$z,w$ are complex numbers (not zero) that are in the first quarter of Gauss's plane such that:
$$z+\frac{4}{z}=4\sin\theta$$
$$w+\frac{4}{w}=4\sin\beta$$
Need to prove that: $$zw+\frac{16}{zw}=-8\cos(\theta+\beta)$$
I thought to find $z$ and $w$ by quadratic equation but then i don't know how to choose the sign of the answers.
for example, I don't know if $\sin\theta$ is positive or negative.
Thanks.
EDIT: the problem is that $ \theta, \beta \in (0,\pi/2)$ and also $\theta, \beta \in (0,2\pi)$ so both solutions $z=2(\sin\theta \pm i\cos\theta)$ are true but only one true for the proof $z=2(\sin\theta+i\cos\theta)$.
Can anyone explain it?
how about we turn $z + 4/z = 4\sin \theta$ into a quadratic equation $z^2 - 4\sin \theta z + 4 = 0.$ we can complete the squares by writing $(z - 2\sin \theta)^2 = 4 \sin^2 \theta - 4 = (\pm 2i\cos \theta)^2.$ so $$z = 2(\sin \theta \pm i\cos \theta), \frac{1}{z} = \frac{1}{2} (\sin\theta \mp i\cos \theta), w = 2(\sin \beta \pm i\cos \beta), \frac{1}{w} = \frac{1}{2} (\sin \beta \mp i\cos \beta)$$
from this you $$zw = 4[-\cos(\theta + \beta) \pm i \sin(\theta + \beta)], {16 \over zw} = 4 [-\cos(\theta + \beta) \mp i\sin(\theta + \beta)]$$.
adding the last two formula gives $zw + \frac{16}{zw} = -8\cos(\theta + \beta)$ which is at least correct when you put $z = w = 2 \ and \ \theta = \beta = \pi/2.$