Complex Number Inequality $|2+z|\leq 2$

Question

If I have a complex number $z = a + ib$, how do you interpret the inequality

$|2+z|\leq 2$?

I believe the answer is a circle in the complex plane, but where I am getting confused is understanding the inequality due to the use of imaginary numbers on the left hand side and no imaginary numbers on the right hand side.

Thanks in advance.

Answer 1

It is a disc, centered at point $-2$ on the complex plane and with radius $2$. The non existence of imaginary numbers in the inequality is only natural: there is no "order" in $\mathbb{C}$, you cannot compare two complex numbers. No such relation is defined. If you want to get some more intuition on that stuff, try expressing complex numbers as numbers of $\mathbb{R}^2$. How would you write this equation for $z=x+iy$? simply as $\sqrt{(x+2)^2+y^2}\leq2$, which is precisely the object we described: a disc center at point $(-2,0)$ and of radius $2$.

Answer 2

Write this as $|z-(-2)| \leq 2$. In that case, this would mean that the distance of your point z from the point (-2,0) is always less than or equal to 2. Or, make the circle centered at (-2,0) radius 2, this gives you all points on the boundary and inside the circle.

Answer 3

$$|2+z|\leq 2 \iff |z-(-2)|\leq 2 \iff d(z, -2)\leq 2$$

That is the closed ball with center at $-2$ and radius $2$

Answer 4

I will ake for granted that you know the equation of the circle.

We know that $z=x+iy$, hence collecting real and imaginary parts together we have $$\left|[x-(-2)]+iy\right|\leq2$$ or

$$(x+2)^2+y^2\leq4$$

Clearly this represents a closed disk of radius $2$ with a centre $(-2,0)$.