Complex numbers problem, solving for Z1 and Z2

Question

If Z1,Z2 are part of C such that Im(z1) = sqrt(3), |z2| = 2 and z1*z2 = 4(cos( 5pi/6 ) + sin( 5pi/6)i ). Find the values of Z1 and Z2.

I was told that i could use the property of |z1*z2| = |z1||z2| but I am still lost as to how I proceed, thank you.

Answer 1

Alexander:

Use the identity z₁ = r₁ * e^{θ1 * i} = r₁*(cos(θ₁) + i*sin(θ₁))

z₂ = r₂ * e^{θ2 * i} = r₂*(cos(θ₂)+i*sin(θ₂))

z₁*z₂ = r₁*r₂* e^{(θ1 + θ2)*i}

We know |z₂| = 2, so r₂ = 2

We also know z₁*z₂ = 4 * (cos(5pi/6)+i*sin(5pi/6)). So r₁*r₂=4; since r₂=2, r₁ = 2.

Since Im(z₁) = √3 and r₁ = 2 , we must have 2*(sin(θ1)) = √3 or sin(θ1)=√3/2; take θ1 = pi/3

Now since θ₁+θ₂ = (5pi/6), we must have θ₂ = (5pi/6)-(pi/3) = pi/2

We know have all the information needed to solve the problem

z₁ = r₁*(cos(θ₁)+i*sin(θ₁)) = 2* (cos(π/3)+i*sin(π/3))

z₂ = r₂*(cos(θ₂)+i*sin(θ₂)) = 2*(cos(π/2)+i*sin(π/2))

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