Question

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?

Answer 1

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`

Answer 2

`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`