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

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frxBest ResponseYou've already chosen the best response.0
But what does that say on a graph?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
says it lines in the region inside a circle with radius 2 and center 0,2
 one year ago

frxBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=ai%3C2
 one year ago

frxBest ResponseYou've already chosen the best response.0
Is it because the Im(z) is constant at 1 and 2< Re(z) < 2, so it just varies in the Re(z) plane?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
so you want z such that it's imaginary part is constant?
 one year ago

frxBest ResponseYou've already chosen the best response.0
No i think I'm a bit wrong in my thoughts
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

frxBest ResponseYou've already chosen the best response.0
Is it not Re betweem 2 and 2 ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry ... sqrt(3) .. lol
 one year ago

frxBest ResponseYou've already chosen the best response.0
Ah, okey :) So how do I get the argument from that?
 one year ago

frxBest ResponseYou've already chosen the best response.0
Just to try sqrt3i and sqrt3i ?
 one year ago

frxBest ResponseYou've already chosen the best response.0
But they are not included in the interval so instead should i take some random x value inside the interval?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah you can get that ... x must be sqrt(3)'s ...just put it up and see what you get.
 one year ago

frxBest ResponseYou've already chosen the best response.0
So i should use i as Im, right?
 one year ago

frxBest ResponseYou've already chosen the best response.0
If so the argument is \[\frac{ 11\pi }{ 6 }\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

frxBest ResponseYou've already chosen the best response.0
\[z=x+\frac{ 11\pi }{6 }i, \] where \[\sqrt{3}<x<\sqrt{3}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no ... find the value of x in similar fashion
 one year ago

frxBest ResponseYou've already chosen the best response.0
Sorry but i have to go for now. But thank you for your help and patience, will look back at this later.
 one year ago
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