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.!