Suppose, for a contradiction, that $f$ has no zero in the open disk. The strict boundary inequality implies no boundary zero either. Thus $1/f$ is holomorphic on a neighborhood of the closed disk: compactness and continuity keep $f$ nonzero in a sufficiently small neighborhood. Apply the $k=0$ Cauchy estimate to $1/f$:This contradicts the assumed strict inequality. If $f(z_0)=0$, the conclusion is immediate without taking a reciprocal. Therefore $f$ has a zero in the open disk. This is a local form of the minimum modulus principle.

On the circle, the integrand in the Cauchy derivative formula has modulus at most $M/r^{k+1}$, and the circle has length $2\pi r$. The ML inequality therefore givesThe maximum $M$ exists because the modulus of the continuous function $f$ attains a maximum on the compact circle. These are the Cauchy estimate.

For the positively oriented circle $C=\{z:|z-z_0|=r\}$, the Cauchy derivative formula statesThe hypothesis that the closed disk lies in the open holomorphy domain ensures the formula applies on and inside this circle.

The complex line integral along the smooth parametrized curve isFor the reversed curve, the chain rule gives ${\tilde{\gamma }}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$. Substitution $s=a+b-t$ yieldsConsequently reversing orientation negates the complex line integral.

A set is compact if every open cover of it has a finite subcover. Let $K_1,\dots ,K_m$ be compact and let $\mathcal{U}$ cover their union. For each $j$, the same family covers $K_j$, so compactness supplies a finite subfamily ${\mathcal{U}}_j$ covering $K_j$. The union of these finitely many finite subfamilies is finite and covers ${\bigcup }_j{K}_j$. Hence a finite union of compact sets is compact.

An arbitrary union need not be compact. Each singleton $\{n\}$, $n\ge 1$, is compact, but their union $\mathbb{N}$ has the open cover $\{(n-1/3,n+1/3):n\ge 1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.

Interpret diagonalizability over the specified underlying field $\mathbb{R}$. Write $A=P\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)P^{-1}$ with real eigenvalues. Invariance of the trace under similarity givesIf $A\ne 0$ is diagonalizable, at least one eigenvalue is nonzero, proving $\phi (A,A)>0$. Real diagonalizability is essential: the rotation block from part (c) is diagonalizable over $\mathbb{C}$ but has eigenvalues $i,-i$ and negative trace of its square.

For $n\ge 2$, take the matrix whose leading $2\times 2$ block is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and whose other entries are zero. Its square has leading block $-I_2$, soThis contradicts the positive definiteness required of an inner product. Even the nonzero matrix unit $E_{12}$ has $\phi (E_{12},E_{12})=0$.

The printed assertion has an exception: when $n=1$, $\phi (a,b)=ab$ is an inner product on $\mathbb{R}$. Thus the requested failure requires $n\ge 2$, a restriction absent from the printed question.

Expand the trace by matrix entries:Hence $\phi (A,B)=\phi (B,A)$. Combined with part (a), this makes $\phi$ a symmetric bilinear form.

For real scalars $\alpha ,\beta$ and matrices $A,C,B$, distributivity of matrix multiplication and linearity of the trace giveThus $\phi$ is linear in its first argument.

The polynomial $X(X-1)$ annihilates $E$ by idempotence, so the minimal polynomial divides it. If $E=0$, its monic polynomial of smallest positive degree is $X$. If $E\ne 0$, both eigenvalues zero and one occur, so both factors must divide every annihilating polynomial. Therefore

If $Ev=\lambda v$ with $v\ne 0$, applying $E^2=E$ gives ${\lambda }^{2}v=\lambda v$, hence $\lambda \in \{0,1\}$. The kernel is nonzero since $E\ne I$, so zero always occurs. One occurs exactly when the image is nonzero. Thus the eigenvalue set is $\{0\}$ if $E=0$, and $\{0,1\}$ otherwise. Their algebraic multiplicities are $N-r$ and $r$ respectively.

In the adapted basis from part (a), the matrix has $r$ diagonal entries equal to one and $N-r$ equal to zero. With the characteristic polynomial convention ${\chi }_E(X)=\det (XI-E)$,The characteristic polynomial is independent of the choice of basis.

Every vector decomposes as $v=Ev+(v-Ev)$, where $Ev\in \mathrm{im}E$ and $E(v-Ev)=0$. If $w$ lies in both the image and kernel, write $w=Eu$ to obtain $w=E^{2}u=Ew=0$. Thusand $E$ is the projection onto its image along its kernel. It is the identity on its image and zero on its kernel. Let $N=\dim V$ and $r=\mathrm{rank}E$. A basis adapted to this direct sum gives matrix $\mathrm{diag}(I_r,0_{N-r})$. Since $E$ is not the identity, $N-r>0$, and therefore

The modulus map is continuous, the denominator $1+|z|$ is positive everywhere, and sums, products and quotients with nonzero denominator preserve continuity. Hence $f$ is continuous throughout $\mathbb{C}$.

For $g$, $|g(z)|=|z|$, so $g(z)\to 0=g(0)$ as $z\to 0$. The chosen argument is continuous locally away from the negative real axis, making $g$ continuous there as well. At a point $-r$, $r>0$, approach along $z=r{e}^{i\theta }$. From $\theta \uparrow \pi$, $g(z)\to ir$; from $\theta \downarrow -\pi$, $g(z)\to -ir$. The value assigned on the cut is $ir$. Thus $g$ is discontinuous at every strictly negative real point, and therefore is not a continuous map on the entire complex plane. Its radial factor is $r$, as printed, rather than the $\sqrt{r}$ of a square root.