Simple inequality with complex numbers

Question

I can't solve inequalities with complex numbers. For example this one:

$|w-(\sqrt{2}-i)|<1$

The field of complex numbers isn't ordered, then is it possible to solve inequalities?

Answer 1

If you prefer to solve it algebraically:

Let $w=x+iy$

$|x+iy-(\sqrt{2}-i)|<1$

$|(x-\sqrt{2})+i(y-1)|<1$

$(x-\sqrt{2})^2 + (y-1)^2 < 1$

What can you say about this inequality then?

It is simply this:

Answer 2

Your inequality is an inequality between real numbers. The set of numbers that satisfy it is just the open disk centered at $\sqrt2-i$ and radius $1$.

Answer 3

The inequality represents all the complex numbers inside the circle centered at $\sqrt2 -1$ and radius R=1, boundary excluded since we have a strictly inequality.

Answer 4

Let $w=re^{i\theta}+\sqrt2-i$, which can cover the whole complex plane (so no solution lost). The locus condition becomes

$$|re^{i\theta}|<1$$ which is simply $r<1$.

Answer 5

Note $|z|$ is the modulus i.e. the length of the complex number and is real so the inequalities make sense.

In words that allows one to understand what is happening, your inequality is saying the distance of all complex numbers $w$ from $\sqrt{2}-i$ is less than $1$, i.e. all $w$ within the circular disc of radius 1 centred at $\sqrt{2}-i$.

Note the boundary of the disc is excluded, hence it is open.

In general $|z-\alpha| \lt \beta$ represents an open disc centered at $\alpha$ with radius $\beta$.