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Integrating ${\varphi }_x=2xy$ gives $\varphi =x^{2}y+h(y,z)$. Matching ${\varphi }_y=x^2+3y^2+z^2$ then gives $h=y^3+y{z}^2+k(z)$. Finally ${\varphi }_z=2yz$ forces $k$ to be constant. A scalar potential isThe initial and final points of the helix are $(b,0,0)$ and $(0,b,cL)$, where $L=5\pi /2$. The fundamental theorem for line integrals givesagreeing with the directly evaluated $I_2$.

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.

Put $L=5\pi /2$. The curve tangent is ${\mathbf{r}}^{\prime }=(-b\sin \theta ,b\cos \theta ,c)$. For the first line integral, the scalar triple product identity gives $(\mathbf{a}\times \mathbf{r})\cdot {\mathbf{r}}^{\prime }=\mathbf{a}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })$. Its cross product isThe needed definite integrals areand ${\int }_0^L(-\theta \sin \theta -\cos \theta )d\theta =[\theta \cos \theta -2\sin \theta {]}_0^L=-2$. HenceFor the second line integral, substitute the curve into ${\mathbf{v}}_2\cdot {\mathbf{r}}^{\prime }$. The cubic terms combine to $b^3\cos \theta$, leavingThe fundamental theorem of calculus therefore givesBoth calculations retain the full path from $\theta =0$ to $5\pi /2$, rather than inadvertently closing the helix after one turn.

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. Thus

Write $f(x,y)=F(u,v)$ in the proposed local coordinates. The multivariable chain rule givesConsequentlyAway from the origin the Jacobian determinant is $u_x{v}_y-u_y{v}_x=4(x^2+y^2)>0$, so the inverse function theorem justifies these local coordinates. Therefore $F_v=0$ and the local general solution isSubstitution verifies the result: $f_x=2x{\Phi }^{\prime }$ and $f_y=-2y{\Phi }^{\prime }$, whose weighted sum vanishes.

There is a global qualification. The coordinate map is two-to-one, and a level set of $x^2-y^2$ has separate characteristic curve branches. A single profile on the entire plane does not follow without a branch-identification assumption. For example, take $f=\exp [-1/(x^2-y^2{)}^2]$ on the wedge $x>0$, $x^2>y^2$, and zero elsewhere. This is smooth, satisfies the equation, and has different values at $(1,0)$ and $(-1,0)$ despite equal $u$. The characteristic components of a hyperbolic first-order equation explain this obstruction. At the origin the coordinate map is singular and the original equation reads $0=0$; the boxed answer is the intended local form on a regular branch, or a global form under an additional identification of branches.

For real $\lambda >0$, squaring the two nonnegative moduli gives $|z-i{|}^2={\lambda }^2|z+i{|}^2$. In Cartesian coordinates this isAssume $\lambda \ne 1$ and put $h=(1+{\lambda }^2)/(1-{\lambda }^2)$. Dividing by $1-{\lambda }^2$ yields $x^2+y^2-2hy+1=0$. Since $z{z}^{\ast }=x^2+y^2$ and $b{z}^{\ast }+b^{\ast }z=2\mathrm{Re}(b{z}^{\ast })$, the requested complex equation hasIndeed these choices make the mixed term $-2hy$. Completing the square givesThe Circles of Apollonius therefore haveThe radius is positive for the stated range. No spurious point at $z=-i$ has been introduced: at that point the undivided equation would require $4=0$. Positive $\lambda <1$ gives a circle above the real axis, while $\lambda >1$ gives one below it.

A quotient with nonzero denominator has zero complex modulus exactly when its numerator is zero. ThusAt that point $z^{\ast }-i=-2i\ne 0$, so the solution is admissible.

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^{\ast }-i|=|z+i|$, the case $\lambda =1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,This locus is the real axis, the perpendicular bisector of the points $i$ and $-i$ in the complex plane. The excluded denominator point does not lie on it.

The polynomial factors asThus the repeated polynomial root is $x=2$, of multiplicity two. Its repetition can also be checked from $p(2)=p^{\prime }(2)=0$ and $p^{\prime \prime }(2)=14\ne 0$.