Solution: Since $f$ is analytic, it must satisfy the Cauchy-Riemann equations, that is, \begin{align*} u_x=v_y \text{ and } u_y = -v_x. \end{align*} Since $u$ is a function of $x$ only and $v$ is a function of $y$ only, we have \[ u_x = v_y \implies u^\prime (x) = v^\prime (y). \] The above will be true only if $u^\prime (x) = v^\prime (y) = a$, for some $a\in \mathbb{R} $. Thus, \begin{align*} u^\prime (x) & = a \implies u(x) = ax + b_1 \text{ and } \\ v^\prime (y) & = a \implies v(y) = ay + b_2 , \end{align*} where $d_1,d_2\in \mathbb{R} $. Therefore, \begin{align*} f(z) & = u(x) + \iota v(y) \\ & = \left( ax + b_1 \right) + \iota \left( ay + b_2 \right) \\ & = az + \left( b_1 + \iota b_2 \right) \\ & = az + b. \end{align*}