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Multiply numerator and denominator by the complex conjugate $5+4i$:The real part and imaginary part are therefore

The characteristic polynomial isIts roots give the two eigenvaluesTheir sum is zero, agreeing with the trace of the matrix.

The first partial derivatives factor asBoth vanish at the origin, verifying that it is a stationary point. Since $2x^2+1>0$, every stationary point has $y=0$; the first equation then gives $x=0,\pm 1$. Thus one requested additional point is $\boxed{(1,0)}$; the other is $\boxed{(-1,0)}$.

The integrand is an even function, since both $x$ and $\sin x$ are odd functions. Using integration by parts,The endpoint cosine terms vanish, leaving the sine contribution.

The first field is ${\mathbf{F}}_1=(z-y,x,-x)$, so direct curl calculation givesIt is not conservative because its curl is nonzero. The second is a radial gradient field:The scalar expression is a smooth function and single-valued on all of space, so ${\mathbf{F}}_2$ is a conservative vector field. Its potential of a conservative vector field is $\Phi =-(1+r^2{)}^{-1/2}$. Parts (a) and (b) check path dependence directly.

Use the real Fourier series convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$. The function is an even function, so $b_n=0$. For noninteger real $\mu$,For $n\ge 1$, the product-to-sum identity givesThus the Fourier series of a noninteger-frequency cosine isIts Fourier coefficients are $O(n^{-2})$, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting $\mu =1/3$ givesEvaluate at $x=0$, where $f(0)=1$, and multiply by $\pi /\sqrt{3}$:Comparing with the required form gives $\boxed{A=3,B=3}$.