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Multiply by the complex conjugate of the denominator:Thus the real part and imaginary part are respectively

Multiply numerator and denominator by the complex conjugate of the denominator:Thus the real part and imaginary part are $\boxed{\mathrm{Re}=-5/13,\mathrm{Im}=-12/13}$.

Multiply numerator and denominator by the complex conjugate:Therefore, away from the origin,

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.

Multiply numerator and denominator by the complex conjugate $2+i$:Thus the real part and imaginary part are

Write $z=X+iY$. Its coordinates are $X=t$, $Y=1/t$, so $XY=1$. For the given positive range of $t$, this traces a segment of the first-quadrant hyperbola, moving rightward and downward:The complex conjugate is $z^{\ast }=t-i/t$. It is the reflection in the real axis, tracing $XY=-1$ in the fourth quadrant, rightward and upward:Neither trajectory crosses either axis, because $t$ and $1/t$ remain nonzero. The left panel of the figure marks the endpoints and directions.

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).

Multiply numerator and denominator by the complex conjugate $4-3i$:Thus the real part is $\boxed{6/25}$ and the imaginary part is $\boxed{-17/25}$.

With $z_1=a+ib$ and $z_2=c+id$, direct multiplication gives $z_1{z}_2=(ac-bd)+i(ad+bc)$. Taking the complex conjugate yields

A counterclockwise rotation through $\alpha$ about the origin in the complex plane sends $z_j$ to $e^{i\alpha }{z}_j$. By part (a), and because the complex conjugate of $e^{i\alpha }$ is $e^{-i\alpha }$,Thus $Q$ is invariant under the common rotation.

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

The characteristic equation of the reduced homogeneous linear differential equation is $m^2-\beta =0$. Its polynomial roots change type with the sign of $\beta$, givingFor $\beta =0$, the zero polynomial root is repeated, so the second independent solution is $t$. For $\beta <0$, the polynomial roots are the complex conjugate pair $\pm i\sqrt{-\beta }$, and the displayed sine and cosine give a real basis of solutions. The constants are arbitrary real numbers. In every case $x(t)=y^{\prime }(t)-\cos t$ reconstructs the corresponding general solution of the coupled system.

For any Hermitian matrix $K$, the scalar $x^{†}Kx$ is real because its complex conjugate equals $x^{†}{K}^{†}x=x^{†}Kx$. Applying this to $H,S$ in part (a) givesThese are the real and imaginary quadratic forms associated with the numerical range of $A$.

For sufficiently rapidly decaying functions, substitute the inverse Fourier transform of $g$ and interchange the integrals:The bracketed integral is the complex conjugate of $\tilde{f}(k)$. This is Parseval's identity for the chosen normalization. One may first prove it for Schwartz functions, where the interchanges are justified, and then extend it to $L^2$ by the Plancherel theorem.

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

The chain rule givesTo make the directions explicit, set $z=x+iy$ and $D=(2+\sin x\cosh y{)}^2+{\cos }^{2}x{\sinh }^{2}y$. Multiplying by the complex conjugate of the denominator gives the logarithmic flow in a sine-mapped half-strip:In the open strip $U>0$. The vertical component is negative near the lower left wall where $\cosh y+2\sin x<0$, and positive above that curve. On the bottom the flow is rightward. On the right wall it is upward, whereas on the left wall it points downward below $z_{\ast }$ and upward above $z_{\ast }$. The field is tangent to the boundaries except at the source and vanishes at the two bottom corners. At large $y$, $U-iV\to -i$, so it tends to a uniform upward field $(0,1)$.

Near $z_{\ast }$, $G^{\prime }(z)\sim 1/(z-z_{\ast })$, showing a radial boundary source. The streamlines are the curves $\mathrm{Im}G=\arg (2+\sin z)=\text{constant}$; the figure shows their orientation without reusing the exam's graphics.

Split the inverse transform into the intervals $|\omega |<\mu$ and $|\omega |>\mu$, and in each case pair the positive-frequency integral with its complex conjugate at negative frequency. The low-frequency pair becomessince $B=0$ and the decaying spatial factor is real. In the high-frequency positive half, replace $A(\omega )$ by $B(-\omega {)}^{\ast }$. The negative half is again its conjugate. Combining the two pieces givesEvery integrand satisfies the massive wave equation because ${\omega }^2=k^2+{\mu }^2$. Linearity, and taking the real part, preserve the equation whenever differentiation under the integrals is justified, or distributionally under the corresponding spectral assumptions. The construction is exactly the real bounded Fourier modes of the Klein-Gordon equation decomposition.

The displayed expression uses ordinary spectral amplitudes. If distributions supported at $\omega =\pm \mu$ are admitted, a further bounded threshold contribution $c_1\cos \mu t+c_2\sin \mu t$, independent of $r$, is possible. It is not a missing propagating or evanescent branch of the open frequency intervals.

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.

The $z$ direction is fixed, giving the eigenvalue one. The planar block has characteristic polynomialThus the full list of eigenvalues isThe last two are real precisely when $\sin \varphi =0$, so all three are real exactly for $\boxed{\varphi \in \pi \mathbb{Z}}$. At even multiples of $\pi$ the matrix is the identity and every vector is fixed. At odd multiples it is a half-turn: the axial eigenvalue remains $+1$, while the whole perpendicular plane has eigenvalue $-1$.

For other angles, the only real eigenvector direction is the rotation axis; the perpendicular plane is rotated rather than preserving an individual real direction. Its complex eigenvectors can be taken as $(1,-i,0{)}^T$ and $(1,i,0{)}^T$ for $e^{i\varphi }$ and $e^{-i\varphi }$, respectively. These represent the two complex conjugate planar eigenvalues.

Let $A=A^{†}$ be a Hermitian matrix and $Av=\lambda v$ with $v\ne 0$. The scalar $v^{†}Av$ equals its own complex conjugate, becauseSince $v^{†}Av=\lambda v^{†}v$ and the squared norm $v^{†}v$ is positive, $\boxed{\lambda \in \mathbb{R}}$.

Now take $Au=\lambda u$ and $Av=\mu v$. The inner product satisfiesSubtracting gives $(\mu -\lambda )u^{†}v=0$. Thus distinct eigenvalues imply $\boxed{u^{†}v=0}$, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.