The Taylor expansions giveand hence $f(z)=b/z+iab+O(z)$. Since $b>0$, zero is a simple pole with residue $b$. It is not a removable singularity, despite the vanishing numerator, because the denominator has a zero of order two.

For $z=x+iy$, $e^{iz}=e^{-y}{e}^{ix}$, so $|e^{iz}|=e^{-y}$. Thereforethe closed upper half-plane. Equality holds on the real axis.

The residue $\mathrm{res}(f;a)$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series at the isolated singularity. Equivalently it is $(2\pi i{)}^{-1}$ times the integral around a sufficiently small counterclockwise circle enclosing $a$.

The residue theorem states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour and its interior except for finitely many isolated singularities $a_j$ inside, thenNo singularity may lie on the integration contour. A more general contour weights each residue by its winding number.

Laurent's theorem states that a function holomorphic on $R<|z-a|<S$ has a unique expansionconverging uniformly on every compact subannulus. Its coefficients arewith counterclockwise orientation. The value is independent of the chosen intermediate radius. The nonnegative powers form a power series convergent inside the outer radius, and the negative powers form the principal part of a Laurent series.

Take $g_n(z)=z^n/n$ and $g(z)=0$, for $n\ge 1$. These functions are entire andso convergence is uniform on the closed unit disc. But $g_n^{\prime }(z)=z^{n-1}$, givingfor every $n$. In particular $g_n^{\prime }(1)=1$ never approaches zero. On every strictly smaller disc the derivative does converge uniformly, consistent with part (iii). This is uniform holomorphic convergence without boundary derivative convergence.

Fix $0<r<1$ and $k\ge 1$. On $|z|=1$ and $|z_0|<r$, the reverse triangle inequality gives $|z-z_0|\ge 1-r$. Subtract the two Cauchy integral formula expressions and apply the estimation lemma, using circle length $2\pi$:Thus every fixed-order derivative converges uniformly on each smaller disc. The right-hand factor tends to zero by the assumed uniform convergence on $D$. Keeping a positive gap $1-r$ from the integration contour is the essential feature of this Cauchy derivative convergence bound. If $r\le 0$, the specified open disc is empty and the claim is vacuous.

For $|z_0|<1$ and $k\ge 0$, the Cauchy integral formula for derivatives isThe formula applies because the closed unit disc lies inside the domain of holomorphy. For $k=0$ this is the ordinary value formula, with $0!=1$.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.

No such entire function exists. If it did, the entire function $h(z{)}^2-z$ would vanish at the distinct points $(n/(n+1){)}^2$, whose accumulation point is one. The identity theorem would give $h(z{)}^2=z$ on all of $\mathbb{C}$.

At zero this forces $h(0)=0$. Differentiating the identity gives $2h(z)h^{\prime }(z)=1$, impossible at zero. Equivalently, a squared holomorphic function has an even-order zero, whereas $z$ has a simple zero. This is the obstruction expressed by no entire square root of the identity function; a square-root branch near one cannot extend holomorphically through zero.

Yes: take the unnormalized sinc function with its removable value at zero,Its power series is ${\sum }_{m=0}^{\infty }(-1{)}^m{z}^{2m}/(2m+1)!$, with infinite radius of convergence, so it is an entire function. At $z=1/n$ it gives $n\sin (1/n)$ as required. The identity theorem also proves this entire solution is unique.

The identity theorem says that two holomorphic functions on a connected domain which agree on a set with an accumulation point inside that domain agree everywhere there.

No such entire function exists. On the subsequence $1/(2m)\to 0$ indexed by even numbers, the proposed function equals $\sin z$. The identity theorem would force $f(z)=\sin z$ throughout $\mathbb{C}$. At the points indexed by odd numbers it is instead required to equal $-\sin z$, a contradiction since $\sin (1/n)>0$ for positive $n$. The accumulation point zero lies inside the assumed domain of holomorphy, which is essential to this argument.

Along $z=e^{i\theta }$ with $0<\theta <2\pi$, the chosen branch has $L(z)-i\pi =i(\theta -\pi )$. Use part (iii) and the circular estimate in part (a):The second step uses $|\theta -\pi |\le \pi$. The single branch-cut point does not affect the integral. Keeping that weight gives the sharper intermediate bound.

Apply the Cauchy integral theorem to $F(z)=f(z)(L(z)-i\pi )$ on the slit plane, which is star-shaped about $-1$. For fixed $\epsilon >0$, first let $\delta \to 0$. The upper bank is traversed from $\epsilon$ to one and the lower bank from one to $\epsilon$, so their limiting sum isThe outer arc tends to the positively oriented unit-circle integral, interpreting the logarithm by its one-sided boundary values at the cut. On the inner arc, $|L-i\pi |\le |\log \epsilon |+\pi$ and $f$ is bounded near zero. Its integral is therefore $O(\epsilon |\log \epsilon |)$ and tends to zero. The sum of all four contour pieces is zero. Letting $\epsilon \to 0$ givesThe integral around the full circle is a piecewise integral of the branch's boundary values; $L$ is not a single holomorphic function across the cut-crossing point. This is the centred logarithmic keyhole identity.

The Cauchy integral theorem for a star-shaped domain says: if $U$ is open and star-shaped and $F$ is holomorphic on $U$, then ${\int }_{\gamma }F(z)dz=0$ for every closed contour $\gamma$ in $U$. Equivalently, $F$ has a complex antiderivative on $U$. Star-shaped means that there is one point of $U$ from which every straight segment to another point of $U$ remains in $U$.

Use the continuous argument $0<\arg z<2\pi$ on the positive-axis slit plane and defineThis analytic logarithm on the positive-axis slit plane is holomorphic, satisfies $e^{L(z)}=z$ and $L^{\prime }(z)=1/z$, and has $L(-1)=i\pi$. That value fixes the branch uniquely on the connected domain. For positive real $x$, the two boundary limits areThe jump, rather than a pole of the integrand, is what the keyhole argument exploits.

For $\varphi (t)=r{e}^{it}$, ${\varphi }^{\prime }(t)=ir{e}^{it}$ and $|{\varphi }^{\prime }(t)|=r$. The triangle inequality for an integral applied to the parametrized contour integral givesThis is the circular-contour instance of the estimation lemma.

For a piecewise continuously differentiable contour $\gamma :[a,b]\to U$, define its contour integral byadding the integrals over the finitely many smooth pieces if necessary. Continuity of $f$ along the contour makes each integrand integrable. The derivative supplies both the direction and the complex line element; replacing it by its modulus would give a different integral with respect to arc length.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.

Suppose a path-connected space had a separation $X=U\cup V$ as above. Choose $u\in U$, $v\in V$, and a continuous path $\gamma$ joining them. The nonempty disjoint sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$ would separate $[0,1]$, contradicting connectedness of an interval. Hence path-connectedness implies connectedness. Equivalently, the connected path image from part (i) cannot meet both sides of a separation.