i need help find the rectangular form for the problem below

Question

z1=1/2(cisπ/3) and z2=3(cosπ/6+isinπ/6)

Answer 1

Remember rectangular form just means a+ib. You should be able to just replace the trig functions with their values on the unit circle.

z₁ = 1/2 cis(π/3) = 1/2 cos(π/3) + i 1/2 sin(π/3) (definition of cis)

Remember

cos(π/3) = 1/2

sin(π/3) = √3/2

The using substitution

z₁ = 1/4 + i √3/4 (If you have a TI-84 you can check this by putting in 1/2*exp(i*pi/3) )

Similarly

z₂ = 3( cos(π/6) + i sin(π/6)) = 3√3/2 + i 3/2

Answer 2

z₁ = (1/2)[cis(π/3)]

z₁ = (1/2)[cos(π/3)+i·sin(π/3)]

z₁ = (1/2)(1/2 + i√3/2)

z₁ = 1/4 + (√3/4)i

z₂ = 3[cos(π/6)+i·sin(π/6)]

z₂ = 3[√3/2 + (1/2)i]

z₂ = 3√3/2 + (3/2)i

Hope this helped! Make sure to use the unit circle to assist you in these problems.