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Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

For a fixed pole away from the saddle point, the smooth amplitude there is $(i-z_0{)}^{-1}$. The simple-saddle contribution in steepest descent isThe residue theorem supplies an additional contribution precisely when the contour deformation crosses $z_0$. The region swept above the original contour has positive orientation: the original contour from left to right followed by the reversed saddle contour encloses it counterclockwise. Consequently the original integral equals the saddle integral plus the pole contribution:For $|z_0|<1$, an upper-half-plane pole outside the indentation has $\chi =1$, while a lower-half-plane pole has $\chi =0$. The entire upper unit half-disk lies below the saddle parabola, so there is no further case inside that half-disk. With a fixed indentation radius $\delta$, a pole inside the upper semicircle, $|z_0|<\delta$, is also below the original contour and has $\chi =0$; real points in the indentation gap are excluded as well. Taking $\delta <|z_0|$ for a fixed nonzero pole gives the usual upper/lower classification. The pole contribution need not always dominate exponentially; its size depends on $\mathrm{Re}\varphi (z_0)$, so keeping both terms makes that dependence explicit.

A pole on the original contour requires a stated Cauchy principal value or an indentation prescription. A pole parameter on the negative imaginary branch cut still defines the integral: the contour stays on its fixed sheet, and only the denominator uses $z_0$. Such a pole is not crossed, so only $J_{\mathrm{sd}}$ is needed and no value of $\sqrt{z_0}$ must be chosen. The residue exponential above is evaluated only when $\chi =1$. At $z_0=0$ the original indentation excludes the singular point, so no additional residue is crossed. These conventions matter before taking any limiting pole position.

A Newtonian reflector uses a concave primary generated by a parabola, followed by a flat secondary inclined at $45^{\circ }$ to the optical axis. Parallel marginal rays converge toward the primary focus; the secondary intercepts that converging beam and folds it sideways to an accessible focal plane.

The Newtonian reflector needs only one powered mirror, so it is relatively simple to fabricate and inexpensive. A parabolic primary has no on-axis spherical aberration. Its principal wide-field limitation is coma, accompanied by field curvature and off-axis astigmatism. The diagonal and its supports obstruct the entrance pupil and introduce diffraction; the tube is long compared with a folded two-powered-mirror design, and heavy instruments at the side focus can be awkward to support. These comments also answer the unheaded design-comparison clause on the next PDF page.