Complex analysis function exercise

Question

$f(z)=u(x,y)+iv(x,y)$ analytic. On $y=0$ we have: $$ u(x,0)=(\sin x)^2, v(x,0)=0. $$ Find $f(z)$ at the point $ 5+i8$.$$ $$ Really have no idea how to solve this. The only thing I thought was about solving 2 Laplace's equations on a plane, one for $u$ and one for $v$ since they are both harmonic, but I'm sure that there is a more "complex" approach.

Answer 1

Let $g(z)=\sin^2(z)$. Then, if $x\in\mathbb R$, $g(x)=\sin^2(x)=u(x,0)+iv(x,0)=f(x)$. So, by the identity theorem, $g=f$. Therefore, $f(5+8i)=\sin^2(5+8i)$.