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?

  • $\begingroup$ $z-2i$ 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 $\endgroup$
    – Henry
    Oct 25, 2015 at 11:34
  • $\begingroup$ 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? $\endgroup$ Oct 25, 2015 at 11:52
  • $\begingroup$ No, perhaps I am wrong, looking at the picture of the answer of Zoli $\endgroup$ Oct 25, 2015 at 11:57

1 Answer 1


To Henry's comment here is the drawing:

enter image description here

  • $\begingroup$ but do we take the magnitude of the complex number or we don't? Looking at the picture, I understand that there is no real part or, better to say, real part = 0 but shouldn't -2i lies in the lower quadrant? What are exactly the steps I have to follow for computation and drawing? $\endgroup$ Oct 25, 2015 at 12:01
  • 1
    $\begingroup$ @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 $\sqrt{x^2+(y-2)^2}\le 2.$ $\endgroup$
    – zoli
    Oct 25, 2015 at 12:08
  • $\begingroup$ I am not sure I understand. Should I write exactly as you and Henry wrote? Namely, the the modulus of the vector - complex number with x and y?...No, I think you wrote like this for showing me the idea... the modulus would be written as square root of z^2 + (-2i)^ and, I as have understood, this modulus is lesser or equal than 2. Now I am wondering what instructions do I have to follow in order to depict it - I know you put the figure but I want to understand the thoughts/steps behind it. 1. Why is it a circle, for instance? How to establish where to collocate it on the plane? $\endgroup$ Oct 26, 2015 at 14:59
  • $\begingroup$ Ok. Maybe I have got it. I don't need to find any modulus, actually, because the inequality tells me that the modulus is lesser or equal than 2 (?) And for about collocation on the plane...I have still to think a bit @zoli $\endgroup$ Oct 26, 2015 at 15:13
  • $\begingroup$ Maybe I have to look up at graphs of functions. I know that the format x^2+y^2=1 stands for a circle but I did not know that also with the radical sign above is still a circle and thought that, in order to have a circle, we had to have also the square of the constant...but playing with graphing calculator now, I see that it isn't as I thought $\endgroup$ Oct 26, 2015 at 15:23

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