# challenging inequality with complex numbers

The statement of the problem: Let $$n \in \mathbb N$$ \ {0} and $$z_1,z_2, ... z_n \in \mathbb C$$. Prove that $$\sum_{i=1}^n |z_i||z-z_i| \ge \sum_{i=1}^n |z_i|^2$$ holds for any $$z \in \mathbb C$$ $$\iff$$ $$z_1+z_2+...z_n=0$$

My approach : I tried to prove the inequality using Cauchy's inequality and the modulus inequality, but I reached some ambiguous results, from which it does not appear that their sum would be 0. I think that the solution can also be a geometric one given the condition that z1 +z2+... zn = 0 it turns out that the vertices of the polygon with these affixes is regular, but I'm not sure.

Any and all proofs will be helpful. Thanks a lot!

• Use \ge to get $\ge$ in MathJax Feb 21 at 14:01
• The inequality holds for $n=1$, $z=0$, and $z_1=1$, while $\sum_{i=1}^nz_1=1\neq0$. Feb 21 at 14:26
• @JohnBentin: As I understand it, the task is to show that $$\left( \forall z \in \Bbb C: \sum_{i=1}^n |z_i||z-z_i| \ge \sum_{i=1}^n |z_i|^2 \right) \iff z_1+z_2+\cdots + z_n=0$$ Feb 21 at 14:30
• OK, I get it. Put that way, I can't misread it. Feb 21 at 19:50

## 3 Answers

Here's a proof using Triangle Inequality and Cauchy-Schwarz

(i.) when $$\sum_{k=1}^n z_k = 0$$
$$\sum_{k=1}^n \vert z_k\vert^2= \Big \vert \sum_{k=1}^n \overline z_k \cdot z_k - \overline z_k\cdot z\Big\vert = \Big \vert \sum_{k=1}^n \overline z_k (z_k-z)\Big\vert$$
$$\leq \sum_{k=1}^n \Big \vert \overline z_k (z_k-z)\Big\vert=\sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert$$
by triangle inequality

(ii.) now suppose the inequality remains true when $$\sum_{k=1}^n z_k =\lambda\neq 0$$, then Cauchy-Schwarz tells us
$$\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert\leq \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)^\frac{1}{2} \cdot \Big(\sum_{k=1}^n \Big \vert z-z_k\Big\vert^2\Big)^\frac{1}{2}$$
$$\implies\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big(z -\overline z_k\big)\overline{\Big(z - \overline z_k\Big)}= n \vert z\vert^2 + \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)- 2\cdot \text{Re}\Big(z \sum_{k=1}^n z_k\Big)$$
$$\implies 2\cdot \text{Re}\Big(z\cdot \lambda\Big)\leq n \vert z\vert^2$$
which is obviously false, e.g. set $$z:=\frac{\overline \lambda}{2n}$$

• Thank you so much ! Feb 22 at 7:57
• @LastX I think the last part of the derivation is in error so that the last line should be $2\Re(\bar z\lambda)\le n|z|^2$. This has however no impact on the conclusion.
– user
Feb 22 at 11:21
• Since the inequality holds for all $z \in \mathbb C$ then $\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z-z_k\Big\vert^2\iff \sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert \overline z- z_k\Big\vert^2\iff \sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z-\overline z_k\Big\vert^2$ so there is nothing inacurate in what I wrote. Sometimes the scaffolding used to get to the solution persists into the answer I post. Feb 22 at 17:20

For convenience, we will use engineering notation $$z^*$$ for the complex conjugate of $$z$$ and statistical notation $$\bar z$$ for the arithmetic mean of the $$z_i\;$$ ($$i=1,...,n$$).

First, let us assume $$\sum_{i=1}^n |z_i||z-z_i|\geqslant \sum_{i=1}^n|z_i|^2 \quad\text{for all}\;z\in\Bbb C.\qquad(1)$$ We have \begin{align}0&\leqslant\sum_{i=1}^n(|z_i|-|z-z_i|)^2\\ &=\sum_{i=1}^n(|z_i|^2-2|z-z_i||z_i|+|z-z_i|^2)\\ &\leqslant \sum_{i=1}^n(|z_i|^2-2|z_i|^2+|z-z_i|^2)\quad\text{(by ineq. 1)}\qquad(*)\\ &=\sum_{i=1}^n(|z-z_i|^2-|z_i|^2)\\ &=\sum_{i=1}^n[(z-z_i)(z^*-z_i^*)-z_iz_i^*]\\ &=\sum_{i=1}^n(zz^*-z_iz^*-zz_i^*)\\ &=n(x^2+y^2)-2x\sum_{i=1}^nx_i-2y\sum_{i=1}^ny_i\quad\text{(where z=x+\mathrm iy with x,y\in\Bbb R, etc.)}\\ &=n[(x-\bar x)^2+(y-\bar y)^2-\bar x^2-\bar y^2].\qquad(**) \end{align} Overall, we may conclude that $$0\leqslant (x-\bar x)^2+(y-\bar y)^2-\bar x^2-\bar y^2$$ for all real $$x$$ and $$y$$. In particular, this is the case when $$x=\frac12\bar x$$ and $$y=\frac12\bar y\,$$: namely, $$0\leqslant-\frac34(\bar x^2+\bar y^2)$$. It follows that $$\bar x=\bar y=0$$.

To prove the converse implication, we will obtain a contradiction by assuming, contrary to inequality $$1$$, that there is $$z_0=x_0+\mathrm iy_0$$, with $$x_0,y_0\in \Bbb R$$, such that $$\sum_{i=1}^n|z_i||z_0-z_i|<\sum_{i=1}^n|z_i|^2\qquad(2)$$ while also $$\sum_{i=1}^nz_i=0$$. We use the same algebraic manipulations as above, but work backwards from line $$**$$ with $$\bar x=\bar y=0$$, replacing $$z$$ by $$z_0$$, and replacing $$\leqslant$$ by $$>$$ (from the contrary inequality $$2)$$ in line $$*$$. Consequently, we get

$$\sum_{i=1}^n(|z_i|-|z_0-z_i|)^2>\sum_{i=1}^n|z_0|^2.$$ Now write this as $$\sum_{i=1}^n[(|z_i|-|z_0-z_i|)^2-|z_0|^2]>0$$ and factorize to get $$\sum_{i=1}^n(|z_i|-|z_0-z_i|-|z_0|)(||z_i|-|z_0-z_i|+|z_0|)>0.$$ However, by the triangle inequality, the first factor of each term of the above sum is non-positive while the second factor is non-negative. So the sum is non-positive: the required contradiction.

You could use the derivative mechanism with $$z$$ and $$z^*$$ (the complex conjugate) treated as independent variables. But if you aren't familiar with it, first study it to convince yourself that this is a consistent procedure for functions that are analytical in $$z$$ and in $$z^*$$, otherwise it would be black magic... We first define, for convenience: $$X = \sum_{i=1}^n|z_i|\ |z-z_i| - \sum_{i=1}^n|z_i|^2 \geq 0$$ We observe that we have $$X=0$$ in the origin ($$z=0$$). We now need to make the expression for $$X$$ analytical in $$z$$ and in $$z^*$$, like this: $$X = \sum_{i=1}^n\left(|z_i|\ \sqrt{(z-z_i)(z^*-z_i^*)} - |z_i|^2\right) \geq 0$$ Note that we do not care about the $$z_i$$, they can remain in the non-analytical |z_i|, we are only going to differentiate w.r.t. $$z$$ and $$z^*$$, giving: $$\frac{dX}{dz^*} = \frac{d}{dz^*} \sum_{i=1}^n|z_i|\sqrt{(z-z_i)(z^*-z_i^*)} = \sum_{i=1}^n\frac{|z_i|\ (z-z_i)/2}{\sqrt{(z-z_i)(z^*-z_i^*)}}$$ $$\mbox{ and similarly} \ \ \ \ \frac{dX}{dz} = \sum_{i=1}^n\frac{|z_i|\ (z^*-z_i^*)/2}{\sqrt{(z-z_i)(z^*-z_i^*)}},$$ and then we look at the result in the origin where $$z=z^*=0$$, giving: $$\left. \frac{dX}{dz^*}\right|_{z=0} = \sum_{i=1}^n\frac{|z_i|\ (-z_i)/2}{\sqrt{z_i z_i^*}} = -\frac{1}{2}\sum_{i=1}^n\ z_i, \ \ \ \ \mbox{and} \ \ \ \ \left. \frac{dX}{dz}\right|_{z=0} = -\frac{1}{2}\sum_{i=1}^n\ z_i^*$$ Now the finishing touch: because $$X=0$$ in the origin and has to be non-negative in any region around the origin it cannot have a finite first-order derivative! This means that the $$\sum_{i=1}^n\ z_i$$ that we encountered in the derivatives must be zero, $$\frac{1}{2}$$QED.

NB: without the originally given restriction, the expansion of $$X$$ around the origin would have been: $$X = c + \alpha z + \alpha^* z^* + \beta z^2 + \beta^* (z^*)^2 + \gamma z z^* + \mbox{higher orders},$$ but in this case we saw that $$c=0$$ and our given inequality then requires the first derivatives to be zero: $$\alpha=\alpha^*=0$$. With some further reasoning you can even find that $$\beta$$ should be $$0$$. We still might have $$\gamma\neq 0$$ (it should then be positive!)

• I haven't learned about derivatives yet, but thanks for the help! Do you know of any other solution to this problem that does not include derivatives ? Feb 21 at 14:54
• Not immediately... (but instead of those complex $z$ and $z^*$ derivatives you could rewrite the whole problem in terms of real 2-vectors with dot products and then use derivatives wrt. the real vector components). Feb 21 at 15:11
• what means z* ? Feb 21 at 15:21
• Complex conjugate! (And for the derivative wrt. $z^*$, if you want to study it look on the web, e.g. here: physicsforums.com/threads/… If you want to go to 2-vectors instead, the derivative will just be wrt. the x and y-components so then it doesn't matter. But it is clumsy to first have to translate to vectors and then doing every derivative for two variables instead of one.) Feb 21 at 15:33
• Why did you replace the inequality with equality?
– user
Feb 21 at 16:05