= Solution
At $z=1$, numerator and denominator both have simple zeroes. Their <derivatives> give the removable value
$$
\boxed{\lim_{z\to1}\frac{\sqrt z-1}{\sin\pi z}=-\frac1{2\pi}.}
$$
At each positive integer $n\ge2$, the numerator is nonzero and the denominator has a simple zero, so these are simple <poles>. At zero the expansion on the principal sheet is
$$
\frac{\sqrt z-1}{\sin\pi z}=-\frac1{\pi z}+\frac1{\pi z^{1/2}}+O(z),
$$
which contains a fractional power. Thus \b[zero is a <branch point> with <pole>-like leading growth], not an ordinary isolated <pole>. The negative real axis is the chosen <branch cut>. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.
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