Question #976bb

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Answer 1 · 2017-03-14T07:40:49

$(i) : 4 c i s 120^{\circ} = - 2 + i 2 \sqrt{3}; (i i) : 9 c i s (3 \frac{π}{2}) = - 9 i .$ $(i) : 4cis120^@=-2+i2sqrt3; (ii) : 9cis(3pi/2)=-9i.$

Recall that, $r c i s θ = r (cos θ + i sin θ) .$ $rcis theta=r(cos theta +i sin theta).$

$(i) 4 c i s 120^{\circ} = 4 ({cos 120}^{\circ} + i {sin 120}^{\circ})$ $(i) 4cis 120^@=4(cos 120^@ + i sin 120^@)$

$= 4 {cos (90^{\circ} + 30^{\circ}) + i sin (90^{\circ} + 30^{\circ})}$ $=4{cos (90^@+30^@)+isin(90^@+30^@)}$

$= 4 {- {sin 30}^{\circ} + i {cos 30}^{\circ}}$ $=4{-sin30^@+icos30^@}$

$= 4 {- \frac{1}{2} + i \frac{\sqrt{3}}{2}}$ $=4{-1/2+isqrt3/2}$

$:. 4cis 120^@=-2+i2sqrt3.$

Similarly, $(ii) : 9cis(3pi/2)=-9i.$

Enjoy Maths.!