Question #976bb

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Answer 1 · 2017-03-14T07:40:49

$(i) : 4cis120^@=-2+i2sqrt3; (ii) : 9cis(3pi/2)=-9i.$

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

$(i) 4cis 120^@=4(cos 120^@ + i sin 120^@)$

$=4{cos (90^@+30^@)+isin(90^@+30^@)}$

$=4{-sin30^@+icos30^@}$

$=4{-1/2+isqrt3/2}$

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

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

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