The chain rule gives $f_x=f_u{u}_x+f_v{v}_x$ and $f_y=f_u{u}_y+f_v{v}_y$. Since $u_x=2x$, $u_y=2y$, $v_x=2x{y}^2$ and $v_y=2x^{2}y$, the answer isEvery partial derivative of $f$ on the right is evaluated at $u=1+x^2+y^2$, $v=1+x^2{y}^2$.

For $a=1$, integrate $Q=2xy\sin x$ with respect to $y$ to obtainIts other partial derivative is $f_x=y^2\sin x+x{y}^2\cos x+C^{\prime }(x)$. Comparing with $P$ gives $C^{\prime }(x)=0$. Thus the most general potential on the plane is

An exact differential form must have $P_y=Q_x$. Direct partial differentiation givesThus $(1-a)xy\cos (ax)=0$ must hold identically. Since $xy\cos (ax)$ is not identically zero for a fixed nonzero $a$, the only possibility isThe explicit potential in part (ii) verifies sufficiency.

For a vertical slice through the triangle, $0\le x\le 2$ and $0\le y\le x/2$. Reversing the order by Fubini's theorem givesThe agreement checks that both orders describe the same region.

With $y$ held fixed, the inner antiderivative givesThe remaining integral is therefore

The inequalities describing the region are $0\le y\le 1$ and $2y\le x\le 2$. It is the closed triangle bounded by $y=0$, $x=2$ and $x=2y$, with verticesA horizontal slice runs from the sloping edge $x=2y$ to the vertical edge $x=2$.

The stated $x$ bounds depend on $y$, so $y$ must be held fixed while the integral over $x$ is performed. The intended iterated integral isThe original PDF places the two sets of bounds beside the opposite integral signs relative to the printed $dxdy$. That placement cannot be read literally as a conventional nested integral; the dependence of the lower $x$ bound on $y$, and the explicit instruction to integrate over $x$ first, specify the intended region unambiguously.

Multiplication by $i$ is a counterclockwise rotation through $\pi /2$: it maps coordinates $(X,Y)$ to $(-Y,X)$. The centre $(2,2)$ therefore becomes $(-2,2)$, and the long axis becomes vertical. For $w=X+iY$, the resulting ellipse isIts horizontal semiaxis is $8$ and its vertical semiaxis is $16$, as shown in the final panel of the figure in part (b).

Adding $4+3i$ is a translation by $(4,3)$ in the complex plane. The ellipse of part (i) therefore keeps its semiaxes and axis directions, with new centre $(2,2)$. Writing $v=X+iY$, its equation isThe third panel in the figure in part (b) shows the translated ellipse.

Write $z=x+iy$ and use the complex conjugate $z^{\ast }=x-iy$. ThenThus the real part is stretched by a factor of two while the imaginary part is unchanged. If $u=X+iY$, substitution into the circle equation givesThis is an ellipse with centre $(-2,-1)$, horizontal semiaxis $16$ and vertical semiaxis $8$. It is the second panel in the figure in part (b).

The inner power takes the values $r_k=e^{-(\pi /2+2\pi k)}$. For each such real exponent, all values of the outer complex exponentiation are obtained from all complex logarithm values of its base $i$:Writing ${\Theta }_{k,m}=(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}$, the answer isAll values lie on the unit circle. The principal choice is $k=m=0$, giving $\cos [(\pi /2)e^{-\pi /2}]+i\sin [(\pi /2)e^{-\pi /2}]$. In particular the two nestings in parts (i) and (ii) are different: complex exponentiation is not associative.

Use the full set of complex logarithm values of $i$:By the definition of complex exponentiation, the inner power can therefore take any value $i^i=\exp [-(\pi /2+2\pi k)]$, a positive real number. Its complex logarithm values are $-(\pi /2+2\pi k)+2\pi im$, where $m\in \mathbb{Z}$. ThusAll possible real parts and imaginary parts are consequentlyChoosing the principal complex logarithm at both stages gives the particular value $-i$. The independent logarithm choices are needed because the question asks for all possible values.

The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. Hence it has one positive and one negative eigenvalue (explicitly $3\pm \sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.

The second partial derivatives are $f_{xx}=4$, $f_{xy}=f_{yx}=3+2y$ and $f_{yy}=2+2x+30y$. Evaluating at the origin gives the Hessian matrix

The matrix trace and determinant of the matrices in part (i) areSince $\mathrm{Tr}X=0$ and $\det X=2$, the extra conditions force $\alpha =0$ and ${\beta }^2=1$. HenceBoth choices are real, commute with $X$, and satisfy the required trace and determinant conditions.

Let $Y=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$. The commutativity condition givesEquality is equivalent to $s=p$ and $q=-2r$. ThereforeEach displayed matrix commutes with $X$, since it is a polynomial in $X$. The centralizer of a nonscalar two by two matrix explains why this two-parameter form is typical.

For $\lambda \ne \pm 2$, the reduced equation gives $y=(3-\lambda )/(4-{\lambda }^2)$. Back substitution into $x=1-\lambda y$ and $z=1-4y$ gives all solutions:The nonzero determinant proves that there are no other solutions at these parameter values.

The determinant is $\det L=4-{\lambda }^2$, so there is exactly one solution whenever $\lambda \ne \pm 2$. A zero determinant does not alone distinguish inconsistency from infinitely many solutions, so check those values separately.

Eliminate $x$ and $z$ using $x=1-\lambda y$ and $z=1-4y$. The remaining equation isFor $\lambda =2$ this reads $0=1$, and for $\lambda =-2$ it reads $0=5$. Both are inconsistent. Consequentlyand each permitted value has a unique solution.

Collecting the coefficients in the given equation order yields the system of linear equations in matrix form

Use Einstein notation, with repeated suffixes summed. The matrix multiplication and transpose giveInterchange the free indices, then relabel the dummy indices $k,\ell$:The negative sign uses $B_{k\ell }=-B_{\ell k}$. Therefore $C$ is skew-symmetric. No orthogonality of $A$ is needed for this part: matrix congruence preserves skew-symmetry.