How do you perform the operation in trigonometric form (6(cos40+isin40))/(7(cos100+isin100))6(cos40+isin40)7(cos100+isin100)?

1 Answer
Sep 17, 2016

3/7(1-isqrt3)37(1i3).

Explanation:

Let z_1=6(cos40+isin40), and, z_2=7(cos100+isin100)z1=6(cos40+isin40),and,z2=7(cos100+isin100)

Knowing that, costheta+isintheta=e^(itheta)cosθ+isinθ=eiθ, we have,

z_1=r_1e^(itheta_1), and, z_2=r_2e^(itheta_2)z1=r1eiθ1,and,z2=r2eiθ2, then,

r_1=6, theta_1=40, r_2=7, theta_2=100r1=6,θ1=40,r2=7,θ2=100.

:." The Expression="z_1/z_2=(r_1e^(itheta_1))/(r_2e^(itheta_2))

=(r_1/r_2)*e^(itheta_1-itheta_2)

=(r_1/r_2)*e^(i(theta_1-theta_2))

=(6/7)*e^(i(40-100))

=(6/7)*e^(i(-60))

=(6/7)(cos(-60)+isin(-60))

=(6/7)(cos60-isin60)

=(6/7)(1/2-isqrt3/2)

=3/7(1-isqrt3).

Enjoy maths.!