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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.

The value $f(\mathbf{a},\mathbf{b},\mathbf{c})$ is twice the signed area of the triangle traversed in that order; it is positive for counterclockwise orientation. Choose points on the parabolaFor $i<j<k$, subtracting the first row in the determinant and factoring givesEach of the three factors is positive, proving the required property for every integer $n\ge 3$.

For $u=1$, the inverse parabolic polar coordinates give $y=v$ and $x=(1-v^2)/2$. For $v=1$, they give $y=u$ and $x=(u^2-1)/2$. Thus the requested curves are the upper halves of the parabolasThe first parabola has vertex $(1/2,0)$ and opens to the left; the second parabola has vertex $(-1/2,0)$ and opens to the right. Both pass through $(0,1)$ and are reflections of one another across the $y$-axis.

Assume $h\ne 0$. The velocity in polar coordinates is $\dot{\mathbf{r}}=\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }$. HenceTaking the dot product of the conserved Laplace-Runge-Lenz vector with $\mathbf{r}$ givesConsequently the polar equation of a conic isOnly angles for which the denominator is positive describe the physical trajectory. Its eccentricity is $A/(mk)$, giving an ellipse, parabola, or hyperbola when this ratio is respectively less than, equal to, or greater than $1$.

If $A=0$, the angle $\gamma$ is undefined, but the same dot product identity gives the circular orbit $\boxed{r=m{h}^2/k}$. If $h=0$, the trajectory is radial and the displayed division degenerates; for $k>0$, $\mathbf{A}=-mk{\mathbf{e}}_r$ and $\gamma =\pi$ on a fixed ray, so the identity does not determine $r$ as a function of a varying angle.

The quadric surface $x^2+y^2+z^2=1$ is the unit sphere centered at the origin. Each section $z=c$, for $|c|<1$, is a circle of radius $\sqrt{1-c^2}$.

The quadric surface $z=x^2+y^2$ is a circular elliptic paraboloid, opening in the positive $z$ direction with vertex at the origin. Its horizontal sections $z=c>0$ are circles of radius $\sqrt{c}$, and its vertical sections through the $z$ axis are parabolas.

For $a{x}^2+y^2=0$, $z$ is unrestricted. If $a>0$, both nonnegative terms must vanish, so the locus is the $z$ axis, a straight line. If $a=0$, it is the plane $y=0$, represented by the repeated factor $y^2$. If $a<0$, factorization givesso the locus is two distinct planes, $y=\pm \sqrt{-a}x$, meeting along the $z$ axis. These are degenerate quadrics, not ordinary curved cylinders.

Expand $u=3x^2-2xy+3y^2$. The Lagrange multiplier equations for $u-\lambda v$ are $(6-2\lambda )x-2y=0$ and $-2x+6y+\lambda =0$, together with the constraint $y=x^2-1/4$. The first two imply $y=(3-\lambda )x$ and $(6\lambda -16)x=\lambda$. The value $\lambda =8/3$ would make the latter equation inconsistent, so division is allowed:Insert these into $x^2-y-1/4=0$ and multiply by $(6\lambda -16{)}^2$. Simplification gives $2(3{\lambda }^3-21{\lambda }^2+48\lambda -32)=0$, henceThe gradient of the constraint $(2x,-1)$ is never zero, so the Lagrange multiplier condition is necessary at every constrained extremum. To establish a global minimum, parametrize the whole parabola by $y=x^2-1/4$. The restricted function becomesIt is coercive: $q(x)\to \infty$ as $|x|\to \infty$, so it attains a global minimum. Moreover $q^{\prime \prime }(x)=36(x-1/6{)}^2+2>0$, making it strictly convex and ensuring that this minimum is unique. This proves more than the permitted assumption of one real multiplier root.

The level sets of $u$ are nested ellipses centered at the origin. In the rotated coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, they have equation $2p^2+4q^2=\text{constant}$, with the longer axis along $x=y$. The smallest one meeting the parabola is tangent to it at the unique minimizing point, approximately $(x,y)=(-0.12654,-0.23399)$, where $u\simeq 0.15307$ and $\lambda \simeq 1.15085$.

The double-angle identity eliminates the parameter: $y=1+\cos 2\theta =2{\cos }^2\theta =2x^2$. As $\theta$ increases from zero to $\pi$, $x$ decreases from one to minus one. The positive area below this parabola is thereforeEquivalently, the parameter integral is $-{\int }_0^{\pi }(1+\cos 2\theta )(-\sin \theta )d\theta$. The leading minus sign corrects for traversal from right to left; the signed integral $\int ydx$ along that traversal would be negative.

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

The point on the parabola is $(1,2)$. Implicit differentiation gives $2ydy/dx=4$, so the tangent line has slope one there. Its perpendicular normal line to a plane curve therefore has slope $-1$. The required line is

If $V_{0x}\ne 0$, the first formula in (a) gives $e^{-\gamma t}=1-\gamma x/V_{0x}$ andSubstitute this and $1-e^{-\gamma t}=\gamma x/V_{0x}$ into $z(t)$:For finite physical times $t\ge 0$, the domain is $0\le \gamma x/V_{0x}<1$, valid for either sign of $V_{0x}$. If $V_{0x}=0$, the trajectory is vertical, $x(t)=0$, and in general no single-valued graph $z(x)$ describes it. As a consistency check, the logarithmic expansion at fixed $x$ cancels the apparent $1/\gamma$ terms and gives $z\to (V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$ as $\gamma \to 0$, the drag-free parabola.

The intersections satisfy $6x-x^2=2x+3$, or $(x-1)(x-3)=0$, so the bounded region has $1\le x\le 3$. Over that interval the parabola is above the line, and their height difference is $(x-1)(3-x)$. Thus its area isUsing upper minus lower ensures a positive geometric area.

The derivative of the parabola is $dy/dx=2+2x$, giving tangent line slope $4$ at $(1,3)$. A normal line to a plane curve has the negative reciprocal slope, since its direction is orthogonal to a tangent vector. HenceThe point $(1,3)$ satisfies the equation, and the product of the two slopes is $-1$.

The parabola opens upward, with its minimum at $(0,-1)$ and zeros at $x=\pm 1$. The quartic graph is symmetric about the vertical axis, has its flat maximum at $(0,1)$, and also has zeros at $\pm 1$.

Their intersections solve $x^4+x^2-2=(x^2-1)(x^2+2)=0$, so the only real intersections are $(-1,0)$ and $(1,0)$. Between them, $1-x^4$ is the upper curve and $x^2-1$ the lower curve.

At an intersection, $x^2=(x-1{)}^2=x^2-2x+1$, so $x=1/2$. Substitution into either parabola givesThe resulting linear equation for $x$ has only one solution, so this is the unique intersection.

At $x=3$, the point on the parabola is $(3,8)$ and the tangent line has slope $f^{\prime }(3)=6$. The normal line to a plane curve therefore has slope $-1/6$. Its equation isThus the requested results are

The absolute value is nonnegative, so a real point must satisfy $1-x^2\ge 0$, or $-1\le x\le 1$. On that interval the two branches areThey form a closed lens symmetric about both coordinate axes, meeting at $(\pm 1,0)$ and reaching $(0,\pm 1)$. The parabolas outside the stated interval do not belong to the original locus.

From $p^2=C/(x+C)$, we have $1-p^2=x/(x+C)$. Substitution into the original relation gives $y=2(x+C)p$, soThese are real parabola branches, with either sign of $C$ allowed when the branch is real. For $C\ne 0$, differentiating gives $p=2C/y$ and verifies both the relation and the slope. Use graph intervals with $y\ne 0$ and $x\ne 0$, so the slope is finite and $p\ne \pm 1$. The separate constant-slope case $p=0$ gives $y=0$, also represented by the degenerate choice $C=0$. Squaring has not imposed a sign on $y$, so both real graph branches must be included.

For $a=-1$, the local minimum at the origin has $f=1$. Its full level set isThe connected component containing the minimum is just the isolated origin. The same level also has the outer branches $y=\pm \sqrt{x^4/2-x^2}$ for $|x|\ge \sqrt{2}$; these are different connected components, not curves approaching the origin.

Both saddle points $(\pm 1,0)$ have $f=e^{1/4}$. Their common level set factors asIts explicit equations are thereforeThe two parabolas cross at each saddle, explaining the change of contour connectivity there. Each displayed equation can be restricted to the square of part (ii) if only that sketch window is wanted.