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$1/2 (cos((2pi)/3)+i sin((2pi)/3)) * 4(cos(pi/2)+isin(pi/2)) $

Use the formula $r_1 (cos⁡theta_1+i sin⁡ theta_1 )∙r_2 (cos⁡θ_2+i sinθ_2 )= r_1 r_2 [cos⁡(θ_1+θ_2 )+isin(θ_1+θ_2 )]$

$= 1/2 *4 [ cos (((2pi)/3)+pi/2) +i sin (((2pi)/3)+pi/2)]$

$=2[cos ((7pi)/6) + i sin ((7pi)/6)] $

$=2[-sqrt3/2+i(-1/2)]$

$=-sqrt3-i$

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