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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.