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anonymous
 3 years ago
Draw the set S of complex numbers such that
\[z2i<2,,, \Im (z)=1\]
What is the argument when \[z \in S\]
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But what does that say on a graph?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1says it lines in the region inside a circle with radius 2 and center 0,2

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1so you want z such that it's imaginary part is constant?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No i think I'm a bit wrong in my thoughts

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1put z = x + i in that equation x + i  2i < 2 x^2 + 1^2 < 4 x^2 < 3 x < sqrt(3) 3 < x < 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it not Re betweem 2 and 2 ?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... sqrt(3) .. lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, okey :) So how do I get the argument from that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just to try sqrt3i and sqrt3i ?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So i should use i as Im, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If so the argument is \[\frac{ 11\pi }{ 6 }\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1put 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is that what you meant?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no ... find the value of x in similar fashion

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry 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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