#### Answer

$a=1, b=0, \text { and } c=1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the fact that a quadratic equation with roots $r_1$ and $r_2$ has a factored form of $(x-r_1)(x-r_2)=0$. Then use the FOIL Method to convert the equation in the form $ax^2+bx+c=0.$
$\bf{\text{Solution Details:}}$
The factored form of the quadratic equation with the given roots $\{ i,-i \},$ is \begin{array}{l}\require{cancel} (x-i)(x-(-i))=0 \\\\ (x-i)(x+i)=0 .\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} (x)^2-(i)^2=0 \\\\ x^2-i^2=0 .\end{array}
Since $i^2=-1$, the expression above becomes \begin{array}{l}\require{cancel} x^2-(-1)=0 \\\\ x^2+1=0 .\end{array}
Hence, $
a=1, b=0, \text { and } c=1
.$