# absolute value inequality with complex number

Strangely, I don't find easily on the internet sources about inequalities with complex numbers. In this moment, I am interested to absolute value inequalities with complex numbers but would be good having a wider perspective with all sorts of inequalities with complex numbers.

I have here an exercise which asks: Describe the set $C=\{z: |z−2i|≤ 2\}$ as a subset of the complex plane. Draw a picture.

It doesn't look hard but I have some questions. Usually, am I wrong if I say that we operate this way in R:

$-2 <= |z−2i|≤ 2$ then, we should isolate the variable within the absolute value - let's say that "-2" within the absolute value is a real constant - and we would end up with: $0 <= |x| <= 4$ which would be the interval solution for our inequality. This is because the absolute value theory in R states that we calculate the distance from the origin of the number line.

Anyhow, we are in C now. As I have recently learnt, we consider the magnitude of the complex number seen as a vector - which is quite intuitive and thus easy to learn. However, I don't see in this case the real part of the complex number - am I wrong if I say that we are dealing only with the imaginary part? As z ∈ C, so I am still not sure whether there is a real part or not...

Since this is the first time that I try to solve an inequality with a complex number and I don't find enough material on the internet, I ask here. This is my guess, however, based on logical thinking about inequalities: do I have consider as interval solution of the inequality all the values which don't make the magnitude of the vector zero or negative?

• $z-2i$ is in general a complex number, so you should not need to consider $-2$ in the inequality. Better to let $z=x+yi$ with $x$ and $y$ real, so you want $|x+(y-2)i|\le 2$ i.e. $\sqrt{x^2+(y-2)^2}\le 2$. This should help with sketching Oct 25, 2015 at 11:34
• x^2 would be z^2 (so, the coefficient 1 of z is the real part, right?) + (-2i)^2 (imaginary part given by the product of a real part by the imaginary number). Do I understand well? Oct 25, 2015 at 11:52
• No, perhaps I am wrong, looking at the picture of the answer of Zoli Oct 25, 2015 at 11:57

• @Alwayslearning: If $z=x+iy$ (corresponding to a vector $(x,y)$) then $z-2i=x+i(y-2)$ corresponds to the vector $(x,y-2)$ whose length is $\sqrt{x^2+(y-2)^2}$. Considering that the length of the corresponding vector equals the absolute value of the complex number we have $|z-2i|=\sqrt{x^2+(y-2)^2}$. So, the condition $|z-2i|\le2$ describes the inside and the edge of the circle $\sqrt{x^2+(y-2)^2}\le 2.$