frx
Draw the set S of complex numbers such that
\[z2i<2,,, \Im (z)=1\]
What is the argument when \[z \in S\]



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frx
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\[ai<2\]

frx
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But what does that say on a graph?

frx
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2<ai<2

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


frx
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It's no circle?

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

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

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

experimentX
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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

frx
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Is it not Re betweem 2 and 2 ?

experimentX
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sorry ... sqrt(3) .. lol

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

frx
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Just to try sqrt3i and sqrt3i ?

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

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

frx
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So i should use i as Im, right?

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

experimentX
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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

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

frx
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Is that what you meant?

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

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