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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.

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).

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.

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.

The given complex numbers have positive real parts, so $\arg z_1=\arctan A$ and $\arg z_2=\arctan B$. Their product isBy the assumed complex argument ranges, this product also has positive real part, hence $1-AB>0$. Its complex argument is consequently $\arctan [(A+B)/(1-AB)]$, with no extra multiple of $\pi$. Since multiplication adds complex arguments,This proves the arctangent addition formula under the stated branch restriction. For arbitrary real $A,B$, the principal inverse tangent formula can require adding or subtracting $\pi$ when $AB>1$.

Write $z_j=x_j+i{y}_j$. Multiplication givesTherefore the real part is the dot product of the corresponding vectors:For nonzero vectors, this is $|z_1||z_2|\cos (\arg z_1-\arg z_2)$, measuring their alignment.

For $\theta \notin 2\pi \mathbb{Z}$, use the finite geometric series with ratio $e^{i\theta }$:The second equality follows by factoring $1-e^{it}=-2i{e}^{it/2}\sin (t/2)$ in numerator and denominator. Taking real parts proves the finite trigonometric sum identityWhen $\theta =2p\pi$, every cosine is $1$, so the sum is $\boxed{N}$. This also gives the continuous limiting value at the excluded arguments.

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.

Using the complex cosine identity,we findIts real part is exactly the prescribed function. Thus a harmonic conjugate is $v=y\cos x\cosh y-x\sin x\sinh y$, and the general answer isTo check the only freedom, two analytic functions with this real part differ by a function with zero real part. The Cauchy-Riemann equations then force both partial derivatives of its imaginary part to vanish, so it is constant on the connected domain. No normalization was supplied to fix $C$.

For the upper Apollonius circle, let its constant ratio be $P>1$. Matching the center and radius from part (i) givesSubtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. AlsoThe relevant inverse hyperbolic cosine is positive. Now $\log |(z+ib)/(z-ib)|$ is the real part of a local holomorphic logarithm, hence a harmonic function away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.

On $y=0$ the numerator and denominator have equal modulus, so the logarithm is zero. On the circular boundary its value is $\log P=\mathrm{arcosh}(y_0/R)$. Therefore the harmonic potential between a grounded line and a circular boundary isThis verifies Laplace's equation and both prescribed Dirichlet boundary conditions. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.

Use the principal complex logarithm:The excluded nonpositive real ray allows a continuous choice of argument, and $L(1)=0$. For $z=x+iy$, its real part is $u=\frac{1}{2}\log (x^2+y^2)$. Differentiating the chosen local argument givesThese continuous partial derivatives satisfy the Cauchy-Riemann equations, proving holomorphy on the slit domain. Its derivative is

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.