= Solution
A <holomorphic function> is complex differentiable throughout an open region; equivalently it is locally represented by a convergent complex power series. For $f=u+iv$, the <Cauchy-Riemann equations> are $u_x=v_y$, $u_y=-v_x$. With continuous first partial derivatives, these equations imply holomorphicity and $f'=u_x+iv_x$.
For $e^z=e^x\cos y+i e^x\sin y$, the equations hold and $u_x+iv_x=e^x(\cos y+i\sin y)=e^z$. For $\sin z=\sin x\cosh y+i\cos x\sinh y$, they also hold, and $u_x+iv_x=\cos x\cosh y-i\sin x\sinh y=\cos z$. Therefore
$$
\boxed{\frac{d}{dz}e^z=e^z,\qquad\frac{d}{dz}\sin z=\cos z.}
$$
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