Question

Show that the equation
$\frac{A^2}{x-a}+\frac{B^2}{x-b}+\frac{C^2}{x-c}+.....+\frac{H^2}{x-h}=k$ has no imaginary roots.

Answer 1

Let us suppose that the equation posses an imaginary root $\alpha +i\beta$.

Since the coefficients are all real numbers, the conjugate of the root must also be another root of the equation, that is $\alpha \pm i\beta$ are roots of the equation.

Substituting both into the equation one after another, we get:

$\frac{A^2}{(\alpha -a)+i\beta }+\frac{B^2}{(\alpha -b)+i\beta }+...+\frac{H^2}{(\alpha -h)+i\beta }=k...\left(1\right)$

$\frac{A^2}{(\alpha -a)-i\beta }+\frac{B^2}{(\alpha -b)-i\beta }+...+\frac{H^2}{(\alpha -h)-i\beta }=k...\left(2\right)$

Subtracting both sides of the equations, we get:

(Consider calculation only for the first term, other terms follow the similar pattern)

$\frac{A^2}{(\alpha -a)+i\beta }-\frac{A^2}{(\alpha -a)-i\beta }=\frac{A^2\left(\right(\alpha -a)-i\beta )-A^2\left(\right(\alpha -a)+i\beta )}{(\alpha -a{)}^2+{\beta }^2}=-2i\beta \frac{A^2}{(\alpha -a{)}^2+{\beta }^2}$

$\Rightarrow \left(1\right)-\left(2\right)=-2i\beta (\frac{A^2}{(\alpha -a{)}^2+{\beta }^2}+\frac{B^2}{(\alpha -b{)}^2+{\beta }^2}+...+\frac{H^2}{(\alpha -h{)}^2+{\beta }^2})=0$

Since the term within the parentheses is a sum of positive terms, the only way the above identity holds is $\beta =0$

This contradicts our earlier assumption that the equation posses an imaginary root. Hence the given equation does not have any imaginary roots.