# refinement of basic triangle inequality

Let $z_1,z_2$ be two complex numbers. It is well known $|z_1+z_2|\le |z_1|+|z_2|$. Is there any refinement of this basic inequality?

Modified.

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This already fails remarkably for the positive reals, so how can you hope to find such a constant $c$? – t.b. Jun 24 '11 at 0:17
It is always a good idea to try some examples before making a conjecture. Do some simulations if the situation is complicated, which is not the case here. – SteveH Jun 24 '11 at 0:35
(-1) because the question has been changed to expel the answers from their original context. – The Chaz 2.0 Jun 24 '11 at 1:38

As you have already seen from the previous answers of joriki and Isaac, there cannot be a naive refinement. As an inequality on the complex numbers, the statement of the triangle inequality is sharp, meaning that you can find pairs $z_1, z_2$ such that $|z_1 + z_2| = |z_1| + |z_2|$, and so you cannot refine the inequality by changing constants.

However, there is at least one way of refining the triangle inequality. But this requires taking in "additional" information. And I am sure you have learned this refinement already: it is called the Law of Cosines. Written relative the the complex numbers

$$z_1 = r_1 \exp i\theta_1 \qquad z_2 = r_2 \exp i\theta_2$$

in polar representation, this says that (I'll first write it in the subtractive form so it is intuitively clearer)

$$|z_1 - z_2|^2 = r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\theta_1 - \theta_2)$$

Now replace $z_2 \to -z_2$ ($\theta_2 \to \theta_2 + \pi$) we have the additive form

$$|z_1 + z_2|^2 = r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\theta_1 - \theta_2 + \pi)$$

Since shifting cosine by $\pi$ amounts to multiplying it by -1, we get

$$|z_1 + z_2|^2 = (r_1 + r_2)^2 + 2 r_1 r_2 \left[\cos(\theta_1 - \theta_2) - 1\right]$$

The classical triangle inequality uses the fact that the term inside the square brackets is non-positive, so we can omit it and take square roots on both sides and re-obtain the triangle inequality. Now, depending on the size of $\theta_1 - \theta_2$, one can obtain more precise estimates of the size of $|z_1 + z_2|^2$ using different polynomial expressions in the angle difference. One may think that this is somewhat silly thing to do, since we already have the exact expression available. But it turns out for many purposes, polynomials are a lot cleaner to handle than trigonometric polynomials. As an example, these types of refinements are very important when studying resonances in nonlinear partial differential equations.

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+1 for your nice answer. Can this kind of refinement be generalized to $n$ complex numbers? – Sunni Jun 24 '11 at 2:34

There can't be such a refinement since equality holds in the triangle inequality whenever $z_1$ and $z_2$ differ only by a positive real factor, whereas $|z_1-z_2|$ isn't $0$ (unless the factor is $1$).

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Let $z_1=2$ and $z_2=1$ so that $$|z_1+z_2|=|2+1|=3$$ and $$|z_1|+|z_2|-c|z_1-z_2|=|2|+|1|-c|2-1|=3-c.$$

In order for $|z_1+z_2|\le |z_1|+|z_2|-c|z_1-z_2|$ to hold, $$3\le 3-c,$$ so $$c\le0,$$ which contradicts your premise that $c$ is positive. So there cannot be such a refinement.

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Related to Willie Wong's answer and others, although the inequality cannot be strengthened, it is possible to specify exactly when the equality holds and exactly when you will have a strict inequality.

Specifically, for real numbers we have:

For all real numbers $a$ and $b$, $|a+b|\leq |a|+|b|$, and equality holds if and only if $a$ and $b$ have the same sign (that is, $ab\geq 0$).

For complex numbers, one likewise has:

For complex numbers $z_1$ and $z_2$, $|z_1+z_2|\leq |z_1|+|z_2|$, with equality if and only if $z_1$ and $z_2$ have the same argument.

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Thanks for your comment. – Sunni Jun 24 '11 at 12:10