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:05












  • $begingroup$
    Thankyou sir. Also can i solve similar problems in this way?
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:10










  • $begingroup$
    See the answer below, it explains it well.
    $endgroup$
    – Winther
    Dec 9 '18 at 17:11








1




1




$begingroup$
Welcome to Maths SX! What don't you understand? The result of the division?
$endgroup$
– Bernard
Dec 9 '18 at 16:57




$begingroup$
Welcome to Maths SX! What don't you understand? The result of the division?
$endgroup$
– Bernard
Dec 9 '18 at 16:57












$begingroup$
I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 17:02






$begingroup$
I don't understand the method being used. Like is it similar as dividing i^n by i^4 to find the actual value of i^n. Can that same concept even be used in the a complex number equation. In general i need someone to explain it to me like why and what is going on.
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 17:02














$begingroup$
The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
$endgroup$
– Winther
Dec 9 '18 at 17:05






$begingroup$
The given $z$ satisfy a second order equation with rational coefficients ($z^2 = Az + B$ for some rational numbers $A,B$). This means that any polynomial expression with rational coefficients can be reduced to the form $Cz+D$ with $C,D$ rational. The way to do this is by polynomial division as you point out. In this particular case $C$ turns out to be $0$ and $D=-1$ so you get the simple answer $-1$.
$endgroup$
– Winther
Dec 9 '18 at 17:05














$begingroup$
Thankyou sir. Also can i solve similar problems in this way?
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 17:10




$begingroup$
Thankyou sir. Also can i solve similar problems in this way?
$endgroup$
– HOME WORK AND EXERCISES
Dec 9 '18 at 17:10












$begingroup$
See the answer below, it explains it well.
$endgroup$
– Winther
Dec 9 '18 at 17:11




$begingroup$
See the answer below, it explains it well.
$endgroup$
– Winther
Dec 9 '18 at 17:11










2 Answers
2






active

oldest

votes


















4












$begingroup$

You surely get $(z-4)^2=-7$, which indeed expands to $z^2-8z+23=0$.



Then you can do the long division of $z^3-4z^2-9z+91$ by $z^2-8z+23$, which yields a quotient $z+4$ and remainder $-1$; therefore, for any value of $z$, you have
$$
z^3-4z^2-9z+91=(z^2-8z+23)(z+4)-1
$$
If you substitute the given $z$, then you get
$$
0(4+isqrt{7}+4)-1=-1
$$



Yes, problems of this kind can be treated in this way; if $w$ is any complex number and $g(z)$ is a polynomial such that $g(w)=0$, then for every polynomial $f(z)$ one has
$$
f(z)=q(z)g(z)+r(z)
$$
where the degree of $r$ is less than the degree of $g$, so
$$
f(w)=q(w)g(w)+r(w)=r(w)
$$
and the computation of $r(w)$ is easier.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you sir. 😊
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:11



















1












$begingroup$

Alternatively, note:
$$z-4=isqrt7 iff z-3=1+isqrt7 iff z+3=7+isqrt7.$$
You can replace step-by-step:
$$begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\
=&(z+3)(z-3)(z-4)+55=\
=&(isqrt7+7)(1+isqrt7)(isqrt7)+55=\
=&(isqrt7+7)(isqrt7-7)+55=\
=&-7-49+55=\
=&-1.end{align}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    You surely get $(z-4)^2=-7$, which indeed expands to $z^2-8z+23=0$.



    Then you can do the long division of $z^3-4z^2-9z+91$ by $z^2-8z+23$, which yields a quotient $z+4$ and remainder $-1$; therefore, for any value of $z$, you have
    $$
    z^3-4z^2-9z+91=(z^2-8z+23)(z+4)-1
    $$
    If you substitute the given $z$, then you get
    $$
    0(4+isqrt{7}+4)-1=-1
    $$



    Yes, problems of this kind can be treated in this way; if $w$ is any complex number and $g(z)$ is a polynomial such that $g(w)=0$, then for every polynomial $f(z)$ one has
    $$
    f(z)=q(z)g(z)+r(z)
    $$
    where the degree of $r$ is less than the degree of $g$, so
    $$
    f(w)=q(w)g(w)+r(w)=r(w)
    $$
    and the computation of $r(w)$ is easier.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – HOME WORK AND EXERCISES
      Dec 9 '18 at 17:11
















    4












    $begingroup$

    You surely get $(z-4)^2=-7$, which indeed expands to $z^2-8z+23=0$.



    Then you can do the long division of $z^3-4z^2-9z+91$ by $z^2-8z+23$, which yields a quotient $z+4$ and remainder $-1$; therefore, for any value of $z$, you have
    $$
    z^3-4z^2-9z+91=(z^2-8z+23)(z+4)-1
    $$
    If you substitute the given $z$, then you get
    $$
    0(4+isqrt{7}+4)-1=-1
    $$



    Yes, problems of this kind can be treated in this way; if $w$ is any complex number and $g(z)$ is a polynomial such that $g(w)=0$, then for every polynomial $f(z)$ one has
    $$
    f(z)=q(z)g(z)+r(z)
    $$
    where the degree of $r$ is less than the degree of $g$, so
    $$
    f(w)=q(w)g(w)+r(w)=r(w)
    $$
    and the computation of $r(w)$ is easier.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – HOME WORK AND EXERCISES
      Dec 9 '18 at 17:11














    4












    4








    4





    $begingroup$

    You surely get $(z-4)^2=-7$, which indeed expands to $z^2-8z+23=0$.



    Then you can do the long division of $z^3-4z^2-9z+91$ by $z^2-8z+23$, which yields a quotient $z+4$ and remainder $-1$; therefore, for any value of $z$, you have
    $$
    z^3-4z^2-9z+91=(z^2-8z+23)(z+4)-1
    $$
    If you substitute the given $z$, then you get
    $$
    0(4+isqrt{7}+4)-1=-1
    $$



    Yes, problems of this kind can be treated in this way; if $w$ is any complex number and $g(z)$ is a polynomial such that $g(w)=0$, then for every polynomial $f(z)$ one has
    $$
    f(z)=q(z)g(z)+r(z)
    $$
    where the degree of $r$ is less than the degree of $g$, so
    $$
    f(w)=q(w)g(w)+r(w)=r(w)
    $$
    and the computation of $r(w)$ is easier.






    share|cite|improve this answer









    $endgroup$



    You surely get $(z-4)^2=-7$, which indeed expands to $z^2-8z+23=0$.



    Then you can do the long division of $z^3-4z^2-9z+91$ by $z^2-8z+23$, which yields a quotient $z+4$ and remainder $-1$; therefore, for any value of $z$, you have
    $$
    z^3-4z^2-9z+91=(z^2-8z+23)(z+4)-1
    $$
    If you substitute the given $z$, then you get
    $$
    0(4+isqrt{7}+4)-1=-1
    $$



    Yes, problems of this kind can be treated in this way; if $w$ is any complex number and $g(z)$ is a polynomial such that $g(w)=0$, then for every polynomial $f(z)$ one has
    $$
    f(z)=q(z)g(z)+r(z)
    $$
    where the degree of $r$ is less than the degree of $g$, so
    $$
    f(w)=q(w)g(w)+r(w)=r(w)
    $$
    and the computation of $r(w)$ is easier.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 17:07









    egregegreg

    182k1485204




    182k1485204












    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – HOME WORK AND EXERCISES
      Dec 9 '18 at 17:11


















    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – HOME WORK AND EXERCISES
      Dec 9 '18 at 17:11
















    $begingroup$
    Thank you sir. 😊
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:11




    $begingroup$
    Thank you sir. 😊
    $endgroup$
    – HOME WORK AND EXERCISES
    Dec 9 '18 at 17:11











    1












    $begingroup$

    Alternatively, note:
    $$z-4=isqrt7 iff z-3=1+isqrt7 iff z+3=7+isqrt7.$$
    You can replace step-by-step:
    $$begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\
    =&(z+3)(z-3)(z-4)+55=\
    =&(isqrt7+7)(1+isqrt7)(isqrt7)+55=\
    =&(isqrt7+7)(isqrt7-7)+55=\
    =&-7-49+55=\
    =&-1.end{align}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Alternatively, note:
      $$z-4=isqrt7 iff z-3=1+isqrt7 iff z+3=7+isqrt7.$$
      You can replace step-by-step:
      $$begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\
      =&(z+3)(z-3)(z-4)+55=\
      =&(isqrt7+7)(1+isqrt7)(isqrt7)+55=\
      =&(isqrt7+7)(isqrt7-7)+55=\
      =&-7-49+55=\
      =&-1.end{align}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Alternatively, note:
        $$z-4=isqrt7 iff z-3=1+isqrt7 iff z+3=7+isqrt7.$$
        You can replace step-by-step:
        $$begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\
        =&(z+3)(z-3)(z-4)+55=\
        =&(isqrt7+7)(1+isqrt7)(isqrt7)+55=\
        =&(isqrt7+7)(isqrt7-7)+55=\
        =&-7-49+55=\
        =&-1.end{align}$$






        share|cite|improve this answer









        $endgroup$



        Alternatively, note:
        $$z-4=isqrt7 iff z-3=1+isqrt7 iff z+3=7+isqrt7.$$
        You can replace step-by-step:
        $$begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\
        =&(z+3)(z-3)(z-4)+55=\
        =&(isqrt7+7)(1+isqrt7)(isqrt7)+55=\
        =&(isqrt7+7)(isqrt7-7)+55=\
        =&-7-49+55=\
        =&-1.end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 18:15









        farruhotafarruhota

        20.3k2739




        20.3k2739






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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