Question

Prove that the equation
$\frac{A^2}{x-a}+\frac{B^2}{x-b}+\frac{C^2}{x-c}+...\frac{K^2}{x-k}=x+1$, has no imaginary roots, where $A,B,C,....,K,a,b,c,...,k$ are real.

Answer 1

Assume $\alpha +i\beta$ is an imaginary root of the given equation then conjugate of this root $\alpha -i\beta$ is also root of this equation.
Putting $x=\alpha +i\beta$ and $x=\alpha -i\beta$ in the given equation, then
$\frac{A^2}{\alpha +i\beta -a}+\frac{B^2}{\alpha +i\beta -b}+\frac{C^2}{\alpha +i\beta -c}+...\frac{K^2}{\alpha +i\beta -k}=\alpha +i\beta +1$ ......(1)

$\frac{A^2}{\alpha -i\beta -a}+\frac{B^2}{\alpha -i\beta -b}+\frac{C^2}{\alpha -i\beta -c}+...\frac{K^2}{\alpha -i\beta -k}=\alpha -i\beta +1$ ......(2)

Subtracting (2) from (1), we get
$\Rightarrow -2i\beta [\frac{A^2}{{(\alpha -a)}^2+{\beta }^2}+\frac{B^2}{{(\alpha -b)}^2+{\beta }^2}+\frac{C^2}{{(\alpha -c)}^2+{\beta }^2}+...\frac{K^2}{{(\alpha -k)}^2+{\beta }^2}]=2i\beta$

$\Rightarrow 2i\beta [1+\frac{A^2}{{(\alpha -a)}^2+{\beta }^2}+\frac{B^2}{{(\alpha -b)}^2+{\beta }^2}+\frac{C^2}{{(\alpha -c)}^2+{\beta }^2}+...\frac{K^2}{{(\alpha -k)}^2+{\beta }^2}]=0$
The expression in bracket $\ne 0$
$\therefore 2i\beta =0\Rightarrow \beta =0(\because i\ne 0)$
Hence all roots of the given equation are real.