Let $z\in \mathbb{C}, n \geq 2$. Show this complex inequality $$|z^n-1|^2\le |z-1|^2\left(1+|z|^2+\dfrac{2}{n-1}\Re{(z)}\right)^{n-1}$$

For $n=2$ the inequality is easy to prove: $$|z^2-1|^2\le |z-1|^2\left(1+|z|^2+2\Re{(z)}\right)$$ $$\Longleftrightarrow |z+1|^2\le 1+|z|^2+2\Re{(z)}$$ $$\Longleftrightarrow z+\overline{z}\le 2\Re{(z)}$$ which is in fact an equality, thus is exact.

for $n=3$. $$\Longleftrightarrow |z^2+z+1|^2\le (1+|z|^2+\Re{(z)})^2$$ $$\Longleftrightarrow(z^2+z+1)(\overline {z^2}+\overline{z}+1)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$ $$\Longleftrightarrow |z|^4+|z|^2+|z|^2(z+\overline{z})+1+(z+\overline{z})+(z^2+(\overline{z})^2)\le |z|^4+2|z|^2+1+\Re^2{z}+2\Re{(z)}+2|z|^2\cdot\Re{(z)}$$ $$\Longleftrightarrow |z|^2+(\Re{(z)})^2\ge z^2+(\overline{z})^2=(z+\overline{z})^2-2|z|^2$$ $$\Longleftrightarrow 3|z|^2\ge 3(\Re{(z)})^2$$ It is clear

Is it true for a general $n$?

  • 1
    $\begingroup$ ...the last two sentences are not sentences. $\endgroup$ Apr 28, 2016 at 7:22
  • $\begingroup$ $$|z+1|^2=(z+1)(z'+1)=|z|^2+(z+z')+1$$ $\endgroup$
    – math110
    Apr 28, 2016 at 7:27
  • $\begingroup$ I have tried to improve the presentation of your text using standard terms. Say if you agree. $\endgroup$
    – Jean Marie
    Apr 28, 2016 at 10:12

1 Answer 1


Firstly: $$\frac{z^n-1}{z-1}=z^{n-1}+z^{n-2}+\cdots1 \hspace{2cm}(1)$$

Let $z_k, \; k=0,\cdots,n-1$ be the roots of unity, i.e. the roots of $z^n-1=0$, and take $z_0=1$. Using Vieta's relations for the sum of the roots (https://en.wikipedia.org/wiki/Vieta's_formulas), one has $\sum\limits_{i=0}^{n-1} z_i=0$. Using that $z_0=1$, we get $\sum\limits_1^{n-1}z_i=-1$. With $z_k = \cos\theta_k+I\sin\theta_k$, and separating the real and imaginary parts, we get: $$\sum\limits_{k=1}^{n-1}\cos\theta_k=-1$$ $$\sum\limits_{k=1}^{n-1}\sin\theta_k=0$$

Equation $(1)$ can be written now as: $$\frac{z^n-1}{z-1}=(z-z_{n-1})(z-z_{n-2}\,)\cdots(z-z_1)$$, hence: $$\vert\frac{z^n-1}{z-1}\vert^2=\prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2$$

We use now the inequality between the geometric and arithmetic means (and $z=x+ I y$): $$\left( \prod\limits_{k=1}^{n-1}\vert z-z_k\vert^2 \right)^{\frac{1}{n-1}}\le\frac{\sum\limits_{k=1}^{n-1}\vert z-z_k\vert^2}{n-1}=\frac{\sum\limits_{k=1}^{n-1}(1+x^2+y^2-2x\cos\theta_k-2y\sin\theta_k)}{n-1}=1+x^2+y^2+2\frac{x}{n-1},$$ where we used the above-derived equalities for the sum of $\sin$ and $\cos$.

Putting it all together:

$$\vert\frac{z^n-1}{z-1}\vert^2 \le (1+\vert z\vert^2+2\frac{\Re(z)}{n-1})^{n-1}$$.

We can re-state the problem also as:

"For an arbitrary point in a plane, the squared product of distances to the vertices of any regular polygon inscribed in the unit circle (with $n$ vertices, that contains the point $(1,0)$, which is not to be included in the product), is at most equal to the power $n-1$ of the sum of the squared distance to the point $(-\frac{1}{n-1},0)$ and the constant $1-(\frac{1}{n-1})^2$ ".

The minimum of the right-hand side term reads: $\left[ 1-(\frac{1}{n-1})^2\right]^{n-1}$. Notably is the limit of it, for $n \to \infty$.

  • $\begingroup$ Unless I am overlooking something, $f(z)$ is not analytic. Neither the absolute value nor the real part are analytic functions. $\endgroup$
    – Martin R
    May 6, 2016 at 8:29
  • $\begingroup$ With your argument, the maximum of $1/(1 + |z|^2)$ in the unit disk would be attained at the boundary of the disk, which is not the case. $\endgroup$
    – Martin R
    May 6, 2016 at 8:34
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    $\begingroup$ @MartinR: I have completely changed the proof. Thank you for pointing out the flaws in the previous one. $\endgroup$
    – Chip
    May 9, 2016 at 2:05

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