How do you write the complex number in standard form #[3(cos(pi/3)+isin(pi/3))][4(cos(pi/6)+isin(pi/6))]#?

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Answer 1 · 2016-12-08T07:41:34

i 12

Use #e^(itheta)=costheta +i sin theta#

Here, the expression is

12e^(ipi/3)e(ipi/6)#

#=12e^(i((1/3+1/6)pi)#

#=12e^(ipi/2)#

#=12(cos (pi/2)+i sin (pi/2))#

#=i12#