The number of imaginary roots of $\sum_{n=1}^{100} \frac {n^2}{x-x_{n} }= 101$, where each $x_n$ is real

Question

Determine the number of imaginary roots of the equation

$$\sum_{n=1}^{100} \frac {n^2}{x-x_{n} }= 101$$ where $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$ are all real.

I did this question a few months back, but I am not able to do this question now.

Also, if anyone has seen this question before, please let me know its source.

Answer 1

Elaborating on the answer from @DavidK:

Let $x=a+bi$.

$$\frac1{x-x_n}=\frac1{a+x_n+bi}= \frac{a+x_n-bi}{(a+x_n)^2+b^2}= \frac{a+x_n}{(a+x_n)^2+b^2}+ \frac{-bi}{(a+x_n)^2+b^2}$$

So the imaginary part of the sum equals $$b\sum_{n=1}^\infty \frac{n^2}{(a+x_n)^2+b^2}$$

Since $101$ is real, the above summation equals zero.

Observe that $$\frac{n^2}{(a+x_n)^2+b^2}>0$$

Can you see that $b=0$?

Answer 2

I think the question is ambiguous about whether it asks for purely imaginary roots or merely for complex roots that are not real. But let's take a look to see what kind of roots we can find.

Let $x$ be a complex number. Suppose the imaginary part of $x$ is positive. For $n$ an integer and $x_n$ real, what can you say about the imaginary part of $$ \frac {n^2}{x-x_{n}}? $$

Now add up a hundred terms in that form. What conditions will make the sum equal to $101$?