Power series

Because \(f\) is holomorphic, \(f'\) is holomorphic.

If the zeros of \(f'\) had an accumulation point in \(\Omega\), the identity theorem would give \[ f'\equiv0. \] A holomorphic function with zero derivative on a connected domain is constant, contradicting the hypothesis. Hence the zeros of \(f'\) are isolated.

For the second statement, let \[ h=f-g. \] Then \(h\) is holomorphic and \[ h'=f'-g'=0. \] Therefore \(h\) is constant on the connected domain \(\Omega\). Thus \[ f-g=C \] for some \(C\in\mathbb C\).

Write \(S_N(z) \da \sum_{0\leq k\leq N} c_k (z-z_0)^k\) and \(S \da \lim_{N\to\infty} S_N\). Suppose \(R\da \inverseof{\qty{\limsup_k \abs{c_k}^{1\over k} }}\) is the radius of convergence and let \(r\leq R\), we’ll show \(S_N\to S\) uniformly on any disc \(\abs{z-z_0}< r\).

Use the \(M\dash\)test: \(\sum f_k\) converges if \(\norm{f_k}_\infty\leq M_k\) where \(\ts{M_k}\in \ell^1(\NN)\). Define \(f_k \da c_k (z-z_0)^k\), then \[ \norm{f_k}_\infty =\sup_{\abs{z-z_0}\leq r} \abs{c_k(z-z_0)^k} \leq \abs{c_k} r^k \da M_k .\] Then \[ \sum_{k\geq 0} M_k = \sum_{k\geq 0} \abs{c_k} r^k ,\] and the claim is that this converges.

Note that since \(r\leq R\), we have convergence of \[ \sum_{k\geq 0} c_k r^k .\] Recall that root test: \[ \sum_k a_k \text{ converges absolutely if } \limsup_k \abs{a_k}^{1\over k} < 1 .\] Here we take \(a_k \da c_k r^k\), then \[ \limsup_k \abs{a_k}^{1\over k} &\da \limsup_k \abs{c_k r^k}^{1\over k} \\ &= \limsup_k \abs{c_k}^{1\over k} r \\ &< \limsup_k \abs{c_k}^{1\over k} R\\ &\da \limsup_k \abs{c_k}^{1\over k} \inverseof{\qty{\limsup_{k} \abs{c_k}^{1\over k} }} \\ &= 1 ,\] so \(\sum_k \abs{c_k r^k} < \infty\). Thus \(\ts{M_k}\in \ell^1(\NN)\), and so \(\sum_k f_k\) converges uniformly and absolutely on \(\abs{z-z_0} = r < R\).

Let \(f(z) = \lim_{N\to\infty} S_N(z)\) with \(S_N(z)=\sum_{k\leq N} c_k (z-z_0)^k\), and let \(R>0\) be the radius of convergence. The domain of convergence is the open disk \(\abs{z-z_0}<R\), the interior of the set where the series converges. For \(0<r<R\), the series converges absolutely and uniformly on the closed disk \(\abs{z-z_0}\le r\) (Weierstrass \(M\)-test with \(M_k=\abs{c_k}r^k\), which is summable since \(r<R\)). Each partial sum \(S_N\) is a polynomial, hence continuous, so \(f\) is continuous on \(\abs{z-z_0}\le r\) by the uniform limit theorem. Every point of the open disk \(\abs{z-z_0}<R\) lies in such a closed disk, so \(f\) is continuous on the open disk of convergence.

That \(f\) is continuous is a local question: fixing a point \(z_0\), take a closed disc \(\DD+z_0\) about \(z_0\). By local uniform convergence \(f_k\to f\) uniformly on \(\DD+z_0\), and differentiable \(\implies\) continuous. So each \(f_k\) is continuous, making \(f\) continuous on \(\DD+z_0\) by the uniform limit theorem.

That \(f\) is differentiable is again a local question: fix \(z\) and write \(\gamma \da \bar{\DD + z}\) as the boundary of the disc about \(z\). Define \(g_k(\xi) \da {f_k\over \xi-z}\), so \(g_k \to {f \over \xi-a}\) locally uniformly. Now apply Cauchy’s integral formula at \(z\): \[ f(z) &= \lim_k f_k(z) \\ &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over \xi - z}\dxi \\ &= \lim_k {1\over 2\pi i}\int_\gamma g_k(\xi)\dxi\\ &= {1\over 2\pi i}\int_\gamma \lim_k g_k(\xi)\dxi \\ &= {1\over 2\pi i}\int_\gamma g(\xi)\dxi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over \xi - z} \dxi ,\] where we’ve used uniform convergence on \(\gamma\) to commute the limit and integral. So \(f\) has an integral representation, making it differentiable.

That \(f_k'\to f'\): \[ \lim_k f_k'(z) &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over (\xi - z)^2 }\dxi\\ &= {1\over 2\pi i}\int_\gamma \lim_k {f_k(\xi) \over (\xi - z)^2}\dxi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over (\xi - z)^2}\dxi\\ &= f'(z) .\]

That the convergence is locally uniform: first consider what happens on an closed discs \(K = D\) with \(\gamma \da \bd{D}\). Then for \(z\in D\), \[ \abs{f'(z) - f_k'(z) } &= \abs{{1\over 2\pi i} \int_{\gamma} {f(\xi) - f_k(\xi) \over (\xi - z)^2}\dxi }\\ &\leq {1\over 2\pi}\int_{\gamma} {\abs{f(\xi) - f_k(\xi) } \over \abs{\xi - z}^2} \dxi\\ &\leq {1\over 2\pi}\int_{\gamma } { \sup_{\xi \in \gamma} \abs{f(\xi) - f_k(\xi) } \over r^2 } \dxi\\ &= {1\over 2\pi} { \sup_{\xi \in \gamma } \abs{f(\xi) - f_k(\xi) } \over r^2} \cdot {2\pi r} \\ &= { \sup_{\xi \in \gamma } \abs{f(\xi) - f_k(\xi) }}/r .\] Since \(\gamma\) is compact, using locally uniform convergence of \(f_k\to f\), there exists an \(n_0\) such that \(n\geq n_0\) bounds this \(\sup\) by \(\eps\). For \(K\) arbitrary, cover \(K\) by discs \(D_z\) for every \(z\in K\) and extract a finite cover \(\ts{D_{z_k}}_{k\leq N}\). Produce \(n_0, n_1,\cdots, n_N\) as in the above argument, and take \(n\da \max\ts{n_k}_{k\leq N}\) to obtain uniform convergence on every \(D_{z_k}\) and thus on \(K\).

By Cauchy–Hadamard, for \(\sum c_kz^k\), \[ {1\over R}=\limsup_{k\to\infty}|c_k|^{1/k}. \]

For \(c_k=a^k\), \[ {1\over R}=|a|, \] so \[ R= \begin{cases} \infty,&a=0,\\ |a|^{-1},&a\neq0. \end{cases} \]

For \(c_k=a^{k^2}\), \[ {1\over R} =\limsup_{k\to\infty}|a|^k. \] Hence \[ R= \begin{cases} \infty,&|a|<1,\\ 1,&|a|=1,\\ 0,&|a|>1. \end{cases} \]

For a center \(c\neq0\), the disk \(\abs{z-c}<\abs c\) does not contain \(0\), so it carries a holomorphic branch of \(\sqrt z\) agreeing with the principal branch near \(c\); the Taylor series of the principal branch at \(c\) converges on this disk. It converges on no larger disk: the binomial expansion \[ \sqrt z=\sqrt c\sum_{n\ge0}\binom{1/2}{n}\qty{z-c\over c}^n \] has radius of convergence exactly \(\abs c\), since \(\binom{1/2}{n}\) has ratio \(\abs{1/2-n}/(n+1)\to1\).

For \(z_0=4+3i\), \[ R_0=|4+3i|=\boxed{5}. \]

For \(z_1=-4+3i\), \[ R_1=|-4+3i|=\boxed{5}. \] The branch cut \(\RR_{\le0}\) meets this disk: the sum of the series equals the principal branch only on the part of the disk above the cut, including the disk \(\abs{z-z_1}<3\), and equals its analytic continuation across the cut below it.

Break this up into a principal part at \(z=0\) and a holomorphic part: \[ f(z) = f_1(z) + f_2(z) \da \sum_{k\geq 1} 2^{-k}z^{-k} + \sum_{k\geq 0} 2^{-k}z^k .\]

Using the ratio test: \[ f_1(z) < \infty &\impliedby \limsup_k \abs{2^{-k}z^{-k}}^{1\over k} < 1 \iff \limsup_k \abs{1\over 2z} < 1 \iff {1\over 2}< \abs{z} \\ f_2(z) < \infty &\impliedby \limsup_k \abs{2^{-k}z^{k}}^{1\over k} < 1 \iff \limsup_k \abs{z\over 2}< 1 \iff \abs{z} < 2 .\]

So \(f\) converges on \({1\over 2}< \abs{z} < 2\).

\[ {1\over z(z-1)} = -{1\over z}{1 \over 1-z} = -{1\over z}\sum z^k .\] and \[ {1\over z(z-1)} = {1\over z^2(1 - {1\over z})} = {1\over z^2} \sum \qty{1\over z}^k .\]

Note: once you see that everything is in terms of powers of \((z-z_0)\), you’re essentially done. For \(z=0\): \[ {z+1 \over z(z-1)} &= {1\over z} {z+1 \over z-1} \\ &= -{z+1\over z} {1\over 1-z} \\ &= -\qty{1 + {1\over z}}\sum_{k\geq 0} z^k .\]

For \(z=1\): \[ {z+1 \over z(z-1)} &= {1\over z-1}\qty{1 + {1\over z} } \\ &= {1\over z-1}\qty{1 + {1\over 1 - (1-z)} } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (1-z)^k } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (-1)^k (z-1)^k } .\]

The three regions are

• \(0 \leq \abs{z} < 1\)

• \(1 < \abs{z} < 3\)

• \(3 < \abs{z} < \infty\)

Write \(f\) in terms of its principal parts at \(z=1\) and \(z=3\) by computing the residues:

• \(\Res_{z=1}f(z) = (z-1)f(z)\evalfrom_{z=1} = {1\over z-3}\evalfrom_{z=1} = -{1\over 2}\)

• \(\Res_{z=3}f(z) = (z-3)f(z)\evalfrom_{z=3} = {1\over z-1}\evalfrom_{z=3} = {1\over 2}\)

Thus \[ f(z) = {-1/2 \over z-1} + {1/2 \over z-3} .\]

Now find the two expansions for each term:

\[ {-1/2 \over z-1} &= {1/2 \over 1-z} = {1\over 2}\sum_{k\geq 0} z^k && 0 < \abs{z} < 1 \\ {-1/2 \over z-1} &= -{1\over 2}{\inverseof{z} \over \inverseof{z} - 1} = -{1\over 2z}{1\over 1-\inverseof{z}} = -{1\over 2}\sum_{k\geq 0}z^{-k-1} && 1 < \abs{z} < \infty \\ {1/2\over z-3} &= -{1\over 2}{1\over 3-z} = -{1\over 6}{1\over 1-{z\over 3}} = -{1\over 6}\sum_{k\geq 0}3^{-k} z^k && 0 < \abs{z} < 3 \\ {1/2\over z-3} &= {1\over 2z}{1\over 1-3\inverseof{z}} = {1\over 2z} \sum_{k\geq 0}3^kz^{-k} = {1\over 2}\sum_{k\geq 0}3^k z^{-k-1} && 3 < \abs{z} < \infty .\]

On each region, combine the expansions of the two terms that converge there: \[ 0 \leq \abs{z} < 1 \qquad & f(z) = {1\over 2}\sum_{k\geq 0}z^k - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 1 < \abs{z} < 3 \qquad & f(z) = -{1\over 2}\sum_{k\geq 0}z^{-k-1} - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 3 < \abs{z} < \infty \qquad & f(z) = - {1\over 2}\sum_{k\geq 0}z^{-k-1} + {1\over 2}\sum_{k\geq 0} 3^{k}z^{-k-1} .\]

\[ \cos\qty{z\over3} =\sum_{k=0}^\infty(-1)^k{(z/3)^{2k}\over(2k)!}, \] so \[ z^2\cos\qty{z\over3} =\sum_{k=0}^\infty(-1)^k{z^{2k+2}\over 3^{2k}(2k)!} =z^2-{z^4\over 2!\,3^2}+{z^6\over 4!\,3^4}-\cdots. \] This is an ordinary Taylor series; there are no negative powers.

One way: polynomial long division. \[ {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^2 + \cdots } \\ &= {1\over z}\qty{1 - {1\over 2}z + \qty{-{1\over 6} + {1\over 4} }z^2 + \cdots } \\ &= \inverseof{z} - {1\over 2} + {1\over 12}z + \bigo(z^3) .\] Alternatively, use geometric series. Note that something like \({1\over 1-e^z} = \sum_{k\geq 0} e^{kz}\) won’t converge, and won’t even be calculable since each \(e^{kz}\) contributes a constant term! \[ {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^3 + \cdots } \\ &= {1\over z(1 + {1\over 2}z + {1\over 6}z^2 + \cdots) } \\ &= \inverseof{z} {1\over 1 + q(z) } \qquad q(z) \da {1\over 2}z + {1\over 6}z^2 + \cdots \\ &= \inverseof{z} \sum_{k\geq 0}(-q(z))^k \\ &= \inverseof{z} \qty{1 - q(z) + q(z)^2 - \cdots } \\ &= \inverseof{z}\qty{1 - \qty{{1\over 2}z + {1\over 6}z^2 + \cdots } + \qty{{1\over 2}z + {1\over 6}z^2 + \cdots }^2 - \cdots } \\ &= \inverseof{z}\qty{ 1 - {1\over 2} z + \qty{-{1\over 6} + \qty{1\over 2}^2 } z^2 + \bigo(z^3) } \\ &= \inverseof{z} - {1\over 2} + {1\over 12}z + \bigo(z^2) .\]

Let \(g(z) \da e^z\cos(z)\), an entire function with \(g(1/z ) = f(z)\) for \(z\ne0\). The Taylor series of \(g\) at \(0\), evaluated at \(1/z\), is the Laurent series of \(f\) on \(\abs z>0\): \[ g(z) &= e^{z}\cos(z)\\ &= {1\over 2}e^z\qty{e^{iz} + e^{-iz}} \\ &= {1\over 2}\qty{e^{(1+i)z} + e^{(1-i)z}} \\ &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {z^k\over k!} \\ \implies f(z) &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {1 \over k!z^k } \\ .\]

Write \(g(z) \da \cos(1-z)\), so \(g(1/z) = f(z)\), and expand: \[ g(z) &= \cos(1-z) \\ &= {1\over 2}\qty{e^{i(1-z)} + e^{-i(1-z)}} \\ &= {1\over 2}\qty{e^i e^{-iz} + e^{-i} e^{iz}}\\ &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {z^k \over k!} \\ \implies f(z) &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {1 \over k!z^k} .\]

Taking \(k=1\) yields \[ c_{-1} = {-ie^i + ie^{-i} \over 2} = -i\cdot {e^i - e^{-i}\over 2} = {e^i - e^{-i}\over 2i} = \sin(1) .\]

\[ f(z) = e^{-(1-z)} = e^{z-1} = \inverseof{e} e^z = \inverseof{e}\sum_{k\geq 0} {z^k\over k!} .\] Since \(e^z\) is entire, this converges on \(\CC\).

At \(z=0\), we can use a geometric series approach since \(\abs{e^z} = e^{\Re(z)} \leq 1\) near \(0\). However, we still have to get rid of the leading 1 in the expansion of \(e^z\) in order to get a constant coefficient. \[ {1\over 1 + e^z} &= {1\over 1 + 1 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + \bigo(z^4)} \\ &= {1\over 2 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + \bigo(z^4) } \\ &= {1\over 2} {1\over 1 + {1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!} z^3 + \bigo(z^4) } \\ &= {1\over 2}{1\over 1 - (-p(z)) } \qquad p(z) \da {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo(z^4) \\ &= {1\over 2} \sum_{k\geq 0} (-p(z))^k \\ &= {1\over 2}\Big[ 1 - \qty{{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} } \\ &\qquad + \qty{ {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} }^2 \\ &\qquad - \qty{{1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} }^3 \\ &\qquad - \bigo(z^4) \Big]\\ &= {1\over 2} \Big[ 1 + z\qty{- {1\over 2}} + z^2\qty{- {1\over 2\cdot 2!} + \qty{1\over 2}^2}\\ &\qquad + z^3 \Big( -{1\over 2\cdot 3! } +\left[ \qty{1\over 2}^3 + {1\over 2}{1\over 2\cdot 2!}\right] - \qty{1\over 2}^3 \Big) \\ &\qquad + \bigo(z^4) \Big]\\ &= {1\over 2} - {1\over 4}z + 0z^2 + {1\over 48}z^3 + \bigo(z^4) .\]

Expanding at \(z-i\pi\): quite a bit easier. Let \(\omega \da z-i\pi\), then \[ {1\over 1 + e^z} &= {1\over 1 + e^{z-i\pi}e^{i\pi}} \\ &= {1\over 1 - e^{\omega} } \\ &= {1 \over -\omega - {1\over 2!}\omega^2 - {1\over 3!}\omega^3 - \bigo(\omega^4) } \\ &= -{1\over \omega} {1 \over 1 + {1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3) } \\ &= -{1\over \omega}{1\over 1-(- p(z) ) } \qquad p(z) \da \sum_{k\geq 2}{\omega^{k-1}\over k!} \\ &= -{1\over \omega} \sum_{k\geq 0} (-p(z))^k \\ &= -{1\over \omega} \left[ 1 - \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3)} + \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3)}^2 - \bigo(\omega^3) \right] \\ &= -{1\over \omega} \left[ 1 + \omega\qty{-{1\over 2!}} + \omega^2\qty{-{1\over 3!} + \qty{1\over 2!}^2} + \bigo(\omega^3) \right] \\ &= -{1\over w} + {1\over 2} - {1\over 12}\omega + \bigo(\omega^2) .\]

1. Use that \(\abs{z^k} = 1\) and \(\sum c_kz^k < \infty \implies \abs{c_k} \to 0\), but \(\abs{kz^k} = \abs{k} \to \infty\) here.

2. Use that absolutely convergent implies convergent, and \(\sum \abs{k^{-2} z^k} = \sum \abs{k^{-2}}\) converges by the \(p\dash\)test.

3. If \(z=1\), this is the harmonic series. Otherwise take \(a_k = 1/k, b_k = e^{i k \theta}\) where \(\theta \in (0, 2\pi)\) is some constant, and apply Dirichlet’s test. It suffices to bound the partial sums of the \(b_k\). Recalling that \(\sum_{k\leq N} r^k = (1-r^{N+1}) / (1-r)\), \[ \norm{ \sum_{k\leq m} e^{ik\theta } } = \norm{1 - e^{i(m+1)\theta} \over 1 - e^{i\theta}} \leq {2 \over \norm{ 1- e^{i\theta}}} \da M ,\] which is a constant. Here we’ve used that two points on \(S^1\) are at most distance 2 from each other.

Goal: Show that a real-valued holomorphic function on a domain must be constant.

<1>1. Since \(\{a_n\} \in \ell_1(\ZZ)\), we have \(\sum_{n \in \ZZ} |a_n| < \infty\).

<1>2. Hence \(a_n \to 0\), so for all sufficiently large \(|n|\) we have \(|a_n| < 1/2\).

<1>3. For \(|a_n| < 1/2\), \(|\log(1 + a_n)| \le 2|a_n|\).

<1>4. Hence \(\sum_{n} |\log(1 + a_n)| \le 2\sum_n |a_n| < \infty\).

<1>5. Therefore \(\sum_n \log(1 + a_n)\) converges absolutely, so the product \(\prod_n (1 + a_n)\) converges to a finite nonzero value.

<1>6. Q.E.D.

Write \(f = u + iv\), then \(0 = 2 f' = u_x + iv_x = u_y - iu_y\), so \(\grad u = \grad v = 0\). Show \(f\) is constant along every straight line segment \(L\) by computing the directional derivative \(\grad u \cdot \vector v = 0\) along \(L\) connecting \(p, q\). Then \(u(p) = u(q) = a\) some constant, and \(v(p) = v(q) = b\), so \(f(z) = a+bi\) everywhere.

• Strategy: show \(f'=0\).

• Write \(f = u + iv\). Since \(f\) is analytic, it satisfies CR, so \[ u_x = v_y && u_y = -v_x .\]

• Similarly write \(\bar f = U + iV\) where \(U = u\) and \(V = -v\). Since \(\bar f\) is analytic, it also satisfies CR , so \[ U_x = V_y && U_y = -V_x \\ \\ \implies u_x = -v_y && u_y = v_x .\]

• Add the LHS of these two equations to get \(2u_x = 0 \implies u_x = 0\). Subtract the right-hand side to get \(-2v_x = 0 \implies v_x = 0\)

• Since \(f\) is analytic, it is holomorphic, so \(f'\) exists and satisfies \(f' = u_x + iv_x\). But by above, this is zero.

• Since \(\Omega\) is connected and \(f'=0\) on \(\Omega\), \(f\) is constant.

Here \(\Omega\) is a region, that is, open and connected. Write \(f=u+iv\) with \(u,v\) real.

<1>1. (1) If \(u\) is constant on \(\Omega\), then \(f\) is constant.

<1>2. (2) If \(v\) is constant on \(\Omega\), then \(f\) is constant.

<1>3. (3) If \(\abs{f} = c\) on \(\Omega\), then \(f\) is constant.

<2>1. If \(c=0\), then \(f=0\).

<2>2. If \(c>0\), then \(\bar f\) is holomorphic on \(\Omega\).

<2>3. Q.E.D.

<1>4. Q.E.D.

Write \(f(z) = \sum_{k\geq 0} c_k (z-a)^k\) on \(\DD_r(a)\), where \(c_k = f^{(k)}(a)/k!\). Recall that \(a\) is a zero of multiplicity \(m\) when \(f(z)=(z-a)^mg(z)\) with \(g\) holomorphic near \(a\) and \(g(a)\neq0\).

\(\impliedby\): Suppose \(f^{(k)}(a)=0\) for \(k\le m-1\) and \(f^{(m)}(a)\ne0\), so \(c_k=0\) for \(k\le m-1\) and \(c_m\ne0\). Then \[ f(z) = \sum_{k\geq m} c_k (z-a)^k = (z-a)^m \sum_{k\geq m} c_k (z-a)^{k-m} \da (z-a)^m g(z) ,\] where \(g\) is holomorphic on \(\DD_r(a)\) with \(g(a) = c_m \neq 0\), making \(a\) a zero of \(f\) of multiplicity \(m\).

\(\implies\): If \(a\) is a zero of multiplicity \(m\), write \(f(z) = (z-a)^m h(z)\) with \(h\) holomorphic near \(a\) and \(h(a)\neq0\), and expand \(h(z)=\sum_{j\ge0}d_j(z-a)^j\) with \(d_0=h(a)\). Then \(f(z)=\sum_{j\geq0}d_j(z-a)^{j+m}\), and by uniqueness of power series coefficients, \(c_k=0\) for \(k\le m-1\) and \(c_m=d_0\neq0\). That is, \(f^{(k)}(a)=0\) for \(k\le m-1\) and \(f^{(m)}(a)=m!\,h(a)\neq0\).

By contradiction: after translating, suppose \(0\in\Omega\), \(f(0) = 0\) and \(f'(0)=0\). An injective \(f\) is nonconstant, so \(0\) is a zero of \(f\) of some finite order \(m\), and \(m\geq2\) since \(f(0)=f'(0)=0\). Write \(f(z)=z^mg(z)\) with \(g\) holomorphic near \(0\) and \(g(0)\neq0\). On a small disk \(D\) about \(0\), \(g\) is nonvanishing and has a holomorphic \(m\)-th root \(h\), so \(f=\varphi^m\) with \(\varphi(z)\da zh(z)\). Since \(\varphi'(0)=h(0)\neq0\), \(\varphi\) maps a neighborhood \(U\subseteq D\) of \(0\) biholomorphically onto a disk \(\abs w<\delta\). For \(0<\eps<\delta\) and \(\zeta=e^{2\pi i/m}\neq1\), the distinct points \(w_1=\eps\) and \(w_2=\zeta\eps\) satisfy \(w_1^m=w_2^m\), so their preimages under \(\varphi\) are distinct points of \(U\) with the same image under \(f\), contradicting injectivity.