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Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

A function is holomorphic on an open set when it is complex differentiable at every point of that set. Explicitly, for every $z\in U$, the limitexists as a finite complex number. The limit is independent of the direction in which the complex increment approaches zero; existence of derivatives only along the real axis would not suffice.

Put ${\alpha }_j=1-{\beta }_j$, so the interior angle at the $j$th polygon vertex is $\pi {\alpha }_j$. A conformal map straightens that corner by a power: locally,This follows by composing with the power $1/{\alpha }_j$ that maps the corner to a half-plane and using Schwarz reflection along its straight edges. The reflected map has a nonzero first derivative in the straightened coordinates.

For an upper-half-plane branch of the complex logarithm, defineAlong each open interval of the real axis, $P$ has constant argument. Crossing $x_j$ from left to right decreases $\arg (z-x_j)$ from $\pi$ to zero, so the argument of $P$ increases by ${\beta }_j\pi$. This is exactly the exterior turning angle of the polygon boundary. Thus $f^{\prime }/P$ has one constant boundary argument on all the intervals. The corner asymptotics remove its apparent singularity at each prevertex.

For a bounded polygon, the sum of exterior angles gives ${\sum }_j{\beta }_j=2$. Since every vertex has a finite prevertex, infinity maps to an ordinary point of an edge. In the local coordinate $1/z$, Schwarz reflection gives $f(z)=f(\infty )+d/z+O(z^{-2})$ with $d\ne 0$. Hence $f^{\prime }/P$ has a finite nonzero limit at infinity. After multiplication by one unit complex number, its boundary values are real. Reflecting across the real axis extends it to an entire function, including each removed prevertex, and it is bounded at infinity. Liouville's theorem makes it constant. Therefore the Schwarz-Christoffel formula isIntegration introduces one further additive constant.

The Möbius transformations preserving the upper half-plane have three real parameters:with common nonzero scaling irrelevant. They can send any cyclically ordered triple of boundary points to another such triple. Consequently three prevertices may be fixed independently; the other $n-3$ are constrained by the polygon's shape, through its side-length ratios. The constants of integration and multiplication account for translation, rotation and scale of the image, rather than further prevertex freedom.

If the last prevertex is infinity, use $1/z$ as its local coordinate. Its factor is absorbed into the overall normalization, leavingFor a finite image vertex at infinity's preimage, this has asymptotic order $z^{{\beta }_n-2}$, because $f(z)-f(\infty )$ is of order $z^{-(1-{\beta }_n)}$. Thus the omitted factor is not evaluated by substituting infinity into the finite product; its angle is encoded by the behavior at infinity.

A group is a nonempty set $G$ with an operation $G\times G\to G$, $(g,h)\mapsto gh$, satisfying associativity, an identity element $e$ with $eg=ge=g$, and an inverse element $g^{-1}$ with $g{g}^{-1}=g^{-1}g=e$ for every $g$. These are the group axioms. An Abelian group additionally satisfies $gh=hg$ for every pair. A subgroup is a subset that is itself a group under the restricted operation; equivalently, it is nonempty and closed under $h{k}^{-1}$.

Let $R:G\to \mathrm{GL}(V)$ be a nonzero, finite-dimensional complex irreducible representation of an Abelian group. For each fixed $g$, commutativity givesThus $R(g)$ is an intertwining operator from $R$ to itself. By Schur's lemma over the complex numbers, $R(g)={\lambda }_g{I}_V$. Every line in $V$ is therefore an invariant subspace. Since $R$ is irreducible, this forcesThis is the finite-dimensional convention used throughout the paper. Without it, the literal assertion for arbitrary Abelian groups needs qualification. A concrete counterexample is $G=\mathbb{C}(t{)}^{\times }$ acting by multiplication on the complex vector space $V=\mathbb{C}(t)$ of rational functions. Any nonzero invariant complex vector subspace containing $f$ contains $gf$ for every nonzero rational $g$, hence contains every rational function. This is an infinite-dimensional irreducible representation. The scalar conclusion of Schur's lemma used above depends on finite dimensionality.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

A group representation on a vector space $V$ over a field $K$ is a group homomorphism $R:G\to \mathrm{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g{)}^{-1}$. Choosing a basis identifies it with a matrix representation.

Two group representations $R$ on $V$ and $S$ on $W$ are equivalent representations when an invertible linear map $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of basis.

A nonzero group representation is a reducible representation if there is a vector subspace $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an irreducible representation if it has no such invariant subspace. Reducibility alone does not assert the existence of an invariant complementary vector subspace; that stronger conclusion says that the representation is a semisimple representation.

The Schur lemma states that an intertwiner $T$ between irreducible representations is either zero or an isomorphism. Consequently an intertwiner between inequivalent irreducible representations is zero. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:For the first assertion, the kernel and image of $T$ are invariant subspaces, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has a nonzero kernel, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the real numbers, as a real planar rotation through $\pi /2$ is an irreducible representation of the cyclic group of order four, while the rotation itself is a commuting nonscalar endomorphism.

Let $T(v\otimes w)=w\otimes v$ on the tensor product $V\otimes V$. The linear map $R(g)\otimes R(g)$ commutes with $T$, so preserves its $+1$ and $-1$ eigenspaces, namely the symmetric square $S^{2}V$ and the exterior square ${\wedge }^{2}V$. Restricting the tensor product representation to these invariant subspaces defines $S^{2}R$ and ${\wedge }^{2}R$. The representation identity follows by restricting $(R(g)\otimes R(g))(R(h)\otimes R(h))=R(gh)\otimes R(gh)$.

For a finite group, $g$ has finite order $m$, so $R(g{)}^m=I$. Over the complex numbers, the polynomial $t^m-1$ has distinct roots; hence $R(g)$ is diagonalizable. Choose an eigenbasis $v_1,\dots ,v_n$ with eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. A basis of the symmetric square consists of $v_i\otimes v_i$ and $v_i\otimes v_j+v_j\otimes v_i$ for $i<j$. A basis of the exterior square consists of $v_i\otimes v_j-v_j\otimes v_i$ for $i<j$. Their eigenvalues are respectively ${\lambda }_i^2,{\lambda }_i{\lambda }_j$ and ${\lambda }_i{\lambda }_j$. Taking traces,But $\chi (g{)}^2={\sum }_i{\lambda }_i^2+2{\sum }_{i<j}{\lambda }_i{\lambda }_j$ and $\chi (g^2)={\sum }_i{\lambda }_i^2$. Thus the characters areAt $g=e$, these also give the dimensions $n(n+1)/2$ and $n(n-1)/2$.

Schur's lemma states that an intertwiner $T:V\to W$ between irreducible representations is either zero or an isomorphism. In particular, inequivalent irreducible representations have no nonzero intertwiners. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:Indeed, the kernel and image of an intertwiner are invariant subspaces, proving the first statement. For the second, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so it must vanish. The scalar conclusion uses an algebraically closed coefficient field: over the real numbers, planar rotation through $\pi /2$ gives an irreducible real representation of ${\mathbb{Z}}_4$ and commutes with itself without being a real scalar matrix.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$: $R(gh)=R(g)R(h)$ and $R(e)=I$. A reducible representation has a nonzero proper invariant subspace $W\subset V$, meaning $R(g)W\subseteq W$ for every $g$. Over the complex numbers, a finite-dimensional representation of a finite group is a direct sum of irreducible representations, by Maschke's theorem.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. The cyclic property of the trace makes it a class function:Thus a character is specified by one number per conjugacy class, much less information than all the representation matrices. Moreover,Here character orthogonality says that $m_r$ is the multiplicity of the irreducible representation $r$. Consequently characters determine finite-dimensional complex representations up to equivalence of group representations and compute tensor product of group representations decompositions without constructing a new basis explicitly.

Use the mathematical Lie algebra convention $U(t)=I+tX+O(t^2)$. Differentiating $U^{†}U=I$ and $\det U=1$ at $t=0$ gives $X^{†}=-X$ and $\mathrm{tr}X=0$. Conversely, for such an $X$, the matrix exponential is unitary and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. ThusThe four diagonal entries are purely imaginary, and the six entries above the diagonal are arbitrary complex numbers determining the entries below it. This gives $4+2\cdot 6=16$ real parameters before imposing the one real trace constraint, so ${\dim }_{\mathbb{R}}\mathfrak{su}(4)=15$.

A Cartan subalgebra of this compact Lie algebra is a maximal commuting Lie subalgebra; chooseA simple basis is $i{H}_1,i{H}_2,i{H}_3$, whereAny matrix commuting with every diagonal element is itself diagonal, which proves maximality. Commuting skew-Hermitian matrices can also be simultaneously diagonalized, so no larger commuting Lie subalgebra is possible. ConsequentlyThe Hermitian quantum generator convention instead uses the Hermitian $H_a$ and $U=\exp (i{\theta }^a{H}_a)$. The two conventions describe the same Lie algebra.

Write $x=(x_1,x_2,x_3{)}^T$, $v=(x_4,x_5,x_6{)}^T$ and $z=x+iv$. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. ThenIndeed $d{z}_{k}d{\bar{z}}_k=d{x}_k^2+d{x}_{k+3}^2$. A transformation $z^{\prime }=Uz$ preserves this metric whenever $U^{†}U=I$, becauseThus the entire unitary group $U(3)$ preserves it, including its special unitary group $SU(3)$.

To obtain a real matrix embedding, write $U=A+iB$ with $A,B$ real. ThenThe unitarity equations imply $A^{T}A+B^{T}B=I$ and $A^{T}B=B^{T}A$, hence $\rho (U{)}^T\rho (U)=I$. Multiplication of the block matrices gives $\rho (UV)=\rho (U)\rho (V)$, and the blocks recover $U$, proving injectivity. FinallyFor this last identity, over the complex numbers the change of basis from $(x,v)$ to $(z,\bar{z})$ conjugates $\rho (U)$ to $\mathrm{diag}(U,\bar{U})$. This proves that the image lies in the special orthogonal group.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I$. A nonzero group representation is an irreducible representation if its only invariant subspaces are $0$ and $V$. Here the representations and characters are finite-dimensional over the complex numbers.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. Three useful properties are:

In particular, ${\chi }_R(e)=\dim V$ and ${\chi }_R(g^{-1})={\chi }_R(g{)}^{\ast }$, the latter following from unitarization of a finite-group representation. The character inner product $⟨{\chi }_r,{\chi }_R⟩$ counts the multiplicity of an irreducible representation $r$ in $R$.

Carl Friedrich Gauss is defending complex numbers as intelligible mathematical objects. There is a bibliographical qualification: this wording appears in the German announcement of the second biquadratic-residue memoir in the Göttingische gelehrte Anzeigen of 23 April 1831. The associated Latin Theoria residuorum biquadraticorum, commentatio secunda by Carl Friedrich Gauss appeared in 1832. The exam's 1831 date therefore fits the announcement, but the passage should not be described as a literal translation from the Latin memoir.^[ref][ref]

The proposed directional vocabulary is mathematically specific. Represent $z=a+bi$ by the point $(a,b)$ in the complex plane. The addition of complex numbers adds components. Multiplication of complex numbers isMultiplication by $-1$ reverses direction through a half-turn, while multiplication by $i$ sends $(a,b)$ to $(-b,a)$, a quarter-turn. Consequently $i^2=-1$ says that two quarter-turns make a half-turn. More generally the moduli multiply and the complex arguments add. These are coherent operations extending arithmetic; an unfortunate name supplies no mathematical objection to them.

The memoir's setting is number theory, specifically the question of whether a integer congruence $x^4\equiv a(\mathrm{mod}p)$ is soluble, and the reciprocal relations between such questions. Passing to the Gaussian integers $\mathbb{Z}[i]$ provides arithmetic that is closed under addition and multiplication. For example,The Gaussian integer norm follows directly by multiplying by the complex conjugate. Here it gives $N(2+i)=5$ and $N(5)=25$, consistent with the factorization. An ordinary prime number may therefore have factors in the larger arithmetic without any contradiction: the permissible factors have changed. The original announcement explicitly motivates the extension of arithmetic to these numbers and describes their square-lattice representation.^[ref]

It is helpful to separate two domains. Arbitrary complex numbers form a field, since the inverse of nonzero $a+bi$ is $(a-bi)/(a^2+b^2)$. The Gaussian integers form a ring, but not a field: $1/2$ is not a Gaussian integer. Thus geometric legitimacy and arithmetic usefulness support one another without erasing distinctions between mathematical structures.

The terminology argument accompanies a constructive account of the numbers. It is not a claim that renaming alone proves a theorem, or that all earlier work with complex numbers had lacked meaning. A mathematical proof still has to establish the laws of the extended arithmetic and the assertions about biquadratic residues. Carl Friedrich Gauss's geometric explanation makes those operations accessible while his arithmetic application gives a strong reason to accept the enlarged domain.

The given permutations satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight elements $r^j,r^{j}s$ for $0\le j<4$ are distinct: $s\notin ⟨r⟩$, whose only element of order two is $r^2=(13)(24)$. Thus $H$ is the order-eight dihedral group, and we may use the matricesThey obey the same relations and describe a quarter-turn and a reflection of a square.

The eigenvalues of $R$ over the complex numbers are $i$ and $-i$, with respective eigenlines spanned by $(1,-i)$ and $(1,i)$. Since $S$ interchanges these eigenlines, neither is an invariant subspace for $H$. Any proper nonzero subrepresentation of this two-dimensional complex representation would be a line invariant under both $R$ and $S$, and an $R$-invariant line must be one of those eigenlines. None qualifies, so $W$ is irreducible.

For the calculation in part (e), its character isIndeed $R^2=-I$, while a quarter-turn or a reflection has trace zero. The character inner product $(2^2+(-2{)}^2)/8=1$ provides a second irreducibility check.

Let $\rho$ afford the character $\chi$, and let $m$ be the order of a group element $g$. Then $\rho (g{)}^m=I$, so its minimal polynomial divides $T^m-1$. Over the complex numbers this polynomial has distinct polynomial roots, since its formal derivative $m{T}^{m-1}$ does not vanish at any of its polynomial roots. Thus $\rho (g)$ is diagonalizable, with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ that are roots of unity.

Each ${\lambda }_j$ has absolute value one, so ${\lambda }_j^{-1}=\overline{{\lambda }_j}$. The eigenvalues of $\rho (g^{-1})=\rho (g{)}^{-1}$ are ${\lambda }_j^{-1}$, givingThis is the character inversion identity.

Proceed by induction on $\dim V$. The result is immediate for dimension zero. In positive dimension, the characteristic polynomial has a root over the complex numbers, so there is a nonzero eigenvector $v$ with eigenvalue $\lambda$. Normalize it to have norm one. Self-adjointness givesso $\lambda$ is real.

The orthogonal complement $v^{\perp }$ is an invariant subspace: if $w\perp v$, thenThe restriction of $T$ to $v^{\perp }$ is again self-adjoint, since the same inner product identity holds on that subspace. By induction it has an orthonormal basis of eigenvectors. Adjoining $v$ givesThe existence of the first eigenvector is the essential use of the complex numbers in this proof of the finite-dimensional spectral theorem.

We construct a cube root of a matrix on each Jordan block. Since $A$ is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block $J_m(\lambda )=\lambda I+N$, where $N=J_m(0)$ is a Nilpotent Jordan block and $N^m=0$. Choose a complex cube root $\beta$ of $\lambda$, so ${\beta }^3=\lambda$.

We need a polynomial $q(z)$ with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$. Once $c_1,\dots ,c_{j-1}$ have been chosen, the coefficient of $z^j$ in $q(z{)}^3$ is $3c_j$ plus an expression depending only on those earlier coefficients. Choose $c_j$ to make this coefficient one at $j=1$ and zero at $2\le j<m$. Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently $c_j=\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)$, where the generalized binomial coefficient is $\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$ and the empty product is one.

Substitute $z=N/\lambda$. The polynomial remainder is a multiple of $N^m$, hence zero. It follows thatFor $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the block diagonal matrix $C={⨁}_r{C}_{m_r}({\lambda }_r)$. Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.

Direct matrix multiplication gives $A^2=0$. Denote the columns by $c_1,\dots ,c_4$; they satisfyThus the matrix rank is at most two. Over the real numbers or complex numbers, the upper-left two-by-two minor is $-3\ne 0$, so the matrix rank is exactly two. Every Jordan block has size at most two because $A^2=0$, and each size-two Nilpotent Jordan block contributes one to the matrix rank. ConsequentlyEquivalently the nullities are $(2,4)$ and the minimal polynomial is $x^2$.

There is a field qualification in the source: Question 2 initially allows an arbitrary field, and part (c) does not specify a new one. The preceding answer holds over every field whose characteristic is not three. In characteristic three, the first row vanishes and the other three rows all become $(2,1,0,2)$, giving matrix rank one while $A^2=0$. Under that reading the answer is instead

Over the complex numbers, the fundamental theorem of algebra makes every irreducible polynomial factor $f_j(x)$ equal to $x-{\lambda }_j$. Part (b)(iv) giveswith distinct eigenvalues. The primary decomposition theorem for linear maps consequently expresses $V$ as the direct sum of the actual eigenspaces $E_{{\lambda }_j}=\ker (T-{\lambda }_{j}I)$, with no higher generalized eigenspaces needed. Part (ii) makes these summands mutually orthogonal.

Apply the Gram-Schmidt process to a basis within each eigenspace. Its linear combinations stay within that eigenspace, so all resulting vectors are still eigenvectors. Their union is orthonormal by part (ii) and spans $V$ by the primary decomposition theorem for linear maps. Thusfor an orthonormal basis $\mathcal{B}$ of eigenvectors. This proves the finite-dimensional complex spectral theorem for normal operators by the requested decomposition argument.

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.