Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $z_1, z_2 \ldots z_m$ be complex numbers, $m \in \mathbb{N}$. Can anybody tell me how to prove the following inequality?

$| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$

In case you're wondering, this is asserted without proof in a paper by Von Neumann, about infinite tensor products of Hilbert spaces.

share|cite|improve this question
@Moron: when upgrading titles with TeX it is better to retain searchable or recognizable words. I restored "complex" and "inequality". – T.. Oct 28 '10 at 17:40
up vote 16 down vote accepted

The inequality in question bounds how far you can get from $1$ by multiplying several complex numbers that may individually not be far from $1$. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction.

Lemma: Suppose $\lvert z_1 - 1\rvert = \alpha_1$ and $\lvert z_2 - 1\rvert = \alpha_2$. Then $\lvert z_1 z_2 - 1\rvert \le (1 + \alpha_1)(1 + \alpha_2) - 1$.

Proof: We know $\alpha_1\alpha_2 = \lvert z_1z_2 - z_1 - z_2 + 1\rvert$. By triangle inequality on the three points $z_1 z_2$, $z_1 + z_2 - 1$, and $1$, we have $$\begin{align} \lvert z_1 z_2 - 1\rvert &\le \lvert z_1 z_2 - z_1 - z_2 + 1\rvert + \lvert z_1 + z_2 - 2\rvert \\ &\le \alpha_1\alpha_2 + \alpha_1 + \alpha_2 \\ &= (1 + \alpha_1)(1 + \alpha_2) - 1. \end{align}$$

Now, for several numbers, $$\begin{align} \lvert z_1 z_2 \cdots z_m - 1\rvert &\le (1 + \alpha_1)(1 + \alpha_{2,\ldots,m}) - 1 \\ &\le (1 + \alpha_1)(1 + \alpha_2)(1 + \alpha_{3,\ldots,m}) - 1 \\ &\vdots \\ &\le (1 + \alpha_1)(1 + \alpha_2)\cdots(1 + \alpha_m) - 1, \end{align}$$ where $\alpha_{2,\ldots,m}$, for example, is my hopefully transparent abuse of notation to denote $\lvert z_2\cdots z_m - 1\rvert.$ Finally, since $1+x \le e^x$ for real $x$, the desired inequality follows.

share|cite|improve this answer
+1: This is what I was about to add! – Aryabhata Oct 28 '10 at 4:04
Very nice, thanks! – Rotwang Oct 28 '10 at 4:11
@Moron: Is this some sort of standard, well-known proof? It feels like a quite general and useful result, that it took me way more work to derive from scratch than it ought to have. (In my first attempt I weakened the inequality too much in the lemma, and couldn't get anywhere.) – Rahul Oct 28 '10 at 4:11
Oops, I meant to type that took me without the it in between. – Rahul Oct 28 '10 at 4:25
I have a feeling it is pretty standard (also supported by the paper stating it as a fact), but I am unable to place where I might have come across it. I had to derive it myself too, though. – Aryabhata Oct 28 '10 at 4:44

EDIT: This answer is wrong.

It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.

Given this inequality, prove by induction that

$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.

Now use $e^x \geq 1 + x$.

share|cite|improve this answer
the first inequality is not true: take $x=3$ and $y=2$ – anonymous Oct 28 '10 at 3:39
No, I've already tried that but I don't think it works. Take e.g. $x = y = 2$. – Rotwang Oct 28 '10 at 3:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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