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For distinct $z,w$ in the convex disc, integrate along the segment:The second term has modulus strictly below $|f^{\prime }(z_0)||z-w|$, so the sum cannot vanish. Thus $f$ is one-to-one.

The first characteristic polynomial factors asThe root condition holds precisely for $0<\alpha <2$: in this range the two reciprocal roots are distinct and on the unit circle; at either endpoint a unit root is repeated, and outside it one root has modulus greater than one. Consistency is given, so these and only these values are convergent. The exceptional order-three value $\alpha =6$ is outside this interval; consequently every convergent case has order two.

Apply the residue theorem to $f(z)/(z-w{)}^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's derivative formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's derivative formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local uniform convergence therefore passes to every derivative.

On $|z-a|=\epsilon$, $g$ has a positive minimum modulus. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

Along real $z=x\uparrow 1$, the modulus is $1/(1+x^2)$ and stays bounded, but along $z=iy$ with $y\uparrow 1$ it is $1/(1-y^2)\to \infty$. Since $\mathrm{limsup}|a_n{|}^{1/n}=1$, $R=1$.

Apply the maximum modulus principle to $f(z)/z$, with its removable value $f\prime (0)$, to obtain $|f(z)|\le |z|$. Equality at a nonzero interior point makes $f(z)/z$ constant, so $f(z)=az$ with $|a|=1$.

The power method repeatedly normalizes $A{q}^{(k)}$ and estimates the eigenvalue with the Rayleigh quotient $(q^{(k)}{)}^{T}A{q}^{(k)}$. It converges when the dominant eigenvalue is unique in modulus and the initial vector has a nonzero component in its eigendirection.

Homotopy invariance of winding number states that a homotopy $H(s,t)$ through closed paths in ${\mathbb{C}}^{\ast }$ preserves winding number. Uniform continuity divides the $s$ interval so successive paths differ pointwise by less than the minimum modulus of the earlier path; the supplied perturbation lemma makes their winding numbers equal. Chaining the subdivisions proves the result.

The amplitude at $y$ is proportional to $e^{2\pi i{x}_{0}y/8}{\sum }_{j=0}^{A-1}{e}^{2\pi ijry/8}$. The geometric sum vanishes unless $ry\equiv 0(\mathrm{mod}8)$, i.e. unless $y$ is a multiple of $A=8/r$. At those $r$ outcomes its modulus is $1/\sqrt{r}$, so each probability is $1/r$. The phase contains $x_0$, but measurement probabilities do not.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For a two-level Fourier recurrence ${\hat{u}}^{n+1}=a(\theta ){\hat{u}}^n+b(\theta ){\hat{u}}^{n-1}$, the amplification matrix isStability requires its powers to be uniformly bounded in the time index and the mesh frequencies. If its two amplification roots have modulus at most one and their separation has a positive mesh-independent lower bound, the eigenvectors $({\xi }_j,1{)}^T$ give a uniformly bounded diagonalization of a matrix, proving stability. A repeated unit-modulus root of this companion matrix instead produces a Jordan block and linear growth in time. Thus checking only the moduli of the roots is insufficient.

A square nonnegative matrix $A$ is primitive if some positive integer power $A^k$ has strictly positive entries. It is therefore an irreducible nonnegative matrix. The Perron–Frobenius theorem gives a leading eigenvalue whose modulus is strictly larger than that of every other eigenvalue. Every strictly positive matrix is primitive; the two-cycle permutation matrix is irreducible but not primitive.