How do you simplify #3(cos(pi/6)+isin(pi/6))div4(cos((2pi)/3)+isin((2pi)/3))# and express the result in rectangular form?

2 Answers
Dec 14, 2016

Answer:

The answer is #=-3/4i#

Explanation:

There are 2 ways for the simplification

#cos(pi/6)=sqrt3/2#

#sin(pi/6)=1/2#

#cos(2pi/3)=-1/2#

#sin(2pi/3)=sqrt3/2#

#i^2=-1#

#(3(cos(pi/6)+isin(pi/6)))/(4(cos(2pi/3)+isin(2pi/3)))#

#=3/4(sqrt3/2+i*1/2)/(-1/2+isqrt3/2)#

#=3/4((sqrt3/2+i*1/2)(-1/2-isqrt3/2))/((-1/2+isqrt3/2)(-1/2-isqrt3/2))#

#=3/4(-sqrt3/4-i3/4-i/4+sqrt3/4)/(1/4+3/4)#

#=3/4(-i)#

We can also use #costheta+isintheta=e^(itheta)#

#cos(pi/6)+isin(pi/6)=e^(ipi/6)#

#cos(2pi/3)+isin(2pi/3)=e^(2ipi/3)#

#:.(3(cos(pi/6)+isin(pi/6)))/(4(cos(2pi/3)+isin(2pi/3)))=3/4e^(ipi/6)/e^(2ipi/3)#

#=3/4(e^(ipi(1/6-2/3)))#

#=3/4e^(-ipi/2)#

#=3/4(cos(-pi/2)+isin(-pi/2))#

#=3/4*(0-i)#

#=-(3i)/4#

Dec 14, 2016

Answer:

#3(cos(pi/6)+isin(pi/6))-:4(cos(2pi/3)+isin(2pi/3))=-3/4i#

Explanation:

Given two complex numbers #z_1=r_1(cosalpha+isinalpha)# and #z_2=r_2(cosbeta+isinbeta)#

#z_1-:z_2=r_1/r_2(cos(alpha-beta)+iin(alpha-beta))#

#3(cos(pi/6)+isin(pi/6))-:4(cos(2pi/3)+isin(2pi/3))#

= #3/4(cos(pi/6-(2pi)/3)+isin(pi/6-(2pi)/3))#

= #3/4(cos(pi/6-(4pi)/6)+isin(pi/6-(4pi)/6))#

= #3/4(cos(-(3pi)/6)+isin(-(3pi)/6))#

= #3/4(cos(-pi/2)+isin(-pi/2))#

= #3/4(cos(pi/2)-isin(pi/2))#

= #3/4(0-i)#

= #-3/4i#