Question

How do you find the product of z1=8(cos pi/6 + i sin pi/6) and z2=3(cos pi/2 + i sin pi/2) and write the answer in standard form?

Answer 1

`z_1*z_2=-12+12sqrt3i`

Explanation:

`z_1=8(cos(pi/6)+isin(pi/6))=8e^(ipi/6)` and

`z_2=3(cos(pi/2)+isin(pi/6))=3e^(ipi/2)`

Hence, `z_1*z_2=8e^(ipi/6)*3e^(ipi/2)=24e^(i(pi/6+pi/2)`

= `24(cos(pi/2+pi/6)+isin(pi/2+pi/6))`

= `24(-sin(pi/6)+icos(pi/6))`

= `24(-1/2+isqrt3/2)=-12+12sqrt3i`