# Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$

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.

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@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

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.

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+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

It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.
$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.
Now use $e^x \geq 1 + x$.
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