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.
    $endgroup$
    – Alex allee
    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?
    $endgroup$
    – Alex allee
    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.
    $endgroup$
    – Alex allee
    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?
    $endgroup$
    – Alex allee
    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







Alex allee

















asked Dec 9 '18 at 16:49









Alex alleeAlex allee

194




194








  • 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.
    $endgroup$
    – Alex allee
    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?
    $endgroup$
    – Alex allee
    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.
    $endgroup$
    – Alex allee
    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?
    $endgroup$
    – Alex allee
    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.
$endgroup$
– Alex allee
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.
$endgroup$
– Alex allee
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?
$endgroup$
– Alex allee
Dec 9 '18 at 17:10




$begingroup$
Thankyou sir. Also can i solve similar problems in this way?
$endgroup$
– Alex allee
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. 😊
    $endgroup$
    – Alex allee
    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. 😊
      $endgroup$
      – Alex allee
      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. 😊
      $endgroup$
      – Alex allee
      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

    181k1485203




    181k1485203












    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – Alex allee
      Dec 9 '18 at 17:11


















    • $begingroup$
      Thank you sir. 😊
      $endgroup$
      – Alex allee
      Dec 9 '18 at 17:11
















    $begingroup$
    Thank you sir. 😊
    $endgroup$
    – Alex allee
    Dec 9 '18 at 17:11




    $begingroup$
    Thank you sir. 😊
    $endgroup$
    – Alex allee
    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

        20k2738




        20k2738






























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            Quarter-circle Tiles

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            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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