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

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.

• $|2+z|$ is a real number – Mirko Apr 26 '18 at 22:10
• $|z-(-2)|$ is the distance of the complex number $z$ from the complex number $-2+0i$. – John Wayland Bales Apr 26 '18 at 22:12
• The absolute value of an imaginary number is a real number. BTW complex numbers don't have order so you can only use $\le$ with real numbers. – fleablood Apr 26 '18 at 22:20

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 $2$.

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

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

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 $(-2,0)$.

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.

• You meant less than or equal to $2$. – N. F. Taussig Apr 26 '18 at 22:14