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?

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 $costheta+isintheta=e^(itheta)$

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

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

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

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

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

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

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

$=-(3i)/4$