Suppose $z_0=e^{i\theta_0}$ a complexe number as $\theta_0\in ]-\pi,\pi[ \setminus\{0\}$.
For $n\in \mathbb{N}$, we pose $z_{n+1}=\dfrac{|z_n|+z_n}{2}$ and $z_n=r_ne^{i\theta_n}$ with $\left(r_n,\theta_n\right)\in \mathbb{R}^+ \times ]-\pi,\pi]$.
The first question was determine $r_{n+1}$ and $\theta_{n+1}$ using $r_{n}$ and $\theta_{n}$ for all $n\in \mathbb{N}$.
And I found that $r_{n+1} = r_n\cos\left(\dfrac{\theta_n}{2}\right)$ and $\theta_{n+1}=\dfrac{\theta_n}{2}$.
but I need to conclude that
$$\forall n\in\mathbb{N}^*, > r_n=\Pi_{k=1}^n\cos\left(\dfrac{\theta_0}{2^k}\right)$$
The second problem is:
We put $n\in\mathbb{N} y_n=\operatorname{Im}\left(z_n\right)$.
and I need to calculate
$y_{n+1}$ using $y_n$.
Then to conclude that
$$\forall n\in \mathbb{N}^*, > \Pi_{k=1}^n\cos\left(\dfrac{\theta_0}{2^k}\right)=\dfrac{\sin\theta_0}{2^n\sin\dfrac{\theta_0}{2^n}}$$
For the last equality:
Multiply and divide $\Pi_{k=1}^n\cos\dfrac{\theta_0}{2^k}\;$ by $\sin\dfrac{\theta_0}{2^k}$ and use the trigonometric identity $$\sin{2\varphi}=2\sin{\varphi}\cos{\varphi}.$$