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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.

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  • $\begingroup$ 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 $\endgroup$ – Noam Dolovich May 28 '16 at 11:11
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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?

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  • $\begingroup$ I still don't quite understand why the circles centre moves up to 2i.. Feels like I'm missing something obvious right now. Could you explain a bit more? $\endgroup$ – Xelak May 28 '16 at 11:26
  • $\begingroup$ 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? $\endgroup$ – 35T41 May 28 '16 at 11:30
  • $\begingroup$ Now it cleared up for me, thanks a lot! $\endgroup$ – Xelak May 28 '16 at 11:39
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Hint let $z=x+iy$ so equation on squaring becomes $x^2+(y-2)^2\leq 1$. Now do you understand why is it $+2i$

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  • $\begingroup$ I still don't quite understand.. $\endgroup$ – Xelak May 28 '16 at 11:29
  • $\begingroup$ If in real plane equation is $x-a)^2+y-b)^2=r^2$ then what is the centre $\endgroup$ – Archis Welankar May 28 '16 at 11:35
  • $\begingroup$ The centre then should be (0,2), thanks! $\endgroup$ – Xelak May 28 '16 at 11:41

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