Answer: Let $f(z)=\dfrac{1}{2}log(x^2+y^2)+itan^{-1}(\dfrac{py}{x})$

$f(z)=u+iv$

On comparing , we get , 

$u=\dfrac{1}{2}log(x^2+y^2)$ and $v=tan^{-1}(\dfrac{py}{x})$

Differentiating u and v partially with respect to x and y respectively, 

$u_x=\dfrac{x}{x^2+y^2} , \ u_y=\dfrac{y}{x^2+y^2}$

$v_x=-\dfrac{py}{x^2+p^2y^2} , \ u_y=\dfrac{px}{x^2+p^2y^2}$

Since , the given functions are analytic .

$u_x=v_y \ and \ u_y=-v_x$

$\dfrac{x}{x^2+y^2}=\dfrac{px}{x^2+p^2y^2}$ and  $\dfrac{y}{x^2+y^2}=-\dfrac{py}{x^2+p^2y^2}$

which are satisfied only when $p=1$

Also find the derivative means f'(z) do it yourself.