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The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.

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

Expand the square using $i^2=-1$:Thus the real part and imaginary part are

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

From $X=s^2-4$ and $Y=4s$, elimination of $s$ givesThis makes the real part a quadratic function of the imaginary part. There is no sign ambiguity because $Y$ determines $s=Y/4$ uniquely. The calculation illustrates power maps of a reciprocal-coordinate hyperbola.

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

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.

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).

By De Moivre's theorem, $\sin 5\theta$ is the imaginary part of $(\cos \theta +i\sin \theta {)}^5$. The binomial theorem gives, with $s=\sin \theta$,Use ${\cos }^2\theta =1-s^2$ and simplify:The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm 1/2$. The complete list in $0\le \theta <2\pi$ is

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

Direct multiplication, using $i^2=-1$, givesThus the real part is $\boxed{18}$ and the imaginary part is $\boxed{1}$.

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

At a point of complex differentiability, the complex derivative is independent of the direction of approach. Taking increments along the real axis gives $f^{\prime }=u_x+i{v}_x$; taking imaginary increments gives $f^{\prime }=v_y-i{u}_y$. Equality of their real parts and imaginary parts proves the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

For the specified real part, observe that $z^2{e}^z$ has exactly that real part. Equivalently, integrating $v_y=u_x$ and checking $v_x=-u_y$ givesThusThe freedom is only a constant imaginary part: the difference of two such holomorphic functions has zero real part, so its Cauchy-Riemann equations make it constant on each connected domain.

For $f=u+iv$ holomorphic in $z=x+iy$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal vectors to the two level curves obeyAt a point where $f^{\prime }\ne 0$, both gradients are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.

TakeIts real part is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its imaginary part is $v=2xy$, so the second family is $v=2\beta$. Since $\beta \ne 0$, their intersections exclude the origin, where $f^{\prime }(z)=2z$ would vanish. Thus these hyperbolas meet orthogonally as required.

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

Reflecting the solution across $\theta =\pi \alpha$ preserves its Dirichlet boundary conditions and outer boundary data, so uniqueness makes the solution reflection-symmetric. Sinceonly odd angular Fourier modes survive. Put $\rho =(r/R{)}^{1/(2\alpha )}$ and $\psi =\theta /(2\alpha )$. On the symmetry axis $\psi =\pi /2$, comparison of the convergent power series givesThusThis is the imaginary part of $\sin (\rho e^{i\psi })$, so a useful closed form isAt either straight edge $\sin \psi =0$, while on the symmetry axis the formula is $\sinh \rho$. Every term is harmonic in the sector, and the factorial denominator gives convergence up to $r=R$, verifying the full solution.

Consider the counterclockwise unit circle and setThe parametrization $z=e^{i\theta }$ givesThe imaginary part is zero: replacing $\theta$ by $2\pi -\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The real part is precisely the defining Bessel integral, so $I=J_n(x)$.

Both exponential series converge absolutely and uniformly on the circle. Their product may therefore be integrated term by term:Uniform absolute convergence can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The residue is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The Cauchy coefficient formula consequently yieldsThis proves the Bessel function of the first kind series directly from its integral, including the sign and factorial normalization.

List the roots with their multiplicities. By the factor theorem, $Q(z)=a{\prod }_{k=1}^n(z-{\alpha }_k)$ with $a\ne 0$. The product rule, followed by division where $Q(z)\ne 0$, gives its logarithmic derivative:Let $Q^{\prime }(\beta )=0$. If $Q(\beta )=0$ too, then $\beta$ is already one of the roots, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all roots have positive imaginary parts but $\mathrm{Im}\beta \le 0$, thenfor every $k$. Their sum cannot be zero. Thus $\mathrm{Im}\beta >0$.

For the unit disk conclusion, the same sum gives an especially transparent convex combination. Taking complex conjugates and setting $w_k=|\beta -{\alpha }_k{|}^{-2}>0$ givesTherefore, if every $|{\alpha }_k|<1$,The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the Gauss-Lucas theorem: every critical point of a nonconstant polynomial lies in the convex hull of its roots. The common-root case, handled before division, is essential when multiplicities occur.

Write $b_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The given limiting identity and the Cauchy coefficient formula from part (a) give the boundary contour integralThe contour passes through $z=1$, so its value here is an improper integral using the one-sided branch values. Parametrize its open arc by $z=e^{i\theta }$, $0<\theta <2\pi$. Substitution of the analytic branch of a square root calculated in part (b) givesThe factor $e^{-in\theta }$ comes from cancelling $e^{i\theta }$ in $dz$ against $z^{n+1}$; the extra $e^{-i\theta /4}$ comes from the chosen branch. Keeping both phases prevents a change in the half-integer frequency.

Now put $t=\theta /2$. ThenDividing by $i-1$ givesTaking the real part therefore provesAll these improper integrals converge absolutely: the numerators are bounded, while $(\sin t{)}^{-1/2}$ is of order $t^{-1/2}$ at zero and $(\pi -t{)}^{-1/2}$ at $\pi$. Hence the substitutions and passage to the real part are legitimate. The imaginary part also shows that the corresponding sine integral has the same positive value, which checks the phase calculation.