Christopher J. answered • 09/21/20

Berkeley Grad Math Tutor (algebra to calculus)

Alexander:

Use the identity z_{1} =_{ }r_{1} * e^{θ1 * i} = r_{1}*(cos(θ_{1}) + i*sin(θ_{1}))

z_{2} = r_{2} * e^{θ2 * i } = r_{2}*(cos(θ_{2})+i*sin(θ_{2}))

z_{1}*z_{2 }= r_{1}*r_{2}* e^{(θ1 + θ2)*i}

We know |z_{2}| = 2, so r_{2} = 2

We also know z_{1}*z_{2} = 4 * (cos(5pi/6)+i*sin(5pi/6)). So r_{1}*r_{2}=4; since r_{2}=2, r_{1} = 2.

Since Im(z_{1}) = √3 and r_{1} = 2 , we must have 2*(sin(θ1)) = √3 or sin(θ1)=√3/2; take θ1 = pi/3

Now since θ_{1}+θ_{2} = (5pi/6), we must have θ_{2} = (5pi/6)-(pi/3) = pi/2

We know have all the information needed to solve the problem

z_{1} = r_{1}*(cos(θ_{1})+i*sin(θ_{1})) = 2* (cos(π/3)+i*sin(π/3))

z_{2} = r_{2}*(cos(θ_{2})+i*sin(θ_{2})) = 2*(cos(π/2)+i*sin(π/2))

Let me know if you have any questions about what I did above.