Use the exponential definition of the hyperbolic tangent:The complex exponential has modulus one here, so $\mathrm{Re}z=0$. Its arguments satisfy $2\mathrm{Im}z=-\pi /2+2k\pi$. Thus all solutions areThe denominator is then $1-i\ne 0$, so no pole was admitted by cross-multiplication. Substitution also follows from $\tanh (i\theta )=i\tan \theta$ and the period $i\pi$.

The complex number $-1+i$ has complex modulus $\sqrt{2}$ and argument $3\pi /4$. A cube root has modulus $(\sqrt{2}{)}^{1/3}=2^{1/6}$, and its argument is one third of $3\pi /4+2k\pi$. Hence the three distinct roots areTheir displayed arguments are $\pi /4$, $11\pi /12$, and $19\pi /12$; the last is equivalently $-5\pi /12$ in the principal range.

Resolve the input along the three perpendicular eigendirections. The linear map removes the component along $(1,2,1)$, stretches the component along $(1,-1,1)$ by $12$, and reverses and stretches the component along $(1,0,-1)$ by $6$. Every output lies in the plane perpendicular to $(1,2,1)$. A nonzero input in the kernel of a linear map $A$ is sent to zero; other nonzero inputs remain in that output plane. Equivalently, take the orthogonal projection onto this plane, then scale by the two signed eigenvalues along its perpendicular eigendirections.

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.

The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

Expanding the determinant along the first row givesThe matrix trace is the sum of the diagonal entries, $\mathrm{tr}A=1+4+1=6$. Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus $A$ is singular; the determinant alone does not specify the two other eigenvalues.

For the partial fraction decomposition, write $A/(x-4)+B/(x+9)$. Clearing denominators gives $13(x+1)=A(x+9)+B(x-4)$. Substitution at $x=4$ and $x=-9$ yields $A=5$ and $B=8$, hence

The original line meets the horizontal axis at $(-3/2,0)$. A perpendicular line has slope $-1/2$, the negative reciprocal of $2$, soSubstituting $x=-3/2$ verifies the required common point.

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

Write the rational expression as $(x^2+4{)}^{-1}$ and apply the chain rule:The denominator is positive for every real $x$, so the derivative is defined throughout the real line.

The product rule for divergence, $\nabla \cdot \mathbf{r}=3$, and the result in (a) give, for $r>0$,This is a pointwise identity away from the origin, where the radial unit vector is undefined.

For $r>0$, use the gradient of a radial function with $g(r)=1/r$:Equivalently each component is $-x_j/r^3$ by differentiating $r^2=x^2+y^2+z^2$. The origin is excluded from the function's domain.

For the divergence in Cartesian coordinates, differentiate each component with respect to its own coordinate:The third component contributes zero because it is independent of $z$.

First take $F_y=x{e}^{\sin (xy)}\cos (xy)$, then differentiate with respect to $x$. The product rule and chain rule giveThe function is smooth, so the two mixed partial derivatives agree.

Project the radius vector onto the $x$-$y$ plane. The polar angle $\theta =\pi /6$ lies between the positive $z$ axis and $OP$. The azimuth $\varphi =\pi /3$ lies in the $x$-$y$ plane between the positive $x$ axis and the projection of $OP$, measured toward positive $y$.

Use the convention that $\theta$ is the polar angle from positive $z$, and $\varphi$ the azimuth from positive $x$. The radius isThen $\cos \theta =z/r=\sqrt{3}/2$, while $\tan \varphi =y/x=\sqrt{3}$ and both planar coordinates are positive. Thus the spherical coordinates are

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.

For $m>-1$, integrate the base power to obtain $J(m)={\int }_0^1{x}^{m}dx=1/(m+1)$. Since ${\partial }_m^n{x}^m=x^m(\log x{)}^n$, differentiating $n$ times givesHere $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x{|}^n$, whose integral becomes ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$ under $x=e^{-t}$ and is finite. The dominated convergence theorem establishes the needed differentiation under the integral sign and the logarithmic moments of a power on the unit interval.

Under the regularity and integrability conditions for the Leibniz integral rule,The first two terms account for the moving endpoints; the last accounts for the integrand's own parameter dependence.

For a real number $x$, the hyperbolic functions satisfy $\cosh x>0$ and $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$. Hence $\cosh x=1/\sqrt{1-u^2}$, and multiplying it by $\tanh x=u$ givesThe sign of the numerator preserves the sign of $\sinh x$.