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So i was learning complex numbers and i came across this problem. In the solution they have made $z-4=i\sqrt{7}$ 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?

HOME WORK AND EXERCISES
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    Welcome to Maths SX! What don't you understand? The result of the division? – Bernard Dec 09 '18 at 16:57
  • 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. – HOME WORK AND EXERCISES Dec 09 '18 at 17:02
  • 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 $-1$. – Winther Dec 09 '18 at 17:05
  • Thankyou sir. Also can i solve similar problems in this way? – HOME WORK AND EXERCISES Dec 09 '18 at 17:10
  • See the answer below, it explains it well. – Winther Dec 09 '18 at 17:11

2 Answers2

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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 $$ 0(4+i\sqrt{7}+4)-1=-1 $$

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.

egreg
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Alternatively, note: $$z-4=i\sqrt7 \iff z-3=1+i\sqrt7 \iff z+3=7+i\sqrt7.$$ You can replace step-by-step: $$\begin{align}z^3 -4z^2 -9z + 91=&z^2(z-4)-9(z-4)+55=\\ =&(z+3)(z-3)(z-4)+55=\\ =&(i\sqrt7+7)(1+i\sqrt7)(i\sqrt7)+55=\\ =&(i\sqrt7+7)(i\sqrt7-7)+55=\\ =&-7-49+55=\\ =&-1.\end{align}$$

farruhota
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