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The polar form is $z=e^{3\pi i/4}$. Interpreting the fractional complex exponentiation as all its values, a candidate $w$ satisfies $w^3=z^2=-i$. The three roots of a complex number areThey have arguments $\pi /2$, $7\pi /6$ and $11\pi /6$, respectively. Since all have modulus one, their squared distances from $z$ in the complex plane are $|w-z{|}^2=2-2\cos (\arg w-3\pi /4)$, givingThe last is strictly largest, so the required root of a complex number isRestricting complex exponentiation to its principal value would give only $i$; the request to choose among roots requires all three values.

Complete the square in the real-valued complex plane equation:Dividing by $\alpha$ gives center and radiusIf the radicand is zero, the locus is the single point $c$, a degenerate circle. If it is negative, the locus is empty. The absolute value of $\alpha$ is essential even when $\alpha <0$.

For the requested sketch, the five centers and radii are respectively $(-1/2,\sqrt{3}/2)$, $(-1,\sqrt{2})$, $(-2,\sqrt{6})$, $(2,\sqrt{2})$ and $(1,0)$, all centers on the real axis.

The original PDF's map is $f(z)=(\beta /\bar{\beta })\bar{z}$; the converted TeX drops the conjugate on its denominator. Write $\beta =|\beta |e^{i\theta }$, so $f(z)=e^{2i\theta }\bar{z}$. This is the reflection symmetry of a complex circle equation. It fixes $c=\beta /\alpha$ and satisfiesThus it maps every point of the circle into the same circle. Conversely, for any point $w$ on that circle, $z=f(w)$ also lies on it and $f(z)=w$. This proves $f(C)=C$, including the surjectivity required by the question.

For any three nonempty circles, fix $u\in C_3$. The triangle inequality gives, for every $z\in C_1,w\in C_2$,Taking the maximum over $z,w$ proves the requested inequality.

For centers $c_1,c_2$ and radii $r_1,r_2$, the maximum separation of two circles is bounded above by $|c_1-c_2|+r_1+r_2$. If the centers differ, choose $z=c_1+r_{1}v$ and $w=c_2-r_{2}v$, where $v=(c_1-c_2)/|c_1-c_2|$, to attain the bound. If the centers coincide, any unit $v$ gives opposite points attaining $r_1+r_2$. ThereforeThe function $m$ is not a metric on positive-radius circles, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.

Project from the north pole $N=(0,0,1)$ onto the equatorial plane, identifying that plane with the complex plane. The line from $N$ to $P=(X,Y,Z)$ meets the plane inThis is stereographic projection. Its inverse isRotation through $\theta$ about the $z$-axis changes $X+iY$ to $e^{i\theta }(X+iY)$ and preserves $Z$, so it becomes the Möbius transformation $u\mapsto e^{i\theta }u$.

Let $B$ be the rotation matrix supplied in the question. It sends the $z$-axis to the $x$-axis, and therefore $R_x(\theta )=B{R}_z(\theta )B^{-1}$. If $b=\varphi B{\varphi }^{-1}$ is its given Möbius transformation, thenwhich is again a Möbius transformation. Such $z$- and $x$-axis rotations can send any chosen point of the sphere to the south pole: first rotate its horizontal projection onto the $y$-axis, then rotate in the $yz$-plane. This supplies the rotations needed below.

The antipodal stereographic coordinate relation is $\varphi (-P)=-1/\overline{\varphi (P)}$, as follows immediately from the inverse formula. Fix the cross-ratio conventionWe claim the required ordering isFor the antipodal cross-ratio and spherical distance identity, rotate $P$ to the south pole. Its coordinate becomes zero and its antipode becomes infinity. A point at angular spherical distance $d$ from the south pole has vertical coordinate $-\cos d$ and horizontal magnitude $\sin d$, so its new projected coordinate $v^{\prime }$ has modulus $\sin d/(1+\cos d)=\tan (d/2)$. By Möbius invariance of the cross-ratio, the displayed expression equalsFor finite nonsingular coordinates the same ratio is $-|u-v{|}^2/|1+u\bar{v}{|}^2$. Zero or infinite antipodal coordinates are interpreted by limits. When $Q=-P$, $d=\pi$ and the ratio has the extended value infinity; the finite real formula applies for $0<d<\pi$.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Put $q=|z{|}^2$ and retain the domain restriction $z\ne 0$. Then $z+1/\bar{z}=(q+1)/\bar{z}$, so the continued fraction simplifies toThe other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomesMultiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first geometric locus in the complex plane is the circleIts center is $-2$ and radius is $\sqrt{2}$; it does not contain the excluded point $z=0$ on its circumference.

The principal complex logarithm has imaginary part equal to the complex argument, so its equation gives the open rayIf one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi /3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.

These latter loci bound a right triangle with verticesThe bounded open region is $0<x<1/\sqrt{3}$, $\sqrt{3}x<y<1$. Its closure is the closed triangle. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier circle is a separate locus and is not a side of this triangle.

The right angle is at $P$, so the circumcenter is the midpoint of hypotenuse $OQ$:Thus its circumcircle has the complex equationTo find the incircle, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt{3}x=0$ is $(1-r-\sqrt{3}r)/2$. Requiring this distance to be $r$ givesThereforeThis tangent circle is indeed the largest contained circle. For any contained disk of radius $\rho$, its center has distances $d_j\ge \rho$ from the three side lines. Splitting the triangle into triangles with that center as vertex gives area $\Delta =\frac{1}{2}{\sum }_j{\ell }_j{d}_j\ge s\rho$, where $s$ is its semiperimeter. Here $\Delta =1/(2\sqrt{3})$ and $s=(1+\sqrt{3})/2$, so $\rho \le \Delta /s=r$. Our circle attains this upper bound, proving maximality rather than just tangency.

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

In a unital Banach algebra, the Neumann series gives $(1-u{)}^{-1}={\sum }_{k=0}^{\infty }{u}^k$ whenever $\Vert u\Vert <1$: the series converges in norm by submultiplicativity and completeness, and multiplication of its partial sums on either side gives $1-u^{N+1}\to 1$. Consequently, if $g$ is invertible and $\Vert g^{-1}\Vert \Vert h-g\Vert <1$, writeThe second factor has a Neumann series inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the group of invertible elements of a Banach algebra, proving that group is an open set.

The same series proves continuity of inversion in a Banach algebra, with the useful quantitative estimateThe right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a bijection whose inverse is the same continuous map. Thus inversion is a homeomorphism of the open group of invertibles onto itself.

For the final unheaded request, defineIt is open by the preceding argument and contains $0$. If ${\lambda }_0$ were a boundary point, a sequence ${\lambda }_n\in E$ would converge to ${\lambda }_0$, while openness gives ${\lambda }_0\notin E$. Then $1-{\lambda }_{n}a\to 1-{\lambda }_{0}a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a left inverse of an algebra element nor a right inverse of an algebra element, contradicting the hypothesis. Therefore $E$ has empty boundary; it is both open and closed in the connected complex plane, and $E=\mathbb{C}$.

For every nonzero $\mu$, put $\lambda ={\mu }^{-1}$. The identity $\mu 1-a=\mu (1-\lambda a)$ proves invertibility, so the spectrum of an element $a$ is contained in $\{0\}$. It is nonempty: if the resolvent of an element existed for all $\mu \in \mathbb{C}$, it would be an entire algebra-valued function tending to zero at infinity by the Neumann series. Applying any bounded linear functional and then the Liouville theorem would make it zero; the Hahn-Banach theorem separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence

Write a straight line in Cartesian form $Ax+By+C=0$, where $A,B,C$ are real and $(A,B)\ne (0,0)$. With $z=x+iy$, set $c=(A+iB)/2$ and $r=C$. Using the complex conjugate,Thus the real linear equation of a line in the complex plane isConversely, writing $c=u+iv$ gives $2ux+2vy+r=0$, an ordinary straight line with normal vector $(u,v)$. The nonzero condition on $c$ matters: $c=0$ would give either the whole complex plane or the empty set.

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

For the unit sphere, project from its north pole $N=(0,0,1)$ onto the equatorial plane, identified with the complex plane. The line through $N$ and $(X,Y,Z)$ meets the plane at the stereographic projectionThe north pole is omitted from the plane chart; adding its image infinity gives the Riemann sphere.

For a point whose image and antipodal image are finite, stereographic projection of $(-X,-Y,-Z)$ gives $v=-(X+iY)/(1+Z)$. Thereforeusing $X^2+Y^2+Z^2=1$. Conversely $u\overline{v}=-1$ implies $v=-1/\overline{u}$. Substituting this into the inverse stereographic projection changes the signs of all three coordinates, proving the converse. Thus finite chart images correspond to antipodes exactly when $u\overline{v}=-1$. On the extended chart the exceptional antipodal pair is $0,\infty$, for which the displayed ordinary complex product is not defined.

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives

Write $z=1+iy$ with $y\in \mathbb{R}$. In the complex plane, this is the vertical line through $1$. The complex exponential gives $e^z=e(\cos y+i\sin y)$, so its complex modulus is always $e$. Conversely, every point of that modulus has an complex argument $y$ and is obtained in this way. The image is the entire circle centered at zero with radius $e$, traversed repeatedly as $y$ varies; it is not a disk.

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere $\hat{\mathbb{C}}$, the complex plane $\mathbb{C}$, or the unit disc $\mathbb{D}$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\hat{\mathbb{C}}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The Riemann surfaces uniformized by the complex plane are

Let $U\subseteq \mathbb{C}$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb{C}\to U\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting two values, contrary to the Little Picard theorem. The plane domain with two omitted points is hyperbolic, so

Now put $S=R\setminus \{P_1,\dots ,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities arewhich give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives