ত্রিভুজ অসমতা

There really aren't too many inequalities involving complex numbers. The reason for this is that we don't even have a basic 'ordering' for complex numbers. As a result, almost all inequalities involving complex numbers will actually be inequalities involving the magnitudes. In this post, we will discuss the "trivial" inequalities, the triangle inequalities and applications, and a bit more. We'll start from very basic inequalities, but hopefully the reader has the patience to continue.

In the following, $\Re (z), \Im (z)$ will denote the real and imaginary parts of a complex number, such that $z = \Re (z) + \Im (z) i$. Recall that for any complex number $z$, we have $$z \cdot \bar{z} = |z|^2 = \Re (z)^2 + \Im (z)^2$$

Now we can prove what some people call the "trivial inequalities" for complex numbers:

1: $|z| \ge 0$.

This is true by definition so I will omit the proof.

2: $\Re (z), \Im (z) \le |z|$.

Recall the identity $|z|^2 =\Re (z)^2 + \Im (z)^2$.

Now use $\Im (z)^2 \ge 0$ to obtain $|z|^2 = \Re (z)^2 + \Im (z)^2 \ge \Re (z)^2$.

Since $|z|, \Re (z) \ge 0$ we can take square roots to obtain $|z| \ge \Re (z)$. $|z| \ge \Im (z)$ follows similarly.

Now that we have the trivial inequalities out of the way, we can move on to the very useful triangle inequalities.

The relationship between complex numbers and vectors is very important. But essentially, suppose we have a triangle of the complex plane with vertices at $0,x,x+y$.

Then the triangle inequality tells us that the side connecting $0$ and $x$ and the side connecting $x$ and $x+y$ sum to a length greater than the side length connecting $0$ to $x+y$.

This translates to the first version of the triangle inequality- $|x|+|y| \ge |x+y|$ for complex $x,y$.

Now, I will state the second version of the triangle inequality- $|x|-|y| \le |x-y|$.

These are the two versions of the triangle inequality. I don't want to go into too much detail, but this has lots of applications especially in polynomial root-bounding: Often we want to show, given a condition on the coefficients of a polynomial, that all its roots lie within a certain range (such as they are all within the unit circle).

You can read more about these applications here in this post about polynomial irreducibility

I will just show one quick application of the triangle inequality- it is quite tricky and involves a trick that experienced students or those who have taken calculus may find familiar.

Problem Let $w,x,y,z$ be complex numbers inside the unit circle. Show that $|wx-yz| \le |w-y|+|x-z|$.

Indeed, the inequality is not homogeneous in $w,x,y,z$. We can guess that this is because we are given the condition $|w|,|x|,|y|,|z|\le 1$. But we can also guess that some kind of homogenization will solve the inequality.

In particular, we have inequalities like $|w-y| \ge | w(w-y)|, |x (w-y)|, |y(w-y)|, |z(w-y)|$ and so on. We can try using these and see if they allow for an application of the triangle inequality.

Indeed, this works magically! We have $$|w-y|+|x-z| \ge |wx - xy| + |xy-yz| \ge |wx-yz|$$ by the triangle inequality. This concludes the proof.

I should mention one last thing regarding the triangle inequality- we can prove it algebraically. The proof is not very interesting but a good exercise.

First we can try squaring to deal with absolute values more easily- We want to show $$|x|^2 +2|x||y| + |y|^2 \ge |x+y|^2$$ But from $|z|^2 = z\bar{z}$ this is equivalent to $$x\bar{x}+2|x||y|+y\bar{y} \ge (x+y)\bar{ x+y} \\= (x+y)(\bar{x} +\bar{y}) =x\bar{x} +y\bar{y} + x\bar{y}+y\bar{x}$$ Thus it suffices to show $2|x||y| \ge x\bar{y} + y\bar{x}$.

Now recall the identity $2\Re (z) = z + \bar{z}$. Since $\bar{ x\bar{y}} = y\bar{x}$ we know $$x\bar{y}+y\bar{x} = 2 \Re (x\bar{y})$$

Thus we want to show $2|x||y| \ge 2\Re (x \bar{y} )$ but this is just the trivial inequality because $| x\bar{y}| =|xy|=|x||y|$.

This concludes our discussion of all the "standard" inequalities. But there are still some very strange ones! The field of complex inequalities is endless.

But I would like to mention something that perhaps will be more enlightening. It seems that any inequality in the reals can be replaced by some kind of complex number equivalent if we put enough magnitude signs in the right places. This behavior generalizes- the magnitude operator is an example of a more general absolute value of integral domains, where our integral domain here is the field $\mathbb C$.

You can read more about this at wikipedia

Although very simple, the triangle inequality can sometimes be very useful when working with vectors. It states

If $x$ and $y$ are vectors in $\mathbb{R}^n$, then $$|x + y| \le |x| + |y|.$$

This can be proven by noting that $|x + y|^2 = (x+y)\cdot (x+y)$. Expanding the right side gives $$|x+y|^2 = |x|^2 + 2x\cdot y + |y|^2.$$ Now we use a very important inequality which states that $x\cdot y \le |x||y|$, also known as the Cauchy Schwarz Inequality. This is pretty obvious when you use the geometric definition of a dot product. We now have $$|x + y|^2 \le (|x| + |y|)^2 \implies |x+y|\le |x| + |y|.$$