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2 Answers
sjc
Dec 9, 2017

option #C#

Explanation:

#z/w=(6(cos((2pi)/7)+isin((2pi)/7)))/(3(cos((5pi)/7)+isin((5pi)/7))#

#=2(cos((2pi)/7)+isin((2pi)/7))(cos((5pi)/7)+isin((5pi)/7))^-1#

by Demoivre's theorem

#color(blue)((costheta+isintheta)^n=cosntheta+isintheta)#

the second bracket can be simplified as

#=2(cos((2pi)/7)+isin((2pi)/7))(cos((-5pi)/7)+isin((-5pi)/7))--(1)#

but

#color(blue)((cosalpha+isinalpha)(cosbeta+isinbeta))#

#color(blue)(=cos(alpha+beta)+isin(alpha+beta))#

#(1)rarr#

#=2(cos((2pi-5pi)/7)+isin((2pi-5pi)/7))#

#=2(cos((-3pi)/7))+isin((-3pi)/7))#

option #C#

George C.
Dec 9, 2017

C #" "2(cos(-(3pi)/7)+i sin(-(3pi)/7))#

Explanation:

Note that:

#r(cos theta + i sin theta) = re^(itheta)#

So when we multiply two complex numbers in polar form, we find:

#r_1(cos alpha + i sin alpha) * r_2(cos beta + i sin beta)#

#=r_1e^(ialpha) * r_2e^(ibeta)#

#=r_1 r_2 e^(i(alpha+beta))#

#=r_1 r_2 (cos (alpha + beta) + i sin (alpha + beta))#

Similarly, when we divide two complex numbers in polar form, we find:

#(r_1(cos alpha + i sin alpha))/(r_2(cos beta + i sin beta))#

#=(r_1e^(ialpha)) / (r_2e^(ibeta))#

#=r_1/r_2 e^(i(alpha-beta))#

#=r_1/r_2 (cos (alpha - beta) + i sin (alpha - beta))#

So in our example, we have:

#z/w = (6(cos((2pi)/7)+i sin((2pi)/7))) / (3(cos((5pi)/7)+isin((5pi)/7)))#

#color(white)(z/w) = 6/3(cos((2pi)/7-(5pi)/7)+i sin((2pi)/7-(5pi)/7))#

#color(white)(z/w) = 2(cos(-(3pi)/7)+i sin(-(3pi)/7))#

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