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I would like an elementary proof that the real part of $$ f(z) = \frac{z^{n+1} - n z - z + n}{(z-1)^2} $$ is greater than or equal to $n/2$ for any $z \in \mathbb C$, $|z| \le 1$, where $n$ is a natural number. I can prove this for $|z| = 1$, and then deduce it is true for $|z| \le 1$ using the maximum principle. But I am looking for a direct proof.

Incidentally $$ \text{Re}(f(e^{i\theta})) = \frac n2 + \frac{\sin^2(n \theta/2)}{2\sin^2(\theta/2)} ,$$ which establishes the result if $|z| = 1$.

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  • $\begingroup$ Is $n$ confined to the positive integers? $\endgroup$ May 23 at 2:51
  • $\begingroup$ @user2661923 Yes it is. I'll add that assumption. $\endgroup$ May 23 at 4:07

1 Answer 1

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This works by direct computation:

$h(z)=f(z)-\frac{n}{2}=\frac{z^{n+1} - n z - z + n}{(z-1)^2}-\frac{n}{2}=\frac{n-z-z^2-..-z^n}{1-z}-\frac{n}{2}$ so the sign of $\Re h(z)$ is the sign of $$\Re [2(1-\bar z)(n-z-z^2-..-z^n)-n(1-z)(1-\bar z)]$$ and with $r=|z| \le 1$ one has

$$\Re [2(1-\bar z)(n-z-z^2-..-z^n)-n(1-z)(1-\bar z)]$$ $$=n-2(1-r^2)\Re(z+..+z^{n-1})-2\Re z^n-(n-2)r^2$$

and majorizing all the negative terms by their (negative) absolute values we need to prove that

$g(r)=n(1-r^2)+2r^2-2(1-r^2)(r+...+r^{n-1})-2r^n >=0$ for $0 \le r \le 1$

But $g(r)=n(1-r^2)-2r+2r^{n+1}=[n(1+r)-2r(1+..r^{n-1})](1-r)$

so $g(1)=0$ and for $0 \le r <1$ one has $$\frac{g(r)}{1-r}=(n-r-..-r^n)+(nr-r-..-r^n) >0$$

and we are done!

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  • $\begingroup$ Thank you very much. Somehow a check mark and an upvote doesn't seem enough. $\endgroup$ May 23 at 5:33
  • $\begingroup$ Happy to be of help $\endgroup$
    – Conrad
    May 23 at 12:56

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