If anyone could help me out here, in particular if you could explain what AB^2 is referring too. Thanks!
Let $\displaystyle z=2e^{i\theta_1}$
$\displaystyle w=5e^{i\theta_2}$
we have now
$\displaystyle \frac{w}{z}=2.5e^{i(\theta_2-\theta_1)}$
hence angle between OB and OC is$\displaystyle \theta_1$
which gives$\displaystyle \theta_1=\frac{\pi}{4}$
Now, we have,
$\displaystyle AB^2=39$
I think AB here represents the distance between A and B.
$\displaystyle 5^2+2^2-2.5.2.cos(\theta_2-\theta_1)=39$
$\displaystyle 29-20cos(\theta_2-\frac{\pi}{4})=39$
there is only one unknown, hence it can be solved.
Thank you for your help!
But unfortunately your solution is outside the level of my current course (I started complex numbers and my further course last week), I do not recognise what you are doing with e. Do you know any solutions with more basic math?
Hi
$\displaystyle arg\left(\frac{w}{z}\right) = arg(w) - arg(z)$
But
$\displaystyle arg\left(\frac{w}{z}\right) = (e_x,OC) [2\pi]$
And
$\displaystyle arg(w) - arg(z) = (e_x,OB) - (e_x,OA) = (OA,OB) [2\pi]$
Therefore
$\displaystyle (e_x,OC) = (OA,OB) [2\pi]$
To find (OA,OB) you can use the formula ABČ = OAČ + OBČ - 2.OA.OB.cos(AOB)