Question

If ${{z}_{1}}, {{z}_{2}}, {{z}_{3}}, and \,{{z}_{4}}$ are the complex roots of the equation ${{z}^{4}}+3{{z}^{2}}+1=0$, then what is the value of $\prod\nolimits_{i=1}^{4}{4+({{z}_{i}}^{2})}$ ?
(a) 9
(b) 12
(c) 25
(d) 27

Answer 1

Hint: we need to find $(4+{{z}_{1}}^{2})\times (4+{{z}_{2}}^{2})\times (4+{{z}_{3}}^{2})\times (4+{{z}_{4}}^{2})$ where ${{z}_{1}},{{z}_{2}},{{z}_{3}},and\,{{z}_{4}}$ are the complex roots of the equation ${{z}^{4}}+3{{z}^{2}}+1=0$. So , we need to think of converting the given equation into a new equation whose roots will be $(4+{{z}_{1}}^{2}),(4+{{z}_{2}}^{2}),(4+{{z}_{3}}^{2})and(4+{{z}_{4}}^{2})$ and from there we can easily find out product of the roots.

Complete step-by-step solution:
We have the given bi-quadratic equation as,
${{z}^{4}}+3{{z}^{2}}+1=0\,\,.....(1)$, and
We need to find,
$(4+{{z}_{1}}^{2})\times (4+{{z}_{2}}^{2})\times (4+{{z}_{3}}^{2})\times (4+{{z}_{4}}^{2})$
where ${{z}_{1}},{{z}_{2}},{{z}_{3}},and\,{{z}_{4}}$ are the complex roots of the equation.
So we will try to convert given bi-quadratic equation into a new equation whose roots will be $(4+{{z}_{1}}^{2}),(4+{{z}_{2}}^{2}),(4+{{z}_{3}}^{2})and(4+{{z}_{4}}^{2})$, so we will suppose like this,
${{z}^{4}}+3{{z}^{2}}+1={{({{z}^{2}}+4)}^{2}}+m({{z}^{2}}+4)+c=0$
Where $m$ and $c$ are any arbitrary constant, so
We supposed like this as we know roots of above equation will be ${{z}_{i}}^{2}+4$ where i = 1, 2, 3, 4. Now,
${{({{z}^{2}}+4)}^{2}}+m({{z}^{2}}+4)+c=0$
After further simplifying and rearranging the above equation, we get
$\Rightarrow {{z}^{4}}+{{z}^{2}}(8+m)+(16+4m+c)=0\,\,.....(2)$
Now we will equate the equations (1) and (2), and we get
$\begin{align}
  & 8+m=3\,\, \\
 & m=-5 \\
\end{align}$
And also,
$16+4m+c=1$
Putting m = - 5 in above equation, we get
$c=5$
Hence our new equation becomes,
${{({{z}^{2}}+4)}^{2}}-5({{z}^{2}}+4)+5=0$
Now, we know that the product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $-\dfrac{b}{a}$, hence
Product of roots of ${{({{z}^{2}}+4)}^{2}}-5({{z}^{2}}+4)+5=0$ is = $5$
But we need to understand that it is the product of the quadratic equation only so, what we get is the product of 2 roots of the equation and, we know it is a complex equation and we have two more roots whose product will be also 5, hence
The product of all four roots of the equation will be $5\times 5=25$
Our answer matches with the given option (c) hence, it is the correct answer.

Note: You need to be cautious about that the quadratic equation only has two roots and what we will get as the product of roots will be of only two roots, for all four roots product we have to square the found product.
Students also make mistakes while letting the new equation variable they just assume one arbitrary constant whereas we need to assume two because there are three different degrees 4,2 and 0 of the variable in the equation. Always remember that the number of independent constants will be 1 less than the number of different degrees of the variable.