1. That can be written
8*e^(i*pi/2) * (1/2)e^(i pi/6)
= 4 e^(2i*pi/3)
= 4 [cos(2*pi/3) + i sin (2*pi/3)]
= -2 + 2(sqrt3)i
2. The quotient is
= 16[cos(pi/3) + i sin(pi/3)]
= 8 + 8 sqrt3 i
Thank you very much for showing me the steps but they don't match my answers that are available: Could you possibly further explain-I'm messed up right now on this concept-
the answers for Number 1 are
2. answers for number two are:
cis Theta = cos Theta + i sin Theta
x = r cos Theta
y = r sin Theta
so for example
4 cis 2 pi/3 = 4 cos 2 pi/3 + 4 i sin 2pi/3
but cos 2p/i/3 = -.5 and sin 2pi/3 = sqrt 3/2
x = 4 (-.5) = -2
y = 4 (.5/ sqrt 3) = 2 sqrt3
-2 + 2 i sqrt 3
Your answer is the same as that of WLS
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