If $a_1, a_2, ...., a_n, A_1,A_2,.....,A_n,k$ are all real numbers, then find the number of imaginary roots of the following equation:
$$\frac{A_1^2}{x-a_1}+\frac{A_2^2}{x-a_2}+\frac{A_3^2}{x-a_3}.....+\frac{A_n^2}{x-a_n}=k$$
Given answer is $0$. Could someone give me little hint to proceed in this question?
Hint: Take L.C.M., form a function of $x$ and apply intermediate value theorem.
Assume that order is $a_1
I don't know my idea is true or not, but I pose it, may helps:
Let $$f(z)=\frac{A_1^2}{z-a_1}+\frac{A_2^2}{z-a_2}+\frac{A_3^2}{z-a_3}.....+\frac{A_n^2}{z-a_n}-k=0$$ $f(z)$ is an analytic function with $n$ zeros. Suppose $z_0=x_0+iy_0$ is one of it's zeros where $y_0>0$. Let $C$ is a circle with center $z_0$ and redius $r$ such that it does not include other zeros (All zeros of analytic function are isolated). From winding number $$\int_Cf(z)=\int_C\frac{A_1^2}{z-a_1}dz+\int_C\frac{A_2^2}{z-a_2}dz+\int_C\frac{A_3^2}{z-a_3}dz+.....+\int_C\frac{A_n^2}{z-a_n}dz-\int_Ck~dz=0$$ or $$2\pi i(A_1^2+A_2^2+\cdots+A_n^2)-2\pi k=0$$ this is contradict if $k$ is real.
Someone check.