1wha!!!! i am getting -pi/3 donno where i am stuck.
13do not do tan-1-√3
check in which quadrant it is....
1answer is 2pi/3...confirm it frm some bhaiya...i may b wrong
1z = -2/(1+i√3) = -2(1-i√3)/4
= -1/2 +i√3/2
So arg(z)= 2Ï€/3
1
Kalyan Pilla
·2009-04-09 23:24:00
cosθ = -1/2
sinθ =√3/2
So, θ =2π/3
Or U could even use the quadrant like abhi said.
It is basically ω, and ω=ei2π/3
1357z=-2/(1+i√3)
arg(z)=arg(-2)-arg(1+i√3)
=Ï€-Ï€/3=2Ï€/3
1
JOHNCENA IS BACK
·2009-04-09 23:38:26
oh,got the mistake instead of taking it in 2nd quad. took it in fourth.
