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Doctors Medicines MCQs Q&A

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Note that:

$e^(itheta) = cos theta + i sin theta$

So we have:

$(cos alpha + i sin alpha)(cos beta + i sin beta) = e^(ialpha)*e^(ibeta)$

$color(white)((cos alpha + i sin alpha)(cos beta + i sin beta)) = e^(i(alpha+beta))$

$color(white)((cos alpha + i sin alpha)(cos beta + i sin beta)) = cos(alpha+beta) + i sin(alpha+beta)$

So in our example:

$4(cos(pi/3)+isin(pi/3))*7(cos((2pi)/3)+isin((2pi)/3))$

$= 28(cos(pi/3+(2pi)/3)+isin(pi/3+(2pi)/3))$

$= 28(cos(pi)+isin(pi))$

$= 28(-1+i*0)$

$= -28$

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