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 $|z| = 1$.

  • $\begingroup$ 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


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!

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.