= Solution
\b[Branches and cuts.] For the <square root of z squared minus one>, a <branch point> is a point around which <analytic continuation> along an arbitrarily small closed loop can change the value of a chosen <branch of a multivalued function>. Near $z=1$, the square root behaves as $\sqrt2(z-1)^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1)^{1/2}$. One circuit around either point reverses its sign. Thus the finite <branch points> are $\boxed{z=\pm1}$. Infinity is not a <branch point>: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.
A single <branch cut> can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both <branch points> or neither, so it produces no sign change. Alternatively, use two <branch cuts>, $(-\infty,-1]$ and $[1,\infty)$, each joining a <branch point> to infinity. On the resulting slit region one can choose a consistent <branch of a multivalued function>.
For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, put
$$
a=x^2-y^2-1,\qquad \rho=\sqrt{a^2+4x^2y^2},\qquad U=\sqrt{\frac{\rho+a}{2}},\quad V=\sqrt{\frac{\rho-a}{2}}.
$$
For $y\ne0$, the selected <square root> is
$$
\boxed{f(x+iy)=\operatorname{sgn}(x)U+i\operatorname{sgn}(y)V.}
$$
Here $\operatorname{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the <branch cut> are
$$
\boxed{f(x+i0)=+i\sqrt{1-x^2},\qquad f(x-i0)=-i\sqrt{1-x^2},\quad -1<x<1.}
$$
Outside the cut, $f(x)=\operatorname{sgn}(x)\sqrt{x^2-1}$. Reversing the global <branch of a multivalued function> reverses every displayed value.
\b[Differential equation.] Write $f(\zeta)=\sqrt{\zeta^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the <branch points>, <differentiation under the integral sign> gives
$$
u'(z)=\oint_C\frac{\zeta e^{z\zeta}}{f(\zeta)}\,d\zeta,\qquad u''(z)=\oint_C\frac{\zeta^2e^{z\zeta}}{f(\zeta)}\,d\zeta.
$$
On the contour $f'=\zeta/f$. Therefore
$$
zu''+u'-zu=\oint_Ce^{z\zeta}\left(zf(\zeta)+\frac{\zeta}{f(\zeta)}\right)d\zeta=\oint_C\frac d{d\zeta}\left[e^{z\zeta}f(\zeta)\right]d\zeta=0.
$$
The last <contour integral> vanishes because the expression in brackets returns to its original value after circling both <branch points>. This proves $\boxed{zu''+u'-zu=0}$, including at $z=0$; division by $z$ is unnecessary.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2-finite-branch-cut.png]
{title=Shrinking a counterclockwise contour around the finite square-root cut}
{description=The upper lip runs from 1 to -1, and the lower lip returns from -1 to 1. The endpoint arcs have vanishing integrals. The values shown use the branch that behaves as z at infinity.}
\b[Value at zero.] Near either endpoint $\zeta_0=\pm1$, the integrand at zero is $O(|\zeta-\zeta_0|^{-1/2})$. A small arc of radius $\varepsilon$ has length $O(\varepsilon)$ and hence contributes $O(\sqrt\varepsilon)\to0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. Thus
$$
u(0)=\int_1^{-1}\frac{dx}{i\sqrt{1-x^2}}+\int_{-1}^1\frac{dx}{-i\sqrt{1-x^2}}=2i\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}.
$$
Using $x=\sin t$ evaluates the final <integral> as $\pi$, and so
$$
\boxed{u(0)=2\pi i.}
$$
The same deformation gives $u(z)=2i\int_{-1}^1e^{zx}(1-x^2)^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.
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