Draw the set S of complex numbers such that \[|z-2i|<2,,, \Im (z)=1\] What is the argument when \[z \in S\]

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Draw the set S of complex numbers such that \[|z-2i|<2,,, \Im (z)=1\] What is the argument when \[z \in S\]

Mathematics
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\[|a-i|<2\]
But what does that say on a graph?
-2

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says it lines in the region inside a circle with radius 2 and center 0,2
http://www.wolframalpha.com/input/?i=|a-i|%3C2
It's no circle?
Is it because the Im(z) is constant at 1 and -2< Re(z) < 2, so it just varies in the Re(z) plane?
so you want z such that it's imaginary part is constant?
No i think I'm a bit wrong in my thoughts
put z = x + i in that equation |x + i - 2i| < 2 x^2 + 1^2 < 4 x^2 < 3 |x| < sqrt(3) -3 < x < 3
Is it not Re betweem -2 and 2 ?
sorry ... sqrt(3) .. lol
Ah, okey :) So how do I get the argument from that?
Just to try -sqrt3-i and sqrt3-i ?
But they are not included in the interval so instead should i take some random x value inside the interval?
yeah you can get that ... |x| must be sqrt(3)'s ...just put it up and see what you get.
So i should use -i as Im, right?
If so the argument is \[\frac{ 11\pi }{ 6 }\]
put z = x + i in that equation |x + i - 2i| < 2 x^2 + 1^2 < 4 x^2 < 3 |x| < sqrt(3) -sqrt(3) < x < sqrt(3) put z = x + 11pi/6 i <--- in above
\[z=x+\frac{ 11\pi }{6 }i, \] where \[-\sqrt{3}
Is that what you meant?
no ... find the value of x in similar fashion
Sorry but i have to go for now. But thank you for your help and patience, will look back at this later.

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