Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 4 / iv

At any point on the imaginary axis, approaching from the left gives limit $-1$, while the assigned value and the right-hand limit are $1$. Thus the function is discontinuous on that axis. Complex differentiability implies continuity, so $f$ is not holomorphic on $\mathbb{C}$. It is holomorphic separately in the two open half-planes.