## anonymous 4 years ago Draw the set S of complex numbers such that $|z-2i|<2,,, \Im (z)=1$ What is the argument when $z \in S$

1. anonymous

$|a-i|<2$

2. anonymous

But what does that say on a graph?

3. anonymous

-2<a-i<2

4. experimentX

says it lines in the region inside a circle with radius 2 and center 0,2

5. anonymous
6. anonymous

It's no circle?

7. anonymous

Is it because the Im(z) is constant at 1 and -2< Re(z) < 2, so it just varies in the Re(z) plane?

8. experimentX

so you want z such that it's imaginary part is constant?

9. anonymous

No i think I'm a bit wrong in my thoughts

10. experimentX

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

11. anonymous

Is it not Re betweem -2 and 2 ?

12. experimentX

sorry ... sqrt(3) .. lol

13. anonymous

Ah, okey :) So how do I get the argument from that?

14. anonymous

Just to try -sqrt3-i and sqrt3-i ?

15. anonymous

But they are not included in the interval so instead should i take some random x value inside the interval?

16. experimentX

yeah you can get that ... |x| must be sqrt(3)'s ...just put it up and see what you get.

17. anonymous

So i should use -i as Im, right?

18. anonymous

If so the argument is $\frac{ 11\pi }{ 6 }$

19. experimentX

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

20. anonymous

$z=x+\frac{ 11\pi }{6 }i,$ where $-\sqrt{3}<x<\sqrt{3}$

21. anonymous

Is that what you meant?

22. experimentX

no ... find the value of x in similar fashion

23. anonymous

Sorry but i have to go for now. But thank you for your help and patience, will look back at this later.