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Let $D=\mathrm{diag}(A)$. The Jacobi method isso $H=I-D^{-1}A$ and $v=D^{-1}b$. If $A$ is strictly diagonally dominant, every Gershgorin disc of $H$ is centered at zero and has radiusEvery eigenvalue therefore has modulus below one, so $\rho (H)<1$ and part (i) proves convergence.

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,Its modulus is therefore periodic with period one. The assumed bound for $f$ on $0\le \mathrm{Re}z\le 1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb{C}$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an open set; the identity theorem would then give $f\equiv 0$, a contradiction to that choice. Therefore $f\equiv 0$ and $F=\Gamma$.

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

The vertices satisfyAfter removing the root $z=0$, the product of the other roots has modulus equal to the constant term of $((z+1{)}^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit regular polygon is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving

The eigenvalues of the recurrence matrix have modulus $e^{-\kappa T}<1$, so transients decay and the coefficients approach the fixed pointThusFor $T=k\pi$ and $\kappa k\ll 1$, this isEven multiples deliver impulses in phase and produce resonance; odd multiples alternate the phase and largely cancel.

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 number $i{e}^{i\varphi }$ has modulus one and argument $\varphi +\pi /2$ modulo $2\pi$. Therefore its complex logarithm has valuesEvery value has real part zero. For the principal complex logarithm, choose the unique integer giving imaginary part in $(-\pi ,\pi ]$. The unwrapped value $\varphi +\pi /2$ is the principal imaginary part only when it already lies in that interval.

Write $-8=8e^{i(\pi +2\pi k)}$. Its cube roots have modulus two and arguments $(\pi +2\pi k)/3$, for $k=0,1,2$. ThusEach cube equals $-8$, and a cubic polynomial has no further distinct roots.

In polar form, a cube root of one has modulus one and argument $2\pi k/3$. The three roots of unity areThey are equally spaced on the unit circle.

For $b\ne 0$, write $z=bw$. The equation becomes $w^2+w+1=0$, whose roots are $(-1\pm i\sqrt{3})/2$, both of modulus one. Hence $|z|=|b|$. If $b=0$, the double root is $z=0$, giving the same relation. Therefore

Complex conjugation reverses the argument, reciprocation also inverts the modulus, and multiplication by $-1$ adds $\pi$ to the argument. With all arguments chosen in $[0,2\pi )$ and $0<\varphi <\pi /2$,Thus $z$ is in the first quadrant, $z^{\ast }$ and $z^{-1}$ lie on the same fourth-quadrant ray at radii two and one-half, and $-z$ is diametrically opposite $z$.

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 $\pm [\sqrt{(\sqrt{2}+1)/2}+i\sqrt{(\sqrt{2}-1)/2}]$.