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For this homogeneous tangent differential equation, set $u=y/x$ on an interval not containing $x=0$. Since $y^{\prime }=u+x{u}^{\prime }$, cancellation gives $x{u}^{\prime }=2\tan u$. Away from zeros of $\tan u$, separation and integration giveRestoring a signed constant yields the compact implicit solutionA complete real branch description isHere $C$ is real and the interval must satisfy $x\ne 0$ and $|C{x}^2|<1$. The strict inequality excludes points with $\cos (y/x)=0$, where the original tangent function is undefined. Separating divided by $\tan u$, so its zeros must be checked separately: $u=k\pi$ gives $y=k\pi x$. These are exactly the $C=0$ branches already included. The branch integer is fixed on each connected solution interval; it cannot be changed arbitrarily at a regular point.

A real square orthogonal matrix satisfies $Q^{T}Q=I$. A complex square unitary matrix satisfies $U^{†}U=I$, where $†$ denotes conjugate transpose. A Hermitian matrix satisfies $H^{†}=H$. The distinction between transpose and conjugate transpose matters for imaginary entries.

For the three displayed matrices, in their source order:

For the first matrix, $A^{†}=-A$ and $A^2=-I$, so $A^{†}A=I$ but $A^{†}\ne A$. It also fails $A^{T}A=I$, even if complex orthogonality is considered. The second matrix is real diagonal with entries $1,-1$: its transpose, conjugate transpose and inverse all equal itself. In the third matrix the imaginary off-diagonal entries are conjugate pairs, establishing the Hermitian matrix condition. Its first column has squared norm $1+3=4$, so it is not a unitary matrix; it is not real orthogonal, and the complex bilinear first-column square is $1-3=-2\ne 1$ as well.

Subtracting the first equation from the third gives $4x-4z=6$, so $z=x-3/2$. The second equation then becomes $(\alpha -2)x=\beta -3$. The third fixes $y=7(1-x)/5$. Therefore the three possibilities areFor the first case the coefficient determinant is $20(\alpha -2)\ne 0$, confirming uniqueness. At $\alpha =2$ its matrix rank is two. The augmented matrix has the same matrix rank only for $\beta =3$, giving one free variable in a linear system; otherwise the linear system is inconsistent.

Use parametric equations for the two lines,Their cross product is $\mathbf{u}\times \mathbf{v}=(-3,0,3)\ne 0$. Projection of the separation of the base points onto this common normal gives the shortest distance between skew lines:This projection is a lower bound on every connecting segment because adding either line's direction leaves its normal component unchanged. It is attained: imposing perpendicularity of $\mathbf{p}+t\mathbf{u}-\mathbf{q}-s\mathbf{v}$ to both directions gives $t=2/3$ and $s=7/3$. The corresponding points are $(10/3,11/3,1/3)$ and $(1/3,11/3,10/3)$; their difference is $(3,0,-3)$, perpendicular to both lines and of length $3\sqrt{2}$.

Put $z=x+iy$. The expression whose complex argument is constrained becomesIts principal argument is $\pi /2$ exactly when its real part is zero and its imaginary part is strictly positive. HenceThe locus consists of the upper-right half of the upper hyperbola branch, $y=\sqrt{x^2+1}$ with $x>0$, and the lower-left half of the lower branch, $y=-\sqrt{x^2+1}$ with $x<0$. The vertices $z=\pm i$ are excluded because $z^2+1=0$ there and zero has no complex argument. The remaining halves have negative imaginary part and give argument $-\pi /2$, not $\pi /2$. The right panel of the original illustration marks the excluded vertices with open circles.

Write $27i=27e^{i(\pi /2+2\pi k)}$. Every cube root has complex modulus $3$ and an angle one third of one of these complex arguments:Equivalently the three complex roots areThey lie on the circle of radius $3$ centred at the origin in the Argand plane, spaced by $2\pi /3$. Further integer values of $k$ repeat these three points. Cubing any of them multiplies its angle by three and cubes its complex modulus from $3$ to $27$, giving $27i$.

Squaring first is simpler than expanding the tenth powers. Since $(1-i{)}^2=-2i$ and $(1+i{)}^2=2i$, while $i^4=1$, the requested Cartesian forms areThus the real part is zero in each case, and the imaginary parts are $-32$ and $32$. The results are complex conjugates, as also follows because the two inputs are complex conjugates and the fifth-power polynomial has real coefficients.