given $f$ a holomorphic function in $\mathbb{C}\setminus\{0\}$ satisfies $|f(z)|\le \sqrt{|z|} + \frac{1}{\sqrt{|z|}}$ for all $z\in \mathbb{C}\setminus \{0\}$. I wish to show that $f$ is constant, well what I did so far was to take the limit: $0\le \lim_{z\rightarrow0} |z||f(z)| \le \lim_{z\rightarrow 0} |z|^{1.5}+|z|^{0.5} =0$ so $\lim_{z\rightarrow0} |zf(z)|=0$ thus $\lim_{z\rightarrow0} zf(z)=0$ and by Riemann's continuations theorem $0$ is a removable singularity. So $f$ is can be extended holomorphicaly to $\tilde{f}$. I thought maybe its possible to bound $\tilde{f}$ and then use Liouville's theorem. But couldn't find a way to do so.
38173
Suppose function $f$ is holomorphic on $\mathbb C-\{0\}$ and satisfies $$|f(z)| \le \sqrt {|z|} + \frac{1}{\sqrt {|z|}}.$$ Prove that $f$ is a constant function. I think it's related to Laurent series representation and ML inequality but I have no idea how to prove it. Can anyone give me some hints? Thanks in advance!
87244
If $f:\mathbb{D}\rightarrow\mathbb{D}$ is holomorphic then prove that   $$\frac{|f(0)|-|z|}{1 + |f(0)||z|} \leq|f(z)| \leq\frac{|f(0)| + |z|}{1 - |f(0)||z|} $$   I have been wracking my brain for hours and am out of ideas.  The hypothesis seems to suggest an application of the Schwarz lemma but I cannot find a suitable function - I've tried functions of the form $$\frac{f(z) - f(0)}{\mbox{scale factor}}$$ and none have worked.  I also tried working backwards to obtain an inequality and got $$-|z|\leq|f(z)||1\pm f(0)z| - |f(0)|\leq |z|\;\;\;\mbox{(*) }$$ If I could prove $$|f(z)[1\pm f(0)z] - f(0)|\leq 1$$ then I could obtain (*), but with so few conditions on $f$ I have no idea how to proceed.
70110
$\textbf{Problem.}$ Suppose $f$ is a holomorphic function on $\{z\in\mathbb{C}:|z|&lt;1\}$, the open unit disk, with the property that Re$f(z)&gt;0$ for every point $z$ in the disk. Prove that $|f'(0)|\leq 2\text{Re}f(0)$.  This is a problem that appeared on a qualifying exam in some graduate school. What I tried is, define $\displaystyle\varphi(z)=\frac{z-1}{z+1}$, then this sends the open right half plane onto the open unit disk, so let $\hat{f}=\varphi\circ f(z)$, then its image is contained in the open unit disk. And I wanted to use the Schwarz lemma, but then the origin had to be fixed, so define $\displaystyle\psi(z)=\frac{\hat{f}(0)-z}{1-\overline{\hat{f}(0)}z}$ and let $\tilde{f}=\psi\circ\hat{f}$, then the image of this is still contained in the open unit disk and it fixes the origin so the Schwarz lemma $|(\tilde{f})'(0)|\leq 1$ could be applied. But after all the calculation, I got $$|f'(0)|\leq\frac{|f(0)+\overline{f(0)}+2|f(0)|^{2}|^{2}}{2|f(0)+1|^{2}|f(0)|}$$ but, for example, if the magnitude of the imaginary part of $f(0)$ is quite bigger than the magnitude of the real part of $f(0)$, then this doesn't result the desired inequality but actually means that the desired inequality is wrong instead. I also thought about composing some different function to the RHS of $\hat{f}$ and use the Schwarz lemma but it didn't seem to work.  Maybe I should try something else, but what could be tried instead?
41450
Suppose $f$ is continuous on the closed unit disc $\overline{\mathbb{D}}$ and is holomorphic on the open unit disc $\mathbb{D}$. Must $f$ have finitely many zeros in $\mathbb{D}$? I know that this is true if $f$ is holomorphic in $\overline{\mathbb{D}}$ (by compactness of the closed unit disc), but I'm not sure of what happens when I just consider $\mathbb{D}$.
86893
Let $f: \mathbb C \to \mathbb C$ be an analytic function, such that for all $z\in\mathbb C$, $$| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.$$   I have to show that $f$ is constant.   I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated.
13160
How can I prove the following assertion?    Let $f$ be a holomorphic function such that $|f|$ is a constant.  Then $f$ is constant.    Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.
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.
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.
94122
I have to solve this problem but i have no idea how to do ...Thanks!  Let be $f:\mathbb{C}\rightarrow \mathbb{C}$ a holomorphic function. Show that if $f\circ f=f$ then f is either constant or $f(z)=z \ \forall\ z\in\mathbb{C}$
37097
I'm learning Complex Analysis and need to verify my work to this problem since my textbook does not provide any solution:     If $f$ is entire and $\lim_\limits{z\to\infty} \dfrac{f(z)}{z} = 0$ show that $f$ is constant.   My work and thoughts:  From the $\varepsilon$ — $\delta$ definition of the limit we have that  $$\forall{\varepsilon} &gt; 0, \exists{n_0} \in \mathbb{N} : \forall{\left|z\right|} \geq n: \left| \frac{f(z)}{z} \right| &lt; \varepsilon \iff \frac{\left| f(z) \right|}{\left| z \right|} &lt; \varepsilon\iff \left| f(z) \right| &lt; \varepsilon \left| z \right|.$$   Now let $C_R = \{z \in \mathbb{C} : \left| z \right| = R \}$.   For every $\left| z \right| &lt; R$, by Cauchy's integral formula for derivatives we have that   $$ \left| f'(z) \right| = \frac{1}{2 \pi } \left| \int_{|\zeta|=R} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta \right|= \frac{1}{2 \pi } \left| \int_{0}^{2\pi} \frac{f(\zeta)}{(\zeta - z)^2} \, \zeta'(t) dt \right| \le$$   $$\le \frac{1}{2\pi} \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} 2\pi R = \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} R.$$  Thus, letting $R \rightarrow \infty$ yields the desired result, that is  $$\left| f'(z) \right| \leq \lim_{R \to \infty} \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} R = 0 \implies f(z) = c \;\; \text{with} \; c \in \mathbb{C}.$$    Is my work correct? Are there parts of the proof that need improvements? I'm also looking for other (possibly quicker) solutions using the  "big guns" theorems. The only one I'm familiar with is Picard's little theorem but it doesn't apply here.
105860
How would I show that an entire function $f$ with the property $|f(z)| \geq 1$ must be constant?    I'm aware of Liouville's theorem, just not sure how to apply it here.
40890
Let $f$ be holomorphic in the unit disk $D(0,1)$, and $f(0)=0$. Suppose $|\Re f(z)|\leq 1$ for all $|z|&lt;1$. Show that  $$|\Re f(z)|\leq\frac{4}{\pi}\arctan|z|;$$ $$|\Im f(z)|\leq \frac{2}{\pi}\ln\frac{1+|z|}{1-|z|}.$$  What I could do is just to observe that $$\frac{e^{i\frac{\pi}{2}f(z)}-1}{e^{i\frac{\pi}{2}f(z)}+1}:D(0,1)\to D(0,1)$$ with $0$ being fixed. And Schwarz lemma applies. However, I could not get any information on the real and imaginary part of $f$.
71420
Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint?
14888
I've been thinking how to prove that an analytic function $f$ in the domain $D$ is a constant if  $f'(z)=0$ in every point in $D$, but I haven't figured it out yet.  What I was thinking is to use Cauchy-Riemann equations, but it didn't work well...  If this is not true, I would like to know the counterexample...  Here is what I tried:  Let $f(z)=u(x,y)+iv(x,y)$  $\color{blue}{(1)}\text{(Cauchy-Riemann})\begin{cases} u_y=-v_x\\ u_x=v_y \end{cases}$  $\color{blue}{(2)}\lim\limits_{\Delta z \to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z }=0$ for all $z_0$ in $D$  I'm stuck here...
13726
Let $D\subseteq \mathbb{C}$ be open and connected and    $f,~g:D \rightarrow \mathbb{C}$ holomorphic functions such that    $|f|+|g|$ is constant on $D.$ Prove that $f,~g$ are constant on $D.$   Attempt. I noticed first that this problem is already set before, and is considered  to be duplicate () and we are sent directly to problem . However, the last reference deals with the sum $|f|^2+|g|^2$, which I do not see how is connected to the sum $|f|+|g|$, as stated in our title (for example, the equality $|f|^2+|g|^2=(|f|+|g|)^2-2|f||g|$ is not helpful here).  Thanks in advance!
1695
Let $f$ be a non-constant and analytic on a neighborhood of closure of the unit disk such that $|f(z)|=\text{constant}$ for $|z|=1$. Prove $f$ has at least one zero inside unit disk.  I thought of using Rouche's somehow. Using $f(z)-z$, and taking constant is less than $1$, I can actually conclude from Rouche's theorem that the equation $f(z)-z=0$ have a fixed point inside the unit disk. I am stuck for other constants greater than equals to one and exactly zero. I hope there should be a little trick I am missing here. It will be awesome to see if maximum principle can be applied to conclude the result.
52230
Show that if $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$, then the function $f(z)$  is constant in $D$.     Here is what I try so far.  So we are given that $|f(z)|$ is constant in $D$, so that means $|f(z)|^2 = u^2 + v^2$ is also constant.  If either $u = 0$, or $v = 0$, then $f(z)$ is constant.  $$\frac{\partial|f(z)|^2}{\partial x} = 2u \frac{\partial u}{\partial x} + 2v \frac{\partial v}{\partial x} = 0$$  $$\frac{\partial|f(z)|^2}{\partial y} = 2u \frac{\partial u}{\partial y} + 2v \frac{\partial v}{\partial y} = -2u \frac{\partial v}{\partial y} + 2v \frac{\partial u}{\partial x} = 0 $$  So what do I do to finish this proof? Would I have to use the Cauchy Riemann equation?