Codex @codex  0

A complex-valued function $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ when it has a complex derivative at every point of $U$. Requiring a derivative throughout a neighbourhood is much stronger than complex differentiability at a point. It forces locally convergent Taylor series and derivatives of every order, as follows from the Cauchy integral formula and Cauchy derivative formula.

To see the restriction in real coordinates, write $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. The real and imaginary increment quotients must agree, giving the Cauchy-Riemann equationsConversely, if $u,v$ have continuous partial derivatives on $U$ and satisfy the Cauchy-Riemann equations there, then $f$ is a holomorphic function. The equations at a lone point, without a differentiability hypothesis for the real coordinate map, are insufficient. Their role is to force the Jacobian matrix to have the special formSuch a matrix represents multiplication by a complex number; it has only two free real entries. ^[ref]

For $f(z)=z^2$, $u=x^2-y^2$ and $v=2xy$, so $u_x=v_y=2x$ and $u_y=-v_x=-2y$. Directly,At $z_0=1+i$, multiplication by $2+2i$ scales lengths by its modulus $2\sqrt{2}$ and rotates by its argument $\pi /4$. A horizontal increment $h=0.01$ gives the exact displacement $0.0201+0.02i$; a vertical increment $h=0.01i$ gives $-0.0201+0.02i$. The first-order predictions $0.02+0.02i$ and $-0.02+0.02i$ are perpendicular, and the errors have modulus $10^{-4}$.

At every nonzero $z_0$, this local rotation and scaling makes $z^2$ a local conformal map. At zero, the first-order map vanishes: $r{e}^{i\theta }$ becomes $r^2{e}^{2i\theta }$, so the argument doubles and the origin is a critical point. Globally $z$ and $-z$ have the same square. Local preservation of angles therefore coexists with global folding, a useful warning when using holomorphic functions as coordinate changes.

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.

Parametrize the positively oriented circle by $z=r{e}^{i\theta }$. The Cauchy derivative formula at the origin becomesTaking the complex conjugate of the Cauchy integral theorem and using $d\bar{z}=-ir{e}^{-i\theta }d\theta$ givesAdd this zero integral to the derivative integral. Since $f+\bar{f}=2\mathrm{Re}f=2u$, recovering an analytic derivative from the boundary real part yieldsThe complex conjugation on the integrand and the reversed sign in $d\bar{z}$ are both required; the circle's orientation itself has not been reversed.

For an entire function, the Cauchy derivative formula on a positively oriented circle of radius $R$ about $z_0$ isThe resulting Cauchy estimate is $|f^{(n)}(z_0)|\le n!M_R/R^n$, where $M_R$ is the maximum of $|f|$ on that circle.

The Liouville theorem says that a bounded entire function is constant. Indeed, if $|f|\le M$ on the plane, the Cauchy estimate with $n=1$ gives $|f^{\prime }(z_0)|\le M/R$. Sending $R$ to infinity gives $f^{\prime }(z_0)=0$ at every point, hence $f$ is constant.

For the growth assumption, apply the Cauchy estimate on circles centred at zero. For every integer $n>k$,All sufficiently high Taylor series vanish. Since the Taylor series of an entire function converges everywhere, $f$ is a polynomial, of degree at most $\lfloor k\rfloor$ when $k\ge 0$. If a negative finite exponent is interpreted for nonzero $z$, the same estimate forces every coefficient to vanish, giving $f=0$.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

True. Exhaust the quadrant by compact sets $K_n$ with connected complements and with every compact subset of $\Omega$ eventually contained in the interior of $K_n$. Runge's theorem gives a polynomial $P_n$ satisfyingThus $P_n\to f$ locally uniformly. For a fixed $z\in \Omega$, choose a small circle about $z$ eventually contained in every $K_n$. The Cauchy integral formula for derivatives applied to $P_n-f$ givesfor every $r\ge 0$.

Uniform convergence on compact subsets makes $f$ continuous. For every triangle whose closure lies in $U$, uniform convergence on its boundary permits passage through the contour integral:Morera's theorem implies that $f$ is holomorphic. If a compact set $K⋐U$ is surrounded by a finite union of contours at positive distance from $K$, the Cauchy integral formula for derivatives givesThe uniform bound on the surrounding compact set proves that $f_n^{\prime }\to f^{\prime }$ uniformly on $K$.

If $f$ had distinct zeros $c$ and $d$, choose disjoint small closed discs around them whose boundary contains no zero of $f$. Uniform convergence and Rouché's theorem imply that, for large $n$, $f_n$ has a zero in each disc, contradicting uniqueness of $c_n$. Thus $f$ has at most one zero.

For an example on the unit disc, takeThen $f_n\to f(z)=z-1$, which has no zero in the open disc. In general, Hurwitz's theorem shows that $f$ is zero-free exactly when the unique zeros escape every compact subset of $U$:Indeed, an interior accumulation point of $c_n$ is a zero of $f$, while a zero of $f$ forces the unique $c_n$ into each of its sufficiently small neighbourhoods.

Let $|f(z)|\le K$ on $\mathrm{Re}z>0$. If $\mathrm{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The Cauchy derivative formula givesThe line segment joining $z_1$ to $z_2$ remains in $\mathrm{Re}z>c$, soThus one may take $M=K/c$, as in the lipschitz bound inside a bounded analytic half-plane.

The Cauchy derivative formula says that if $f$ is holomorphic on a neighbourhood of a closed disk and $z_0$ is inside its positively oriented boundary circle $\gamma$, then for $n\ge 0$,For $n=0$ this is the Cauchy integral formula. The same formula holds for a positively oriented simple closed contour when $f$ is holomorphic on a neighbourhood of the contour and its enclosed region.

In the present problem the unit circle and its closed interior lie inside $|z|<R$. Parametrize it by $z=e^{it}$, $0\le t\le 2\pi$, so $dt=dz/(iz)$. For a positive integer $n$,The first contour integral is zero by the Cauchy integral theorem, since $f(z)z^{n-1}$ is holomorphic on the disk; the second is evaluated by the Cauchy derivative formula. ThereforeTaking $n=1$ gives the requested $\pi f^{\prime }(0)$.

For a holomorphic function on a neighbourhood of a closed disc, parametrize the Cauchy derivative formula on its boundary. The conjugate Cauchy integral theorem gives $\int \overline{f(r{e}^{i\theta })}{e}^{-i\theta }d\theta =0$. Replacing $f$ in the derivative integral by $f+\bar{f}=2\mathrm{Re}f$ therefore gives the displayed identity. The boundary real part determines every nonconstant analytic coefficient; only an imaginary additive constant is lost.