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An orthogonal matrix satisfies $A^{T}A=I$. Taking determinants and using $\det A^T=\det A$ gives $(\det A{)}^2=1$, soThis calculation remains valid over the complex numbers: orthogonality uses the transpose, whereas a unitary matrix uses the conjugate transpose.

Expand the sine of the complex number using hyperbolic functions:Its squared complex modulus isTaking nonnegative square roots proves $\boxed{|\sin z|\ge |\sin x|}$. Equality holds exactly when $y=0$.

The complex number $\sqrt{3}+i$ has modulus two and principal argument $\pi /6$. Therefore its principal complex logarithm isIf the source's logarithm means the full multivalued complex logarithm, rather than its principal value, the complete answer is $\ln 2+i(\pi /6+2\pi n)$ with $n\in \mathbb{Z}$. A branch convention is necessary to obtain a unique imaginary part.

Here $i^3=-i$, so the argument condition concerns $z+i$. Writing $z=x+iy$, the complex number $x+i(y+1)$ must point along the direction $1+i$. ConsequentlyThis is an open ray in the Argand plane, rather than the whole straight line. Its endpoint $z=-i$ is excluded because the argument of zero is undefined; the opposite half-line has a different argument.

The characteristic polynomial isThus the eigenvalues are $i$ and $-i$. There is no real diagonalization, because a real diagonal matrix would have real eigenvalues. Over the complex numbers, the two distinct eigenvalues give linearly independent eigenvectors, so the matrix is diagonalizable.

For $\lambda =i$, the first row gives $(1-i)x+2y=0$. Choose $x=2$, giving $y=-1+i$. Its standard complex norm is $\sqrt{|2{|}^2+|-1+i{|}^2}=\sqrt{6}$; the conjugate eigenvector belongs to $-i$. Hence normalized eigenvectors and a suitable change of basis areDirectly, $AP=P\mathrm{diag}(i,-i)$ and $\det P=-2i/3\ne 0$, soNormalization does not assert that these eigenvectors are orthogonal.

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

For nonzero complex numbers, $x\overline{y}=(x/y)|y{|}^2$. Since $|y{|}^2$ is positive real, the condition is equivalent to $x/y\in \mathbb{R}\setminus \{0\}$. This is an equivalence relation. The ratio $x/x=1$ gives reflexivity; the reciprocal of a nonzero real ratio shows that the relation is symmetric; and $(x/y)(y/z)=x/z$ gives transitivity. An equivalence class is a real line through zero with zero removed, or equivalently all nonzero complex numbers with the same argument modulo $\pi$.

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

De Moivre's theorem says that, for a real angle $\theta$ and an integer $m$,For a complex number $z=\rho (\cos \theta +i\sin \theta )$, the corresponding formula is $z^m={\rho }^m[\cos (m\theta )+i\sin (m\theta )]$, with $z\ne 0$ required for negative powers.

Put $q=e^{i\theta }$. If $\theta \notin 2\pi \mathbb{Z}$, the finite geometric series givesTo extract the real part, multiply numerator and denominator by $1-q^{-1}$. The denominator becomes $2(1-\cos \theta )$, and the numerator is $q-1-q^{n+1}+q^n$. Thus this finite trigonometric sum isAt $\theta \in 2\pi \mathbb{Z}$, the sum equals $n$; the printed quotient is undefined there, though its continuous limiting value is $n$.

Now take $\theta =2p\pi /(n+1)$ with $1\le p\le n$. This is not a multiple of $2\pi$, while $(n+1)\theta =2p\pi$ and $\cos (n\theta )=\cos (2p\pi -\theta )=\cos \theta$. Substitution gives

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

A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

The special unitary group is $\mathrm{SU}(2)=\{g\in M_2(\mathbb{C}):g^{\ast }g=I,\det g=1\}$. The special orthogonal group is $\mathrm{SO}(3)=\{R\in M_3(\mathbb{R}):R^{T}R=I,\det R=1\}$. Conjugation $A\mapsto gA{g}^{-1}$ preserves trace and skew-Hermiticity: for unitary $g$, $(gA{g}^{-1}{)}^{\ast }=g{A}^{\ast }{g}^{-1}=-gA{g}^{-1}$. It therefore defines the representation on the real space $V$ described in the following parts. The bar over the transpose in the PDF is essential; the condition is $A+A^{\ast }=0$, not ordinary antisymmetry over the complex numbers.

An orthonormal basis for the pairing proved below isPut $J_j=\sqrt{2}{E}_j$, so $J_j^2=-I$. For any unit real vector $v$, $J(v)={\sum }_j{v}_j{J}_j$ also squares to $-I$. Thus $g=\cos (\theta /2)I+\sin (\theta /2)J(v)$ belongs to $\mathrm{SU}(2)$. Conjugation fixes $J(v)$ and rotates its perpendicular plane by angle $\theta$ up to the orientation convention of this basis, as follows either by multiplying the displayed matrices or from their anticommutator relation $J(v)J(w)+J(w)J(v)=-2(v\cdot w)I$. Since every three-dimensional proper rotation has an axis and an angle, this proves surjectivity onto $\mathrm{SO}(3)$.

If an element is in the kernel, it commutes with each $E_j$. Commutation with $E_3$ forces it to be diagonal; commutation with $E_1$ forces its two diagonal entries to agree. Thus it is a scalar matrix, and determinant one and unitarity give exactly $\{I,-I\}$. Therefore $\boxed{\mathrm{SO}(3)\cong \mathrm{SU}(2)/\{\pm I\}}$.

The finite-dimensional complex irreducible representations of $\mathrm{SU}(2)$ are ${\mathrm{Sym}}^m({\mathbb{C}}^2)$, $m\ge 0$, of dimension $m+1$. The central element $-I$ acts by $(-1{)}^m$, so precisely the even ones descend. The irreducibles of $\mathrm{SO}(3)$ are $\boxed{{\mathrm{Sym}}^{2\ell }({\mathbb{C}}^2),\ell =0,1,2,\dots }$, viewed as representations of the quotient, with dimensions $2\ell +1$.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.

We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.

The Artin–Wedderburn theorem here says that there are positive integers $d_1,\dots ,d_r$ withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions $d_i$, and the factor sizes are unique up to reordering.

Here is a proof. Decompose the regular left module as ${}_{A}A={⨁}_{i=1}^r{S}_i^{\oplus n_i}$ with pairwise nonisomorphic simple modules $S_i$. These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$ and ${\mathrm{End}}_A(S_i)=\mathbb{C}$. For the latter assertion, an endomorphism has an eigenvalue over $\mathbb{C}$, and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at $1$, so ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{op}}$. Taking opposite algebras and using matrix transposition gives the asserted decomposition with $d_i=n_i$.

For a matrix algebra $M_d(\mathbb{C})$, the matrix units show that its only simple module is ${\mathbb{C}}^d$: the spaces $E_{jj}S$ are isomorphic through $E_{ij}$, and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of $S$. In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

Expanding the commutator immediately gives $[A,A]=0$, $[A,B]=-[B,A]$, and $[A,\lambda B]=\lambda [A,B]$. For its matrix trace, writeConsequently every matrix commutator is traceless: $\boxed{\mathrm{tr}[A,B]=0}$. This is the trace of a matrix commutator identity, valid for arbitrary square matrices over the complex numbers.

The exponential acts on the Fock vacuum: $|\psi ⟩=e^{\zeta {\sum }_n{\psi }_n{a}_n^{†}}|0⟩$. For bosons, ${\psi }_n$ are commuting complex numbers. Since $[a_n,{\sum }_m{\psi }_m{a}_m^{†}]={\psi }_n$, commuting $a_n$ through the exponential gives $a_n|\psi ⟩={\psi }_n|\psi ⟩$. These are unnormalized bosonic coherent states.

For fermions, the labels are independent odd Grassmann variables, which anticommute with one another and with the fermionic operators. For one mode, the fermionic coherent state is $|\psi ⟩=(1-\psi a^{†})|0⟩$. The canonical anticommutation relations give $a(-\psi a^{†})|0⟩=\psi |0⟩=\psi |\psi ⟩$, using ${\psi }^2=0$. The even factors $1-{\psi }_n{a}_n^{†}$ commute between modes, so the same argument applies to every $n$. A Grassmann eigenvalue is a formal extension of the state space, not an ordinary complex eigenvalue of the nilpotent annihilation operator.

With dual state $⟨\psi |=⟨0|e^{\zeta {\sum }_n{a}_n{\bar{\psi }}_n}$, both cases obey $⟨\psi |{\psi }^{\prime }⟩=e^{{\sum }_n{\bar{\psi }}_n{\psi }_n^{\prime }}$.