Plot the numbers z in the complex plane that fufill $|z−2i|≤1$ and $\text{Im}(z){\ge}2$

I have recently started learning more about complex numbers and stumbled upon this problem:

Plot the numbers $z$ in the complex plane that fulfil $|z-2i| ≤ 1$ and $\text{Im} (z) ≥ 2$

I know that $\text{Im}(z) ≥ 2$ means that all imaginary numbers who are positioned somewhere on the horizontal line $\text{Im} = 2$ or above the line can be plotted. However I’m not quite sure what $|z-2i| ≤ 1$ means. I understand that if I only had $|z| ≤ 1$ that would mean I can plot all numbers inside a circle with the radius $1$ including the ones on the circles circumference. But I don’t really understand what the term $-2i$ does to the circle. I compared it to other similar questions and came to the conclusion that the centre should be at $2i$ instead of $0$. My problem is that I don’t quite understand why the circle gets a centre at $2i$ and not at $-2i$. Thankful for any help and explanation.

• you could let z=x+yi and and use the definition of the absolute value of a complex number to find the equation of the circle – Noam Dolovich May 28 '16 at 11:11

Let's write z=x+iy and then the inequality $|z-2i|\leq1$ becomes $|x+(y-2)i|\leq1$ so we get $x^2+(y-2)^2 \leq 1$. Can you see it now?
• The center of the circle $(x-a)^2 + (y-b)^2=R^2$ is (a, b). In this case a=0, b=2. Is that helpful? – 35T41 May 28 '16 at 11:30
Hint let $z=x+iy$ so equation on squaring becomes $x^2+(y-2)^2\leq 1$. Now do you understand why is it $+2i$
• If in real plane equation is $x-a)^2+y-b)^2=r^2$ then what is the centre – Archis Welankar May 28 '16 at 11:35