Elementary proof that $\sin$ is the unique function for which $y'' = -y$ with $y(0) = 0$ and $y'(0)=1$

Question

Defining $\sin$ by means of the exponential function or by power series, it is straightforward to show that $\sin$ satisfies: $$y''=-y, \ \ \ \ y(0)=0, \ \ \ \ y'(0)=1.$$ There is an elementary proof that there is a unique solution to the above on the reals. I wonder if the argument can be extended to, or if there is another elementary proof that shows there is a unique solution in the complex plane.