Solve for complex numbers $z_1$ and $z_2$

Question

Solve for complex numbers of $$\left\{\begin{array}{l} \frac{\left|z_{1}\right|^{2}}{3}+\frac{\left|z_{2}\right|^{2}}{5}=\frac{\left|z_{1}+z_{2}\right|^{2}}{8} \\ 10 z_{1}+z_{2}=7+14 i \end{array}\right.$$

My attempt: Let $z_1=\gamma_1e^{i\theta_1}$ and $z_2=\gamma_2e^{i\theta_2}$ Then we have, $$\frac{\gamma_{1}^{2}}{3}+\frac{\gamma_{2}^{2}}{5}=\frac{\gamma_{1}^{2}+\gamma_{2}^{2}+2 \gamma_{1} \gamma_{2} \cos \left(\theta_{1}-\theta_{2}\right)}{8}$$ $$\Rightarrow \quad \frac{5 \gamma_{1}^{2}}{6}+\frac{3 \gamma_{2}^{2}}{10}=\gamma_{1} \gamma_{2} \cos \left(\theta_1-\theta_{2}\right) \tag{1}$$

Also from the second equation we get, $$\begin{array}{l} 10 \gamma_{1} \cos \theta_{1}+\gamma_{2} \cos \theta_{2}=7 \\ 10 \gamma_{1} \sin \theta_{1}+\gamma_{2} \sin \theta_{2}=14 \end{array}$$ Squaring and adding the above two equations we get, $$100 \gamma_{1}^{2}+\gamma_{2}^{2}+20 \gamma_{1} \gamma_{2} \cos \left(\theta_{1}-\theta_{2}\right)=245$$ Using $\cos(\theta_1-\theta_2)$ from $(1)$ we get, $$\frac{50 \gamma_{1}^{2}}{3}+\gamma_{2}^{2}=35 \tag{2}$$ How to proceed further?

Answer 1

Hint: By Cauchy -Schwarz: $$\frac{|z_1|^2}{3}+\frac{|z_2|^2}{5}\ge \frac{{(|z_1|+|z_2|)}^2}{8}\ge \frac{{(|z_1+z_2|)}^2}{8}$$

For equality to hold we must have $$\frac{|z_1|}{\sqrt{3}}=\frac{|z_2|}{\sqrt{5}}$$

Also we must have $$|z_1+z_2|=|z_1|+|z_2|$$ when $\arg(z_1)=\arg(z_2)$

Thus $$\frac{z_1}{\sqrt{3}}=\frac{z_2}{\sqrt{5}}$$

Can you end it now