Question

Draw the set S of complex numbers such that

\[|z-2i|<2,,, \Im (z)=1\]

What is the argument when \[z \in S\]

Answer 1

\[|a-i|<2\]

Answer 2

But what does that say on a graph?

Answer 3

-2<a-i<2

Answer 4

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

Answer 5

http://www.wolframalpha.com/input/?i= |a-i|%3C2

Answer 6

It's no circle?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Is it not Re betweem -2 and 2 ?

Answer 12

sorry ... sqrt(3) .. lol

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

So i should use -i as Im, right?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Is that what you meant?

Answer 22

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

Answer 23

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