$f$ is entire, satisfying    $$|f(z)|\le \frac1{|\text{Re}(z)|}.$$   Show that $f$ is identically zero.    It suffices to show boundedness. Since $f$ is bounded on the real line, then if $f$ is unbounded on the complex plane, the infinity must be an essential singularity, which means, for an arbitrarily chosen point $z_0\in\Bbb C$, we can pick a sequence $z_n$ whose moduluses tend to infinity with $f(z_n)\to z_0$, and it follows that $|z_0|\le \frac1{|\text{Re}(z_n)|}$. This is the best I can do.   I'm not sure if I'm going in the correct direction. It however seems unlikely for me to deal with the problem on the $y$-strip in this way.
122888
This is a problem from my complex analysis textbook.  The hint is to consider $g(z)=(z-iR)^2(z+iR)^2 f(z)$ and to show that $|g(z)| \leq 8R^2$.  This is what i have tried: Consider $Re z \geq 0$, then $|(z-iR)^2 f(z)|\leq \left(\frac{|z-iR|}{Re z}\right)^2\leq 2$   But then I don't know what to do next. Any help I'd appreciate :)
105695
Let $\Omega$ be a bounded open subset of $\mathbb{C}$ and $f:\Omega\rightarrow\Omega$ be holomorphic in $\Omega$. Prove that if there exists a point $z_0$ in $\Omega$ such that $$f(z_0)=z_0~~~~\text{and}~~~~f'(z_0)=1$$ then $f$ is linear.  Please give some hints. Thanks in advance!
90698
Let $f$ be analytic on $0&lt;|z-a|&lt;R$ for some $R&gt;0$. If $z=a$ is not removable singularity of $f$, show that $e^{f(z)}$ has an essential singularity at $z=a$.  Since it's not removable singularity, it is either a pole of $f$ or essential singularity of $f$. The case of $z=a$ is a pole I can prove it. For the case of essential singularity, I write  $$ f(z)=\sum\limits_{k=0}^\infty c_k(z-a)^k+\sum\limits_{j=1}^{\infty}d_j\frac{1}{(z-a)^j}. $$ Call first series $g$ and second series $h$, then $e^{f(z)}=e^{g(z)}e^{h(z)}$. It is obvious that $e^{g(z)}$ is analytic at $z=a$ and $e^{g(a)}\neq 0$.Thus, it can't write as the form $(z-a)g_1(z)$ where $g_1$ is analytic at $z=a$.  Now, we focus on $e^{h(z)}$.  $$ e^{h(z)}=\sum\limits_{k=0}^\infty \frac{\sum\limits_{j=1}^\infty [d_j\frac{1}{(z-a)^{j}}]^k}{k!} $$ Then, in order to claim $z=a$ has essential singularity of $f$, I tried to write $e^{h(z)}$ as  $$ e^{h(z)}=\sum\limits_{s=0}^\infty e_s\frac{1}{(z-a)^s} (\star) $$  However, it seems that it shall be able to write as the form as $(\star)$, but do we need to worry about whether the series converges and whether we can change the double series? Or, is it possible to prove $z=a$ is essential singularity of $f$ avoiding to face this problem? Thanks.
49608
Let $f,g$ be entire functions such that $g \not\equiv 0.$ If $fg\equiv0$ in $\mathbb{C},$  could anyone advise me how to show $f \equiv0$ in $\mathbb{C} \ ?$  Thank you.
6162
I'm having difficulty with the following exercise in Ahlfors' text, on page 227.     Prove that in any region $\Omega$ the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded? Hint: Consider the functions $e^{-f}$.   Here is what I've tried:  I will start with a remark:  Apparently, Ahlfors wants us to show that the family is "normal in the classical sense". That is, every sequence of functions in the family has a subsequence which converges uniformly on compact subsets or tends uniformly to $\infty$ on compact subsets. In order to see why this is the right definition, consider the sequence $f_n(z)=n$. It is contained in the family but has no appropriate subsequence (in the sense of definition 2 in the text with $S=\mathbb C$).  Now, to the attempt itself:  Let $\Omega \subset \mathbb C$ be a fixed region, and consider the family $$\mathfrak F=\{f: \Omega \to \mathbb C | f \text{ is analytic and } \Re(f) &gt;0 \}. $$ We would like to show that $\mathfrak F$ is normal in the classical sense. Following the hint, we examine the family $$ \mathfrak G=\{e^{-f}:f \in \mathfrak F \}.$$  $\mathfrak G$ is locally bounded (since $|e^{-f}|=e^{- \Re (f)}&lt;1$ for every $f \in \mathfrak F$), thus it is normal with respect to $\mathbb C$ (theorem 15), and obviously, it is normal in the classical sense as well.  Let $\{ f_n \}$ be a sequence in $\mathfrak F$, and consider the sequence $\{ g_n \}=\{e^{-f_n} \}$ in $\mathfrak G$. According to normality it has a convergent subsequence $\{ g_{n_k} \}=\{e^{-f_{n_k}} \}$ which converges uniformly on compact subsets of $\Omega$ to some function $g$ (which is analytic by Weierstrass' theorem).  Since each $\{ g_{n_k} \}$ is nonvanishing, the limit function $g$ is either identically zero, or non vanishing as well (Hurwitz's theorem). In the former case it is easy to show that the subsequence $\{ f_{n_k} \}$, obtained by the same indices, tends to $\infty$ uniformly on compact sets. Hence, we will assume from now that $g(z) \neq 0$ for all $z \in \Omega$.  Up until now, I was trying to show that the subsequence $\{ f_{n_k} \}$ works in all cases, but sadly, this is not the case. Consider the sequence $f_n(z) \equiv 1+2 \pi i (-1)^n \in \mathfrak F$. In that case $g_n(z)=e^{-1}$, and an admissible subsequence is $g_{n_k}=g_k=e^{-1}$. However, $f_{n_k}=1+2 \pi i (-1)^k$ diverges everywhere.  Can anyone please help me finish this proof? Or maybe give me some hints?  Thanks!
81335
If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is an entire function such that $f(z)\neq w$ for all $z\in \mathbb{C}$ and for all $w\in [0,1]\subset \mathbb{R}$, how to prove that $f$ is constant (without using Picard's little theorem).   Any hint would be appreciated.
8820
If $f$ is holomorphic on $D\setminus \{0\}$ and takes no values in $(-\infty,0]$ then $0$ is a removable singularity.   I thought to prove this by elimination, but I can't really tell anything about the behavior of $f$ around $0$. How can one translate the information about the definition of $f$ in semi-open interval $(-\infty,0]$.
95040
If $f$ is ananlytic in $D_r(z_0)$\ {$z_0$} and $Ref(z)&gt;0$ for all $z\in D_r(z_0)$\ {$z_0$} then $z_0$ is a removable singularity. I want to prove this statement but I just cannot seem to find a way. Clearly $f(z)\neq0$ in $D_r(z_0)$\ {$z_0$}. I just feel like this can be proved with this fact. Can someone give me a hint? Assistance is appreciated. Thanks a lot
53651
Let $\Omega$ be a non-empty, open subset of $\mathbb C$. Consider an holomorphic function $f:\Omega \setminus \{z_0\} \to \mathbb C$ and suppose we know $z_0$ is an essential singularity of $f$.   I am wondering what we can say about the function $\tilde{f}:=\frac{1}{f}$ and its singularity in $z_0$. Do you know any theorem that answers to this question?  Actually, I can't prove anything, since I do not know the answer: I've studied some examples. For instance, if you take $f(z)=e^{\frac{1}{z}}$ then $\tilde{f}$ will still have an essential singularity, hasn't it?   On the other side, if we take $f(z)=\sin(\frac{1}{z})$ then I think that $z_0=0$ becomes a limit point of poles for $\tilde{f}$ (so we can't classify it, because it isn't an isolated singularity).   Wha do you think? Do you know any useful theorem concerning this? Thank you in advance.
43334
If $f$ is non-constant and entire, prove that there exists a $z_0 \in C$ such that $f(z_0)$ is a positive real number, without appealing to Picard's theorem.  The obvious approach to this would be to modify a proof of Picard's little theorem to try and prove this, but all of the proofs I have found of little Picard's seem to be higher level than I have - I'm trying to solve this using theorems from a quarter long course in complex. I would really appreciate a hint in the right direction....
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?
80471
Let $f$ be an entire function. Assume that $\mid f(1/n)\mid\le e^{-n}$ for all $n\in \mathbb{N}$. Show that $f$ vanishes identically.
39217
Let $f$ be an entire function and $B$ be a bounded open set in $\mathbb {C} $. Prove that boundary of image of $B$ under $f$ is contained in image of boundary of $B$. Does the same result is true for unbounded open set in $\mathbb C.$
106246
Problem: Let $g:[0,1]\to\mathbb{R}$ be a continuos function such that $g(1)=0$. Show that $\sup\limits_{x\in[0,1]}|x^ng(x)|\to0$ as $n\to\infty$.  My approach: Let $f(x)=x^ng(x)$. Then, $f(x)$ is continuos on $[0,1]$. Using the extreme value theorem we can say, that $|f(x)|$ is bounded on $[0,1]$ and attains its bounds, i.e., $\exists x_0\in[0,1]$ such that $|f(x_0)|=\sup\limits_{x\in[0,1]}|f(x)|$.  Now, $|f(x_0)|=|x_0^ng(x_0)|$. If $x_0=0$, then we are done. So, let us assume $x_0\neq0$. By the Archimedean property of $\mathbb{R}$, $\exists$ a $y\in\mathbb{R}$ such that $y=x_0^n\leq1$, since $x_0\in[0,1]$.   Hence, $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|x^ng(x)|=\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|f(x)|=\lim\limits_{n\to\infty}|f(x_0)|=\lim\limits_{n\to\infty}|yg(y^{\frac{1}{n}})|\leq\lim\limits_{n\to\infty}|g(y^{\frac{1}{n}})|=g(1)=0.$  I will be glad if anyone can tell me if I have done any mistake and if so please help me to correct it.  Thank you.
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
64166
Let $f(z)$ be an entire function whose real part is bounded by a polynomial in $|z|$. Does it follow that $f(z)$ is a polynomial?  Or, without loss of generality and more suggestively $$(f(z)=\sum_{k=0}^\infty a_k z^k \quad \land \quad \operatorname{Re}(f(z))\leq |z|^n)\quad \forall z\in \mathbb{C} \qquad \Rightarrow \qquad f(z)=\sum_{k=0}^n a_k z^k \quad\forall z\in \mathbb{C}$$
45135
I'm having trouble with the following problem:  Let $f$ be holomorphic on the punctured unit disc, $D$. If $\int_D|f(z)|dA(z)&lt;\infty$, then $z=0$ is either a removable singularity or a simple pole of $f$.  A similar problem is to prove or disprove that if $\int_D|f(z)|^2dA(z)&lt;\infty$, then $f$ has a removable singularity at $0$.  I've tried using the Mean Value Property for $f$ and taking the limit as $z\to0$, but I didn't get anywhere.