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For $z\ne 0$, rationalizing the denominators gives $1/z=(x-iy)/(x^2+y^2)$ and $1/\overline{z}=(x+iy)/(x^2+y^2)$. Their imaginary parts cancel, soThis also follows because the summands are complex conjugates.

For $z\ne 0$, write $z=r{e}^{i\theta }$ with $r=\sqrt{x^2+y^2}>0$. The complex logarithm has values $\ln r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. ThereforeFor a specified single-valued branch, choose the corresponding continuous determination of the argument. In particular, the principal value uses $\mathrm{Arg}z\in (-\pi ,\pi ]$; it is not represented correctly by $\arctan (y/x)$ in every quadrant. The complex logarithm is undefined at zero.

Using $\cos z=(e^{iz}+e^{-iz})/2$ and collecting real and imaginary terms gives the complex cosine identity $\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y$. ThusThe sign follows from the factors $e^{-y}$ and $e^y$ in the two complex exponentials.

The complex exponential function factorizes as $e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y)$. HenceThe real exponential multiplies both components; it does not change the argument.

Interchange the two dummy indices $j,k$. The product $a_j{a}_k$ is symmetric under this interchange, while the Levi-Civita symbol changes sign. Thus the contraction equals its negative and is the zero vector, equivalently $\mathbf{a}\times \mathbf{a}=0$.

Compare with $(\mathbf{b}\times \mathbf{a}{)}_i={ϵ}_{ijk}{b}_j{a}_k$. The scalar factors commute, so the answer isThe order of the contracted indices matters because the Levi-Civita symbol is antisymmetric.

The contracted Kronecker delta is ${\delta }_{ll}=3$. The other scalar contraction is $a_i{C}_{ij}{b}_j={\mathbf{a}}^{T}C\mathbf{b}$, while $C_{ik}{d}_i$ is the $k$th component of the matrix transpose applied to $\mathbf{d}$. Hence the resulting vector isBoth terms have the same single free index $k$, as required by the Einstein summation convention.

This is the incorrectly formed expression. In the Einstein summation convention, an index in a term is either free and occurs once, or summed and occurs twice. Here $i$ occurs three times in the same product, so it cannot consistently be classified as either. Renaming a dummy index before multiplying would give a valid different expression, such as ${\delta }_{kk}{\delta }_{ij}=3{\delta }_{ij}$, but does not make the printed expression valid.

The two free indices are the row and column indices of a matrix. The Kronecker delta has value one on its diagonal and zero off it, so the answer is the $3\times 3$ identity matrix $I$.

Suppose an analytic extension $F$ existed on the punctured plane. The identity theorem would extend $F^2=z$ from $D$ to that connected domain, so $F$ would never vanish and differentiation would give $2F^{\prime }/F=1/z$. Integrate around the positively oriented unit circle. The argument principle says $\int F^{\prime }/F$ is $2\pi i$ times the integer winding number of the closed curve $F(e^{it})$ about zero, whereasThis would make that winding number $1/2$, a contradiction. Hence no such analytic extension exists. Equivalently, continuation of a square root once around zero reverses its sign, preventing a single-valued holomorphic function on the punctured plane.

The slit plane is connected and contains $D$. If two holomorphic functions on it extend the same $f$, their difference vanishes on the nonempty open disc $D$. The identity theorem forces their difference to vanish on the entire slit plane. Thus there is at most one analytic extension of the chosen $f$. Choosing the opposite square root initially changes that chosen function; it does not produce two extensions of the same function.

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.

Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial isIt is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\le j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves that $P(x+1)$ is irreducible, and therefore $P$ is irreducible. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an irreducible polynomial arising as a cyclotomic polynomial.

A cubic over a field is reducible exactly when it has a root in that field: a nontrivial factorization must contain a degree-one factor. By the rational root theorem, a rational root of this monic integer polynomial must be an integer dividing its constant coefficient, hence either $1$ or $-1$. Substitution gives $-2$ and $4$, respectively. Neither is zero. Thus the cubic is irreducible over $\mathbb{Q}$.

Use the Eisenstein criterion with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore it is irreducible over $\mathbb{Q}$.

This relative absolute-error loss weights each parameter by $1/\theta$. Its posterior expected loss isThe effective density for the Bayes estimator under weighted absolute loss is proportional to $\pi (\theta \mid x)/\theta$, hence an exponential distribution of rate $b$. Its median solves $1-e^{-ba}=1/2$.

More explicitly, for $a\ge 0$, differentiating the split integrals at $\theta =a$ gives $R_x^{\prime }(a)=b(1-2e^{-ba})$ and $R_x^{\prime \prime }(a)=2b^2{e}^{-ba}>0$. For $a<0$, the loss has derivative $-b$, so its minimum cannot lie there. The unique global minimum is thereforeThis is the weighted posterior median; the ordinary median of the Gamma posterior would minimize unweighted absolute loss and would give a different answer.

Under squared-error loss, the posterior expected loss decomposes asIt is uniquely minimized by the posterior mean. Integrating the Gamma posterior gives $\mathbb{E}[\theta \mid x]=b^2{\int }_0^{\infty }{\theta }^2{e}^{-b\theta }d\theta =2/b$. Hence the Bayes estimator is

Take the Fourier transform in $x$ of the Laplace equation. Writing $U(s,y)$ for the transformed solution gives $U_{yy}-s^{2}U=0$. Its boundary values givewith the multiplier at $s=0$ understood as $1-y$. The particular data simplify this to $U(s,y)=2\sinh (s(1-y))$ for $|s|<1$, and zero outside. Thus, with $a=1-y$, the Fourier solution of the strip Dirichlet problem isThe displayed closed expression applies in the open strip. The integral defines its continuous boundary values, including the removable point $(x,a)=(0,0)$. It gives $u(x,0)=g(x)$ and $u(x,1)=0$. Differentiation under this finite integral verifies $u_{xx}+u_{yy}=0$ directly, since each integrand has opposite second derivatives in $x$ and $y$.

The boundary conditions alone do not specify a unique solution without a growth restriction: adding $C{e}^{\pi x}\sin (\pi y)$ preserves both boundary values and harmonicity. The constructed solution is bounded, and it is unique among bounded solutions continuous on the closed strip. To see this, apply the maximum principle for harmonic functions to a bounded real or imaginary component $w$ of the difference. For $0<\delta <\pi$, the positive harmonic function $B(x,y)=\cosh (\delta x)\cos (\delta (y-1/2))$ has a positive minimum factor in $y\in [0,1]$. For any $\epsilon >0$, the functions $\pm w-\epsilon B$ are nonpositive on the horizontal sides of a sufficiently wide rectangle and on its vertical sides, because $w$ is bounded and $B$ grows there. The maximum principle gives $|w|\le \epsilon B$ at every fixed point. Letting $\epsilon \downarrow 0$ proves uniqueness in this natural Fourier solution class.

With the Fourier transform convention of 5A, use the printed Fourier inversion formula directly. The possible choice of values at $s=\pm 1$ has no effect on the integral. Integration over the support givesCombining the two fractions yieldsIn particular $g$ is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.

At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2{)}^2=12\ne 0$. The holomorphic implicit function theorem states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne 0$, there is a unique holomorphic function $z=\gamma (c)$ near $c_0$, with $\gamma (c_0)=z_0$ and $P(\gamma (c),c)=0$. It therefore supplies an analytic root branch $\gamma (c)$ near zero. Differentiate its equation:It follows that ${\gamma }^{\prime }(0)=i/6$ andFor sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus that root lies above the real axis for all sufficiently small positive $c$.