# Elementary proof that $\text{Re}\big(\frac{z^{n+1} - n z - z + n}{(z-1)^2}\big) \ge \frac{n}2$?

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 $$|z| = 1$$.

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

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!

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