= Solution
Compute the complex derivative along real increments to get $f'=u_x+iv_x$, and along imaginary increments to get $f'=v_y-iu_y$. Equality gives the <Cauchy-Riemann equations>, $\boxed{u_x=v_y,\quad u_y=-v_x}$.
For the specified real part, observe that $iz e^{-z}=e^{-x}[(x\sin y-y\cos y)+i(x\cos y+y\sin y)]$. Thus the most general <holomorphic function> with this real part on the connected plane is
$$
\boxed{f(z)=iz e^{-z}+iC,\quad C\in\mathbb R.}
$$
The difference of two such holomorphic functions has zero real part, so the <Cauchy-Riemann equations> make its imaginary part constant.
Next, $\cosh z=\cosh x\cos y+i\sinh x\sin y$. Its real and imaginary parts satisfy $u_x=\sinh x\cos y=v_y$ and $u_y=-\cosh x\sin y=-v_x$. Their partial derivatives are continuous everywhere, proving that the complex <hyperbolic cosine> is entire.
For the <Laurent series>, set $h=z-1$. Since $z^2-1=h(2+h)$,
$$
\frac{\cosh z}{z^2-1}=\frac{\cosh1+h\sinh1+O(h^2)}{2h}\left(1-\frac h2+O(h^2)\right).
$$
The <regular part at a simple pole> therefore gives
$$
\boxed{a_{-1}=\frac{\cosh1}{2},\qquad a_0=\frac{\sinh1}{2}-\frac{\cosh1}{4}.}
$$
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