For $|z|<1$, differentiate the geometric series to obtain $(1-t{)}^{-2}={\sum }_{k\ge 0}(k+1)t^k$. Since $(z-i{)}^2=-(1-z/i{)}^2$,The convergence region is precisely $0<|z|<1$, the specified punctured disk; the other pole is on its boundary at $i$.

For any $0<r<1$, a Laurent series converges uniformly on the circle $|z|=r$, so termwise contour integration is valid. Multiplication by $z^{-k-1}$ and integration gives, for every integer $k$,Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the principal part.

The complex exponential function $E(z)=e^z$ maps the strip bijectively and holomorphically to the upper half-plane: its modulus is $e^{\Re z}>0$ and its argument is $\Im z\in (0,\pi )$. Its inverse is the holomorphic logarithm $L(w)=\log |w|+i\mathrm{Arg}w$ with $\mathrm{Arg}w\in (0,\pi )$.

Part (iii) applies to both maps. The required inverse correspondences areThey compose to the identity because $L\circ E$ and $E\circ L$ do. This proves injectivity and surjectivity, not merely a map between the two sets.

Write $f=u+iv$ and evaluate all derivatives of $\psi$ at $(u,v)$. The real chain rule givesThe Cauchy–Riemann equations give equal squared lengths $u_x^2+u_y^2=v_x^2+v_y^2=|f^{\prime }{|}^2$ and zero cross term $u_x{v}_x+u_y{v}_y=0$. Part (ii) removes the last two terms. ThereforeThe composition is $C^2$, so it is harmonic. No assumption that $f^{\prime }$ is nonzero is required.

The allowed theorem makes $f^{\prime }$ holomorphic and hence continuous; applying it repeatedly gives continuous derivatives of every order. Since $u_x=\Re f^{\prime }$, $v_x=\Im f^{\prime }$, $u_y=-\Im f^{\prime }$ and $v_y=\Re f^{\prime }$, the real and imaginary parts have continuous second partial derivatives. Differentiate the Cauchy–Riemann equations and commute mixed derivatives:Thus $\boxed{\Delta u=\Delta v=0}$, with the required $C^2$ regularity, proving harmonicity.

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

The zero ideal is a prime ideal because $R$ is an integral domain. Each of $(2),(3),(5)$ is a prime ideal: for $p$ among these three, reduction $a/b\mapsto \overline{a}{\overline{b}}^{-1}$ is a surjective ring homomorphism $R\to {\mathbb{F}}_p$ with kernel of a ring homomorphism $(p)$, so its quotient ring is a field.

Any other nonzero proper ideal in (c) has generator $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ with total exponent at least two. Choose one prime factor $p$ and put $e=d/p$. Then $pe=d\in (d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required prime exponents. Such an ideal is not prime. Thus with the standard proper-ideal convention the answer isThe definition printed in this paper omits properness. Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four ideals together with $\boxed{R}$. Both interpretations are thereby accounted for.

The intersection $J=I\cap \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$, and $nj\in I\cap \mathbb{Z}$ for $n\in \mathbb{Z}$, $j\in J$. Thus it is an ideal of $\mathbb{Z}$, necessarily $d\mathbb{Z}$ for a unique $d\ge 0$.

Because $d\in I$, one has $dR\subseteq I$. Conversely, if $a/b\in I$ with $b$ coprime to $30$, multiplication by $b\in R$ gives $a\in I\cap \mathbb{Z}=d\mathbb{Z}$. Write $a=dk$; then $a/b=d(k/b)\in dR$. Thus $I=dR$.

Every positive integer factors as $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }u$ with $\gcd (u,30)=1$. The factor $u$ is a unit, so it does not change the ideal. The complete distinct list isThey are distinct because the intersection of the displayed nonzero ideal with $\mathbb{Z}$ is $2^{\alpha }{3}^{\beta }{5}^{\gamma }\mathbb{Z}$: admissible denominators cannot cancel any of these prime powers. The triple $(0,0,0)$ gives $R$ itself.

Take a reduced fraction $a/b\in R$. If $a\ne 0$, its only possible inverse in $\mathbb{Q}$ is $b/a$. Since $\gcd (a,b)=1$, this is in lowest terms up to sign and belongs to $R$ exactly when $\gcd (a,30)=1$. Zero is not invertible. ThereforeEquivalently, these are the rational numbers whose valuations at $2,3,5$ are all zero.

Interpret membership as having a representation with denominator coprime to $30$, equivalently having this property in lowest terms. Clearly $0,1\in R$. If $a/b,c/d\in R$ with admissible denominators, thenThe product $bd$ is still coprime to $30$. These are elements of $R$, proving that $\boxed{R\text{ is a subring of }\mathbb{Q}}$.

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

The polynomials $t^c$ and $(t-1{)}^d$ are coprime. Bézout's identity gives polynomials $u,v$ with $u(t)t^c+v(t)(t-1{)}^d=1$. SetFor every $x$, the decompositionhas first summand in $V_0$ and second in $V_1$, since $T^c(T-I{)}^d=0$ and polynomial expressions in $T$ commute. If $x$ belongs to both kernels, the same Bézout identity gives $x=0$. Thus $V=V_0\oplus V_1$. Both vector subspaces are invariant under $T$.

On $V_0$, the restriction of $T$ is nilpotent; on $V_1$, the restriction of $T-I$ is nilpotent. Apply (b) separately, then concatenate the two bases. This yieldswhere $A,N$ are strictly upper triangular. Empty blocks are allowed if one subspace is zero. In particular $C=I+N$ has precisely the required diagonal.