I am currently encountering difficulties while trying to solve the following question, and I would greatly appreciate any assistance you can provide.
Let $a,b,c$ be complex numbers.
The roots of $z^{3} + a^{2} + bz +c = 0$ are: $z_1,z_2,z_3$
All the roots are non zero.
I have to find cubic equation that has the following roots: $\frac{z_1}{z_2}, \frac{z_2}{z_3}, \frac{z_3}{z_1}$
I am aware that I need to apply the fundamental theorem of algebra and utilize symmetric elementary polynomials. Consequently, I have successfully computed the coefficients and exponents:
The coefficient of $z^{3}$ is just $1$
The coefficient of $z^{2}$ : $-b = z_1 + z_2 + z_3$
The coefficient of $z^{1}$: $c = z_1z_2 + z_2z_3 + z_1z_3$
The coefficient of $z^{0}$: $d = z_1z_2z_3$
I have calculated:
$q(z) = (z – \frac{z_1}{z_2})(z – \frac{z_2}{z_3})(z – \frac{z_3}{z_1}) = z^{3} + z^{2}(-\frac{z_2}{z_3} -\frac{z_1}{z_2} -\frac{z_3}{z_1}) + z(\frac{z_1}{z_3} +\frac{z_2}{z_1} +\frac{z_3}{z_2}) – 1$
I have explored several approaches in an attempt to simplify the coefficients using symmetric elementary polynomials. However, despite these efforts, I have not been successful in finding an expression that can be solely expressed in terms of the parameters b, c, and d as the coefficients of the new polynomial.
Best Answer
Note that there has to be more than one answer to this question. That's so because if, say, $z_1=1$, $z_2=2$, and $z_3=3$, then $\left\{\frac{z_1}{z_2},\frac{z_2}{z_3},\frac{z_3}{z_1}\right\}=\left\{\frac12,\frac23,3\right\}$, but, if $z_1=1$, $z_2=3$, and $z_3=2$, then you still have the same roots, but now $\left\{\frac{z_1}{z_2},\frac{z_2}{z_3},\frac{z_3}{z_1}\right\}=\left\{\frac13,\frac32,2\right\}$.
So,$$q(z)=z^3-Az^2+Bz-1,$$with$$A=\frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1}\quad\text{and}\quad B=\frac{z_1}{z_3}+\frac{z_3}{z_2}+\frac{z_2}{z_1}.$$In other words,$$A=\frac{z_1^{\,2}z_3+z_2^{\,2}z_1+z_3^{\,2}z_2}{z_1z_2z_3}=-\frac{z_1^{\,2}z_3+z_2^{\,2}z_1+z_3^{\,2}z_2}c$$and$$B=-\frac{z_1^{\,2}z_2+z_2^{\,2}z_3+z_3^{\,2}z_1}c.$$Therefore\begin{align}c(A+B)&=-z_1^{\,2}z_2-z_2^{\,2}z_3-z_3^{\,2}z_1-z_2^{\,2}z_1-z_3^{\,2}z_2-z_1^{\,2}z_3\\&=-\left(z_1+z_2+z_3\right) \left(z_1 z_2+z_3 z_2+z_1z_3\right)+3z_1z_2z_3\\&=ab-3c\end{align}and$$\begin{align}c^2AB&=z_1^{\,3}z_2^{\,3}+z_2^{\,3}z_3^{\,3}+z_1^{\,3}z_3^{\,3}+z_1^{\,4}z_2z_3+z_1z_2^{\,4}z_3+z_1z_2z_3^{\,4}+3z_1^{\,2}z_2^{\,2}z_3^{\,2}\\&=z_1 z_2 z_3\left(z_1+z_2+z_3\right)^3-6 z_1 z_2 z_3\left(z_1 z_2+z_3 z_2+z_1 z_3\right)\left(z_1+z_2+z_3\right)\\&\qquad+\left(z_1z_2+z_3z_2+z_1z_3\right)^3+9z_1^{\,2}z_2^{\,2}z_3^{\,2}\\&=a^3c-6abc+b^3+9c^2\end{align}$$So, take $A$ and $B$ such that$$\left\{\begin{array}{l}A+B=\frac{ab}c-3\\AB=\frac{a^3}c-\frac{6ab}c+\frac{b^3}{c^2}+9.\end{array}\right.\label{a}\tag1$$As a test, take the example from above: $z_1=1$, $z_2=2$, and $z_3=3$. Then system \eqref{a} becomes$$\left\{\begin{array}{l}A+B=8\\AB=\frac{575}{36},\end{array}\right.$$whose solutions are $(A,B)=\left(\frac{23}6,\frac{25}6\right)$ and $(A,B)=\left(\frac{25}6,\frac{23}6\right)$. Then the solutions of the equation $z^3-Az^2+Bz-1=0$ are: