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The meromorphic continuations of $F$ and the gamma function have the same recurrence and the same value at $1$. Hence their residues at every nonpositive integer are the same. Their difference $f=F-\Gamma$ therefore has removable singularities there and is analytic elsewhere, so it extends to an entire function. Equivalently, $f(z+1)=zf(z)$ and $f(1)=0$ show inductively that the apparent quotient singularities are removable.

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

The Taylor series of the entire function $g$ givesThe coefficient of $(z-z_0{)}^{-1}$, and therefore the residue, isIf $g(z_0)=0$, the apparent pole is a removable singularity and the residue is zero.

Write $a=s+ib$ with real $s,b$, and first translate the real integration variable by $s$. It remains to integrate $e^{-z^2}$ along the horizontal line $\mathrm{Im}z=b$.

For $b>0$, apply the Cauchy integral theorem to the rectangle with vertices $-R,R,R+ib,-R+ib$. The integrand is an entire function, sois the negative of the two vertical-edge integrals. On either vertical edge, $z=\pm R+iy$ andTheir combined absolute value is at most $2b{e}^{-R^2+b^2}$ and tends to zero. The rectangle below the real axis gives the same estimate with $|b|$ when $b<0$; $b=0$ needs no deformation. Absolute convergence follows from $|e^{-(x+ib{)}^2}|=e^{b^2}{e}^{-x^2}$. Taking $R\to \infty$ proves the Gaussian contour translation identity

Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the zeros are exactlyAt each point ${\cos }^{\prime }(z_n)=-\sin z_n\ne 0$, so the cosine has a simple zero. Cubing it givesThe order of a zero of a holomorphic function is therefore three at every zero. There are no finite singularities because this is an entire function.

To classify infinity, examine ${\cos }^3(1/w)$ at $w=0$. The identity ${\cos }^{3}z=(3\cos z+\cos 3z)/4$ shows that its Laurent series contains the nonzero negative-power coefficientsThere are infinitely many terms in the principal part, the criterion for an essential singularity. Therefore $\boxed{\infty \text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial entire function at infinity.

The Cauchy-Riemann equations determine the harmonic conjugate. Firstso $v_y=u_x$ integrates with respect to $y$ toNowand the second Cauchy-Riemann equation, $v_x=-u_y$, forces $h^{\prime }(x)=0$. HenceThe normalization at zero gives $C=0$, so the required entire function is

Put $S={\sum }_{j=0}^{n-1}|a_j|$ and choose $R>\max (1,2S/|a_n|)$. For $|z|>R$, the triangle inequality givesThe reverse triangle inequality therefore yieldsIf $P$ had no root, the function $h=1/P$ would be an entire function. The bound just proved makes $h(z)\to 0$ as $|z|\to \infty$. On the remaining closed disk, continuity and the extreme value theorem make $h$ bounded. Thus $h$ is a bounded entire function, so Liouville's theorem makes it constant. Its limit at infinity forces that constant to be zero, impossible for a reciprocal.

Consequently $P$ has a root in $\mathbb{C}$. This proves the existence part of the fundamental theorem of algebra. The invoked Liouville theorem follows from the Cauchy integral formula: the Cauchy estimate $|h^{\prime }(w)|\le M/r$ on arbitrarily large circles gives $h^{\prime }=0$ when $h$ is bounded by $M$.

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.

The hyperbolic sine satisfies $\sinh z=(e^z-e^{-z})/2$. The complex exponential function is entire, and sums and compositions of entire functions are entire. Therefore

The expression is $F(z)=e^{z^2+1/z}=e^{z^2}{e}^{1/z}$, which is holomorphic for every $z\ne 0$. At zero,has infinitely many nonzero negative-power terms, so it has an essential singularity. Multiplication by the nonvanishing entire function $e^{z^2}$ preserves that classification: if $F$ had a removable singularity or a pole, multiplying by $e^{-z^2}$ would give the same type for $e^{1/z}$, a contradiction. Hence

Suppose $w\in \mathbb{C}$ were omitted by $g$. Thenwould be an entire function. The reverse triangle inequality gives $|g(z)-w|\ge |g(z)|-|w|\to \infty$ as $|z|\to \infty$, so $F(z)\to 0$ there. On a sufficiently large closed disk, $F$ is bounded by continuity and the extreme value theorem; outside it, the limit gives a bound. The Liouville theorem therefore makes $F$ constant, and the limit forces $F\equiv 0$.

That is impossible, since $1/(g(z)-w)$ is nonzero at every finite $z$. No value $w$ can be omitted, and hence

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ is holomorphic on $D_S$ and $h$ is holomorphic on $A_{R,\infty }$. To check their extended regions of convergence explicitly, fix $r\in (R,S)$ and put $M_r={\max }_{|z|=r}|f(z)|$. The Laurent coefficient formula givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; letting $r$ range down to $R$ covers $A_{R,\infty }$. Locally uniform convergence of these power series, in $z$ and $1/z$ respectively, proves the claimed holomorphy. On the original annulus their sum is $f$.

The second estimate also proves $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty }$. The two definitions agree on the overlap, and their domains cover $\mathbb{C}$, so they define an entire function. Moreover, $F(z)\to 0$ at infinity. It is bounded outside a large disk by this limit, and inside by continuity and the extreme value theorem. The Liouville theorem makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

The Riemann mapping theorem gives a conformal bijection $G\to D$ exactly whengiven the hypothesis that $G$ is a simply connected domain. A domain is nonempty, open, and connected, so no extra nonemptiness qualification is needed.

For necessity, a conformal bijection $\mathbb{C}\to D$ would be a bounded nonconstant entire function, contradicting the Liouville theorem. For sufficiency, $G$ is a proper simply connected domain, exactly the hypothesis of the Riemann mapping theorem.