Solution: iven that $f(z)$ is a holomorphic function on the unit disc and $\vert f(z) \vert = 1$ for all $\vert z \vert \leq 1$.

1. We will show that $f$ is constant using the Open Mapping Theorem. Let us recall the Open Mapping Theorem:
   Let $U$ be an open and connected subset of $\mathbb{C} $ anf $f: U \to \mathbb{C} $ be a non-constant holomorphic function, then $f$ is an open map, that is it sends an open subset of $U$ to an open subset of $\mathbb{C} $.
   If $f$ is not constant, then the image of $|z| \lt 1$ will be open by the open mapping theorem. As the set $|z| \lt 1$ is connected, the image must be connected, since $f$ is continuous. As the image of $f$ is the unit circle, this can not be possible, since no connected subset of the unit circle is open in the complex plane. Therefore, $f$ must be constant.

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2. Now we will show the same by using the CR-equations and the identity \begin{equation}\label{eq:27Jun2023-1} \left\vert f(x + \iota y) \right\vert ^2 = u(x,y)^2 + v(x,y)^2 \end{equation} Since $\vert f(z) \vert = 1$, we have $u^2 + v^2 = 1$. Differentiating, we get \begin{align} & u u_x + v v_x = 0, \label{eq:27Jun2023-2}\\ & u u_y + v v_y = 0. \label{eq:27Jun2023-3} \end{align} Now we multiply the equation \eqref{eq:27Jun2023-2} by $u$ and substitute $v_x = -u_y$ (CR equation) to obtain \[ u^2 u_x - uv u_y = 0. \] Similarly, we multiply the equation \eqref{eq:27Jun2023-3} by $v$ and substitute $v_y = u_x$ (CR equation) to obtain \[ uv u_x v^2 u_x = 0. \] Adding the last two equations to get \[ u_x \left( u^2 + v^2 \right) = 0 \implies u_x = 0 \implies v_y = 0. \] Substituting these values in the equations \eqref{eq:27Jun2023-2} and \eqref{eq:27Jun2023-3} we obtain \begin{align*} vv_x = uu_y = 0 & \implies v^2v_x = u^2u_y = 0 \\ & \implies v^2 v_x - u^2 u_y = 0 \\ & \implies v^2 v_x - u^2 (-v_x) = 0\\ & \implies v_x \left( u^2 + v^2 \right) = 0. \end{align*} This implies, $v_x = -u_y = 0$. Hence both $u$ and $v$ are constant on the unit disc. So $f$ is constant ant on an infinite number of points and therefore, constant in all of $\mathbb{C} $.