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Prove that, for integral values of $n\ge 1$, all the roots of the equation $$nz^n=1+z+z^2+...+z^n$$ lie within the circle $\vert z\vert=\frac{n}{n-1}$

Taking modulus on both sides, $$n\vert z\vert^n=\vert1+z+z^2+...+z^n\vert$$ Using triangle inequality, $$n\vert z\vert^n\le 1+\vert z\vert+\vert z\vert^2+...+\vert z\vert^n$$ $$\vert z\vert^n(n-1)\le 1+\vert z\vert+\vert z\vert^2+...+\vert z\vert^{n-1}$$

Using sum of GP formula, $$(n-1)\vert z\vert^n\le\frac{\vert z\vert^n-1}{\vert z\vert -1}$$ $$(n-1)\frac{\vert z\vert^n}{\vert z\vert^n-1}\le\frac{1}{\vert z\vert -1}$$ (I am not sure about the above step because I am multiplying with a number that can be negative.)

How should I proceed?

Ted Shifrin
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Aditya Dev
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2 Answers2

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$z=0$ is never a solution so we can safely rewrite the equation as

$$n=\frac1{z^n}+\cdots+\frac1z+1.$$

If $z$ is a solution with norm strictly greater than $n/(n-1)$ then the first $n$ terms on the right hand side are strictly less (in absolute value) than $(n-1)/n$ so we would have

$$n<n\frac{(n-1)}n+1$$

which is clearly impossible.

Justpassingby
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Assume $|z|>1$ since otherwise there is nothing to prove. Then $|z|^k \leq |z|^{n-1}$ for $k<n$. Start from $$\begin{align}|z|^n (n−1)&≤1+|z|+|z|^2+...+|z|^{n−1} \\ &\leq n |z|^{n-1}\end{align}$$ Divide by $(n-1)|z|^{n-1}$ to get the result.

mlu
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