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mathslover
 2 years ago
Convert into polar form :
(1/2) + i ( \(\frac{\sqrt{3}}{2}\) )
mathslover
 2 years ago
Convert into polar form : (1/2) + i ( \(\frac{\sqrt{3}}{2}\) )

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mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0(1/2) + i ( \(\frac{\sqrt{3}}{2}\) )

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0I got : cos ( 4 pi/3 ) + sin ( 4 pi/3)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0But the book says : cos (  2pi/3 ) + i sin (  2pi/3). Which is also right...

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry typing mistake I meant i sin 4pi/3

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0^ cos ( 4 pi/3 ) + i sin(4pi/3)

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1Find the absolute value of this: \[\left \frac{ 1 }{ 2 } \frac{ i \sqrt{3} }{ 2 }\right\]

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0So which should I prefer more ? 2pi/3 or 4 pi/3

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1So \[=\sqrt{(\frac{ 1 }{ 2 })^2+(\frac{ \sqrt{3} }{ 2 })^2}\]

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1Drawing it gives you a way better picture of what the angle is.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large{\sqrt{ (\frac{1}{2} )^2 + ( \frac { \sqrt{3}}{2})^2}}\] right so it is : \(\1\)

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1You took the other angle

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1You go backwards meaning it will be negative.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0By the diagram I think that the angle theta will surely lie in the III quadrant i.e. 4 pi /3

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.1If it's in the third of fourth quadrant go backwards.
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