Let $f(z)$ be analytic on the unit disc, and suppose that there is an annulus $U = \{z ∈ C \mid r &lt; |z| &lt; 1\}$ such that $f(z)$ restricted to the annulus $U$ is injective. Show that $f$ is injective on the unit disc.  I tried to show that $g(z) = f(z) - f(z_0)$ has only one zero in the unit disk, but how can I use the hypothesis of the annulus?
21389
Is the following statement true?  Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ D $.  P.S. I think I got a solution. But I was wondering if this problem has a trivial solution.  Here is my solution, since nobody has posted a solution yet:-  Suppose, $ \gamma $ is the boundary of $ D $(assuming $ \gamma(t)=e^{2\pi it} $). Then $ f\circ \gamma $ is a simple continuously differentiable curve. So by Jordan curve theorem, $ f\circ \gamma $ divides $ \mathbb C $ into 2 path connected regions $ A_1, A_2 $ such that if $ a\in A_1 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=2\pi i$ and if $ a\in A_2 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=0$. So, $ f(z)=a $ for some $ a\in A_1,z\in D $ implies $ \int_{\gamma} \frac {f'(z)}{f(z)-a} dz=2\pi i $. So $ f(z)=a $ has only one solution in $ D $. Similarly if $ a\in A_2 $, then $\int_{\gamma} \frac {f'(z)}{f(z)-a}=0 $, so $ f(z)=a $ has no solution in $ D $. If $ f(z)=a $ for some $ z\in D, a\in f\circ \gamma $, then by open mapping theorem, $ f(z)=a' $ for some $ z\in D, a'\in A_2 $, which is not possible. So $ f $ is one-one in $ D $.
115418
Let $D \subset \mathbb{C}$ be a domain and let $a \in D$. Suppose $f: D \smallsetminus \{a\} \to \mathbb{C}$ is analytic and that $a$ is an essential singularity of $f$. Show that $f$ cannot be univalent (= injective) in any neighborhood of $a$.  This is a trivial consequence of the Picard theorem. But I don't know if there is any elementary approach.
123954
Let $U$ be an open connected neighborhood of $0$ in $\Bbb{C}$ and $f$ a holomorphic function on $U$. The exercise is to prove that if $f(\frac{1}{n})=\frac{1}{n+1}$ for all $n\ge 1$ then $f(z)=\frac{z}{z+1}$ on $U$.  So I did it using the function $g:\Bbb{C} \setminus\{-1\}\to \Bbb{C}$ such that $g(z)=\frac{z}{1+z}$ so $g$ is holomorphic on $\Bbb{C} \setminus\{-1\}$, now as $U$ is open we can find a disc $D(0,r)\subset U$ for $r$ so that $r&lt;1$. Therefore, using analytic continuation we have $f=g$ on $D(0,r)$. My main problem is now to prove that $U\subset\Bbb{C} \setminus\{-1\}$. Perhaps, I can say that if $-1\in U$ we can find a path from $0$ to $-1$ but not sure how can I continue.
9645
Let $f: B(0,1) \rightarrow \mathbb{C}$ be holomorphic and suppose $\exists\ r \in (0,1)$ such that $f$ is injective in $A = \{z \in \mathbb{C} : r &lt; |z| &lt; 1\}$. Prove that $f$ is injective.  I tried using Rouché theorem, or the identity theorem, but I don't know what to do. Any hints? :)
16796
I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?  Also, more specific questions about the different cases:   Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple? Essential singularity case: Why exactly is $f(\{|z| &gt; r \} \cap f(\{|z|&lt;r\}) \neq \emptyset$? $f(\{|z| &gt; r \})$ is dense, but how does $f(\{|z|&lt;r\}$ being open guarantee that their union is non-empty?
229
I am completely stuck in the following problem:  let f be holomorphic function on the unit disk $\{z:|z|\lt 1\}$ in the complex plane.which of the following is/are necessarily true?  (a) if for each positive integer $n$ we have $f(\frac{1}{n})=\frac{1}{n^2}$,then $f(z)=z^2$ on the unit disc.  (b) if for each positive integer $n$ we have $f(1-\frac{1}{n})=(1-\frac{1}{n})^2$,then $f(z)=z^2$ on the unit disc.  (c) $f$ cannot satisfy $f(\frac{1}{n})=\frac{(-1)^n}{n}$ for each positive integer $n$  (d) $f$ cannot satisfy $f(\frac{1}{n})=\frac{1}{n+1}$ for each positive integer n   what I thought in this problem is the following:  in the first part,by identity theorem we can say that $f(z)=z^2,\forall$z in the unit disc.  but,in the second part ,if we choose $g(z)=z^2$ then,$f(1-\frac{1}{n})=g(1-\frac{1}{n})$ but,the limit point ,i.e, $1$ is not in the unit disc.so,in this case,we cannot use identity theorem to conclude whether $f(z)=z^2$ on the unit disk or not .so,what procedure can we follow in this case???also,explain the options (c) and (d).  please answer in detail..  thanks a lot...
45698
Let $D = B(0,1) \subset \mathbb{C} $ a disc, $f$ holomorphic on $D$. I want to demonstrate that if $$ 2|f^{'}(0)| = \sup_{z, w \in D} |f(z)-f(w)|$$  then $f$ is linear.   I know this is a well-known result. Where can I find the proof ?
97399
Specifically, the question is as follows:     Let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$ and suppose that $f:\bar{\mathbb{D}}\rightarrow\mathbb{C}$ is a continuous function such that both $\Re f$ and $\Im f$ are harmonic.       (a) Show that   $$f(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta-t)dt$$   for all $re^{i\theta}$ in $\mathbb{D}$.       (b) Show that $f$ is holomorphic in $\mathbb{D}$ if and only if   $$\int_{-\pi}^\pi f(e^{it})e^{int}dt=0$$   for all $n\geq1$.   I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:  Let $\mathbb{D}$ be the open unit disk and $f:\partial\mathbb{D}\rightarrow\mathbb{R}$ be continuous. Then there exists a continuous function $u:\bar{\mathbb{D}}\rightarrow\mathbb{R}$ such that $u(z)=f(z)$ if $z\in\partial\mathbb{D}$, $u$ is harmonic on $\mathbb{D}$, and $u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)f(e^{it})dt$ and $u$ is unique.  It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?  For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.
34513
I'm having trouble with the following exercise from Ahlfors' text (not homework)  "If $f(z)$ is analytic in a neighborhood of $\infty$ and if $z^{-1} \Re f(z) \to0$ as $z \to \infty$, show that $\lim_{z \to \infty} f(z)$ exists. (In other words, the isolated singularity at $\infty$ is removable)  Hint: Show first, by use of Cauchy's integral formula, that $f = f_1 +f_2$ where $f_1(z) \to 0$ for $z \to \infty$ and $f_2(z)$ is analytic in the whole plane.  My attempt:  $f$ is analytic in the exterior of some disk $\{|z|\geq R_0 \}$. For large enough $|z|$ the circle $C_1(z)$ around $z$ with radius one lies in the domain of analyticity, and we can apply Cauchy's integral formula:  $$f(z)=\frac{1}{2\pi i}\oint_{C_1(z)}\frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta=\frac{1}{2 \pi i} \oint_{C_1(z)}\frac{\Re f(\zeta)}{\zeta-z} \mathrm{d} \zeta+\frac{1}{2\pi} \oint_{C_1(z)}\frac{\Im f(\zeta)}{\zeta-z} \mathrm{d} \zeta. $$ I want to say that the last expression is exactly the decomposition $f_1+f_2$. Indeed, using Cauchy's estimate (and the fact about the real part) one can show that the first integral tends to zero for large $|z|$, but I can't understand how to define the other one for all $z \in \mathbb C$.  In fact, even if I have such a decomposition, I don't think that it will suffice: For instance $f(z)=f_1(z)+f_2(z)$ with $f_1(z)=0,f_2(z)=\exp(z)$ has an essential singularity at $\infty$.  Thank you for any directions.
61181
How can one show there is no holomorphic function $f$ on the open unit disc $\mathbb{D}$ such that it extends continuously to $\frac{1}{z}$ on $\partial\mathbb{D}$? I mean $f$ takes value $\frac{1}{z}$ when $|z|=1$ and continuous here.  Thank you.
113480
Let $\gamma$ be the circle $\{z \in \mathbb{C}: \lvert z\rvert=1  \}$. Suppose $f$ is a function analytic on an open set containing $\gamma$ and its interior and that $\lvert\, f(z)\rvert&lt;1$ for each $z$ on $\gamma$. Show that $f$ has exactly one fixed point inside $\gamma$   That is, there is exactly one $z$ in the open unit disk with $f(z)=z$.   Is this a result of Louville's theorem?   I don't know how to approach it.
13254
Suppose $f$ is a holomorphic map on the unit disc. Let $d$ be the diameter of the image of $f$. If $ 2|f'(0)|=d $, please show that $f$ is a linear function.   I think maybe I can prove it by using Schwarz lemma, but I failed.
59374
$n$ points $z_1,z_2,\cdots,z_n$ in the unit open disk are given. Prove or disprove that there exists $z$ in the unit circle such that $\prod_{i=1}^n |z-z_i|^i \ge 1$.  I think it can be solved by elementary analytic methods but no idea.
108995
in our online lecture we had the following statement but I don't see exactly why it's true , if anyone could explain me why  it would be appreciated.          We have  $a \in \mathbb{C}$ a complex number and $r \in \mathbb{R}$ with $r&gt;0$. Let $U=\mathbb{D}(a,r)\ _{\diagdown \left\{a \right\}}$ and $f \in O(U)$ a holomorphic function such that $Re(f(z)) \geq 0 \; \forall z \in U$.      And it says that $f$ can be extended ( in a holomorphic function )  all over the disk $\mathbb{D}(a,r)$ and that $a$ isn't a essential singular point.   Thanks in advance for your help .
82184
Suppose I have a function $f$ that is analytic on the unit disk $D = \{ z \in \mathbb{C} : |z| &lt; 1 \}$ that is also continuous up to $\bar{D}$. If $f$ is identically zero on some segment of of the boundary (e.g. $\{ e^{it}, 0 \leq t \leq \pi/2 \}$ ), is it then true that $f$ is identically zero on the entire boundary?   I know that analytic functions that are zero at an accumulation point inside the domain of analyticity are identically zero throughout the entire domain, but I don't know what (if anything) can be said if something similar occurs on the boundary of the domain.
11977
I'm having trouble proving the following:  Let $f$ be analytic on $D(0,R)$ and suppose there is a real constant A for which $\operatorname{Re}f(z) &lt; A$ for all $z \in D(0,R)$. If $f(0) = 0$, show that $|f(z)| \leq \frac{2A|z|}{R-|z|}$.   I was told to first start by showing that $|\frac{f(z)}{2A-f(z)}| \leq 1$ and also use the following result, which I was able to prove:  Let $f$ be analytic on $D(0,R)$ and let $f(0)=0$. If $|f(z)| \leq M$ for all $z \in D(0,R)$, then $|f(z)| \leq M|z|/R$ for all $z\in D(0,R)$.  Thanks in advance!
32283
This comes straight from Conway's Complex Analysis, VII.4, exercise 4.     Find an analytic function $f$ which maps $G:=$ {${z: |z| &lt; 1, Re(z) &gt; 0}$} onto $B(0; 1)$ in a one-one fashion.   $B(0;1)$ is the open unit disc.  My first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \subset B(0;1)$.  Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.  My next thought is to modify $z^2$, something like $a(z-h)^2+k$.  I've yet to work out the details, but my gut tells me this isn't the right idea.  I've been teaching myself conformal maps in preparation for a qualifying exam.  So, if there's a shockingly basic, obvious solution... please patronize me.