Complex Roots Magnitude (Posted on 2021-02-25)

	Submitted by Brian Smith
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Start with proving a lemma: if z is a root of f(z) then z^(n+1) + z^n + ... + z = n+1 Multiply both sides of f(z) = z^n + 2*z^(n-1) + 3*z^(n-2) + ... + (n-1)*z^2 + n*z + n+1 by z-1 to get (z-1)*f(z) = z^(n+1) + z^n + ... + z - (n+1). If z is a root of f(z), then f(z)=0 which implies z^(n+1) + z^n + ... + z = n+1. Lemma proved. We will prove the main claim in the problem by showing that |z| cannot be less than 1 or |z| cannot equal 1. Case 1: if root z has 1>|z| Apply the Triangle Inequality to the Lemma. Then |z^(n+1)| + |z^n| + ... + |z| >= |n+1|. |n+1| is just n+1 but since |z| is less than 1 all the powers |z^m| are also less than 1 which makes the left side sum of inequalities a sum of n+1 positive elements all less than 1, which have a sum less than n+1. However the Triangle Inequality suggests the sum should be at least n+1. This is a contradiction, therefore |z| is not less than 1. Case 2, if root z has 1=|z| Express z in its polar form: z=e^(i*theta) for some theta; then substitute into the Lemma equaiton: e^((n+1)*i*theta) + e^(n*i*theat) + ... + e^(i*theta) = n+1 Take real parts of this equation, which then yields: cos((n+1)*theta) + cos(n*theta) + ... + cos(theta) = n+1. The maximum value of cos is 1, and in order for the equality to hold, all n+1 cos terms must equal 1, in particular cos(theta)=1. But cos(theta)=1 implies z can only be equal to 1. But by direct evaluation f(1) is not zero. Therefore |z| cannot equal 1. Since we have proven |z| cannot be less than 1 nor equal to 1 then |z| must be greater than 1 which is the goal. Proof complete.