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For $\mathrm{Im}z>0$, close the $t$-contour in the upper half-plane. Jordan lemma removes the large semicircle, and the only enclosed pole is the order-$n$ pole at $t=z$. By the residue at a pole of order n,The residue theorem therefore gives

To continue the integral itself, move the $t$-contour downward locally as $z$ approaches and crosses the real axis, always keeping the pole above the contour. This analytic continuation by contour deformation defineswhere ${\mathcal{C}}_z$ passes below $z$. Closing ${\mathcal{C}}_z$ upward continues to enclose the pole, even when $\mathrm{Im}z\le 0$, sothere as well. The right-hand side is an entire function, so it is the unique analytic continuation of $F$ to the whole complex plane.

Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

PutThis slit complex plane is a simply connected domain. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

The inverse is nonconstant; locally at zero,If the assumed meromorphic function $\mathcal{H}$ had no poles, it would be entire. It is bounded on the compact closure of a fundamental parallelogram, and its two periods then make it bounded throughout the complex plane. The Liouville theorem would force it to be constant, a contradiction. Therefore $\mathcal{H}$ has at least one pole, as stated by the general result that a nonconstant elliptic function has a pole.

The uniformization theorem states that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere ${\mathbb{C}}_{\infty }$, the complex plane $\mathbb{C}$, or the open unit disc $\mathbb{D}$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.

For ${\mathbb{C}}_{\infty }$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is

Every conformal automorphism of $\mathbb{C}$ has the form $z\mapsto az+b$. If $a\ne 1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb{C},+)$ has rank zero, one, or two. The corresponding quotients arewhere $\Lambda$ is a rank-two lattice and $\mathbb{C}/\Lambda$ is a complex torus. This proves the completeness of the Riemann surfaces uniformized by the complex plane.

Now let $D\subseteq \mathbb{C}$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb{C}$, the covering map followed by $D\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the Little Picard theorem. The uniformization theorem therefore leaves onlywhich is the plane domain with two omitted points is hyperbolic result.

Finally let $j:{\mathbb{C}}^{\ast }\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the Great Picard theorem in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically toA generic point of the open image $j({\mathbb{C}}^{\ast })$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the Riemann-Hurwitz formula rules out branching and gives the same conclusion. Therefore the compact Riemann surface containing an embedded punctured plane is

The point $1+i$ has modulus $\sqrt{2}$ and argument $\pi /4$ modulo $2\pi$. Taking complex square roots halves the argument and takes the square root of the modulus:The two roots are opposite points in the complex plane, explicitly $\pm [\sqrt{(\sqrt{2}+1)/2}+i\sqrt{(\sqrt{2}-1)/2}]$.

Write $z=x+iy$. The complex modulus condition is equivalent to $(x-1{)}^2+y^2=4$. Thus the locus is a circle of radius two centred at $(1,0)$ in the complex plane. Its extreme points are $(-1,0)$, $(3,0)$, and $(1,\pm 2)$.

For real $\lambda$, the numerator and denominator have equal complex modulus:The denominator is always nonzero. HenceThus the real axis maps into the unit circle in the complex plane.

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^{\ast }-i|=|z+i|$, the case $\lambda =1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,This locus is the real axis, the perpendicular bisector of the points $i$ and $-i$ in the complex plane. The excluded denominator point does not lie on it.

With $z=x+iy$, the expression inside the modulus is $x+3i(y-1)$. Squaring its nonnegative modulus therefore givesThe locus is an ellipse in the complex plane, centred at $(0,1)$, with horizontal semiaxis $2$ and vertical semiaxis $2/3$. Its vertices are $(\pm 2,1)$ and $(0,1\pm 2/3)$.

Adding $4+3i$ is a translation by $(4,3)$ in the complex plane. The ellipse of part (i) therefore keeps its semiaxes and axis directions, with new centre $(2,2)$. Writing $v=X+iY$, its equation isThe third panel in the figure in part (b) shows the translated ellipse.

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.

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong real increments $w$ the first factor $\overline{w}/w$ is $1$, while along purely imaginary increments it is $-1$. Thus complex differentiability of $\overline{h}$ requires $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the complex derivative exists and equals zero.

For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. ConsequentlyThese points form a discrete subset of the complex plane and contain no nonempty open set. Every neighbourhood of each one contains points where $F$ is not complex differentiable. Therefore $F$ is holomorphic on no nonempty open set, even though it has complex derivatives at the specified points.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

Define $F(z)=\overline{f(\overline{z})}$. To apply the identity theorem, we first verify that $F$ is holomorphic; mere complex conjugation of the values of $f$ would not suffice. For $h\ne 0$,Thus $F$ is an entire function, just as $f$ is. At each positive integer $n$, the point $1/n$ is real and $f(1/n)$ is real, so $F(1/n)=f(1/n)$. These distinct points accumulate at $0\in \mathbb{C}$. The identity theorem on the connected complex plane givesThis is conjugate reflection of a holomorphic function, with the reality data forcing equality to its reflection.

The phase step between adjacent slits isFor equal amplitudes $a_0$, the six unmodified contributions form a closed regular hexagon in the complex plane:Reversing one slit at an array minimum replaces slit $j$'s contribution by its negative, so $S=S_0-2a_0{e}^{ij\pi /3}$. ConsequentlyThis is independent of which slit is reversed. The result uses the far-field, narrow-slit and small-angle approximations; retaining $\sin \theta$ changes the phase step only by about $7\times 10^{-7}$ radians.

The cross-ratio permutation orbit is generated by the two operations $\lambda \mapsto 1/\lambda$ and $\lambda \mapsto 1-\lambda$ from part (d). The full list isThese are all the possibilities: the four permutations generated by interchanging both pairs and by exchanging the two pairs leave the cross-ratio unchanged. They form the Klein four-group inside the symmetric group on four symbols, leaving at most $24/4=6$ different values; the displayed operations attain all six rational expressions.

There are fewer than six values exactly when a nonidentity operation in this six-element action fixes $\lambda$: if two operations give the same value, composing one with the inverse of the other gives such a fixed point. For real $\lambda \ne 0,1$, the fixed-point equations areThe other two equations reduce to ${\lambda }^2-\lambda +1=0$, which has no real root. HenceEach exceptional value has the same three-element orbit $\{-1,\frac{1}{2},2\}$. Every other permitted real value gives six different values.

For the geometric interpretation, put $P_0=[1:0]=\infty$ and $P_i=[x_i:1]$ for $i>0$. ThenThe three exceptional cases becomeThus the three finite points are collinear and equally spaced, with one their midpoint, even when the $x_i$ are viewed as points of the complex plane. Equivalently they form an arithmetic progression in some order. These are the harmonic conjugate points configurations with the fourth point equal to the point at infinity of the complex projective line.

For real $x\ne -1$, $w=1-2/(x+1)$ is real. Its finite image is the entire real axis except $w=1$, which is approached only as $x\to \pm \infty$. The point $x=-1$ maps to infinity. On the extended complex plane the extended real axis maps onto itself, with $z=\infty$ mapping to one.