Does there exist a holomorphic function $f:D\to D$ where  $D=\{z\in \mathbb{C} : |z|&lt;1\}$ such that $f(1/2)=-1/2$ and $f'(1/4)=1$?  I tried with , but for that I need the value of $f(1/4)$ which is not given. So kindly tell me ow to do this.  Thanks.
121413
Does there exist a holomorphic function $f:\mathbb{D}\longrightarrow\mathbb{D}$ such that $f(1/2)=-1/2$ and $f'(1/4)=1$, where $\mathbb{D}=\{z:|z|&lt;1\}$?
46389
Let $f$ be a holomorphic function defined on $\Bbb C$ such that $f^{-1}(\Bbb R)=\Bbb R$. Prove that there exists $a,b \in \Bbb R, \; a \neq 0 \;$ such that $f(z)=az+b \;$ for every $z \in \Bbb C$.  I was given a hint: define the function $g(z)=\frac{f(z)-f(0)}{z}$. such $g$ is also holomorphic in $\Bbb C$ since the only singularity is in $0$, and it is a removable singularity since we can define $g(0)=f'(0)$. Next I thought about proving $g$ is bounded, and thus constant (using Liouville's theorem), but pretty much out of ideas. How can I finish the proof/solve in another way?
46080
We want to prove the following:     If $f$ is a holomorphic function on the unit disc $\mathbb{D}$ s.t. $f(z) \neq 0$ for $z \in \mathbb{D}$, then there is a holomorphic function $g$ on $\mathbb{D}$ such that $f(z)=e^{g(z)}$ for all $z \in \mathbb{D}$.   For such a $g$, we have that $\frac{f'(z)}{f(z)}=g'(z)$ on $\mathbb{D}$ since $f$ has non-zero derivative.   Perhaps then we define $g(z)=\int_\gamma \frac{f'(z)}{f(z)}dz$ where $\gamma$ is a path in $\mathbb{D}$ from $0$ to $z$.  Does this work?
72288
I have some trouble solving this problem: Does there exist a holomorphic function $f$ on $\mathbb C\setminus \{0\}$ such that $$|f(z)|\geq \frac{1}{\sqrt{|z|}}$$ for all $z\in\mathbb C \setminus \{0\}$? I don't know where to start. My intuition is that you would get a problem with the singularity near $0$, but I am not sure how to prove it. Any help would be appreciated! Thanks!
68156
Let $D$ be an open set in the complex plane and $f(z)$ be a non-constant holomorphic function on D. Then $|f(z)|$ has no local maximum on D.  I can follow the proof fine - usually if I don't understand a theorem intuitively beforehand, the proof will offer the insight necessary. Here, however, I can't see the reason for the Maximum Principle to hold - or perhaps I was just too shallow in my grasp of the proof.  Does anybody have any shillings of wisdom that they would be willing to offer? Cheers.
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.
8166
This is from an old qualifying examination question.  If f is holomorphic in the unit disk $D$ and $|f(z)|&lt;1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$  I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.   Any helpful hints are greatly appreciated.  Edit: Of course the $a$ above is one of the fixed points.
34897
Prove if $f : D(0,1) \rightarrow \overline{D}(0,1)$ is an holomorphic function in $D(0,1)$ and $f$ has in $z=0$ a zero of order $n$. Then,    $$|f(z)| \leq |z|^n$$   for all $z\in D(0,1)$.   $\textbf{My attempt}$. If $f$ has a zero of orden $n$, then   $$f(0)=f'(0)=\ldots = f^{n-1)}(0)=0 \text{ and } f^{n)}(0)\neq 0$$.  Then, applying the Schwarz's Lemma I get that  $$|f(z)|\leq |z| \; , \; |f'(z)|\leq |z| \text{ and } |f^{n-1)}(z)|\leq |z| $$.  I don't know what to do now to get $|f(z)| \leq |z|^n$.
25402
I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof.
31149
Find all entire functions $f$ such that $|f|=1$ whenever $|z|=1$.  My conjecture : $f=cz^m,|c|=1,m \geq  0$.  But I don't know how to prove it.  I will appreciate your help.
119136
Let $U \subset \mathbb{C}$ be an open set such that {$ z\in\mathbb{C};|z|\leq2$}$ \subset U$ and $f:U\rightarrow\mathbb{C}$ a holomorphic function.   Show that there exists infinitely many natural numbers $n \in \mathbb{N}$ such that $f(\frac{1}{n})\neq(\frac{1}{n+1})$  So i have been told to start by comparing with the function $g(z)=(\frac{z}{z+1})$  however i am still not getting very far with this. Would i have to prove by a contradiction?   Completely stuck and unsure, some help would be fantastic, many thanks!
21547
Let $D \subset \mathbb{C}$ be the open unit disk, $f:D \to \mathbb{C}$ a holomorphic function and $d=\sup \{|f(z)-f(w)|:z,w\in D \}$. How can I show that $|f'(0)|\leq \frac{d}{2}$?  I tried using Schwarz's lemma on the function $g:D \to \mathbb{C}$ defined by $g(z)=\frac{f(z)-f(0)}{d}$, but I am only getting $|f'(0)|\leq d$.
10121
Problem:     Let $D = \{ z \in \mathbb C : |z| &lt; 1 \}$. Suppose $f : D \to \mathbb C$ is a non-constant analytic function such that $\mathrm{Re} f(z) \geq 0$ for all $z \in D$.      (a) Show that $\mathrm{Re} f(z) \neq 0$ for all $z \in D$.      (b) Using a Möbius transformation and Schwarz's lemma, prove that if   $f(0) = 1$, then $$ |f(z)| \leq {1+|z| \over 1-|z|}. $$   What can be said if $f(0) \neq 1$?      (c) Show that if $f(0) = 1$, then $f$ also satisfies $$ |f(z)| \geq {1-|z| \over 1+|z|}. $$   Progress:  (a) We can use the open mapping theorem.  (b) We can compose a Möbius transformation $m$ that sends the right-half plane to $D$ with $f$. Then $h = m \circ f$ satisfies the hypotheses of Schwarz's lemma by construction. Simple manipulations of the inequality from Schwarz's lemma give the result.  (c) Looking at part (a), we know that $\mathrm{Re} f(z) \neq 0$, and so $f(z) \neq 0$. Thus, we can use the function $g = 1/f$ in the statement of part (b).  Question:  What about when $f(0) \neq 1$? I don't think you can get anything from $f(z)-f(0)+1$ unless $\mathrm{Re} f(0) &lt; 1$. And you can't in general rearrange the inputs of $f$ because $1$ isn't necessarily in the image of $f$. Also, I don't think you can get anything from $f(z)/f(0)$ unless $f(0)$ is real, because otherwise the range of $f(z)/f(0)$ could extend into the left-half plane.
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!
96627
Let's say $f:D\to R$ is an injective function on some domain where it is also differentiable. For a real function, i.e. $D\subset\mathbb R, R\subset\mathbb R$, is it possible that $f'(x)\equiv f^{-1}(x)$?  Intuitively speaking, I suspect that this is not possible, but I can't provide a reasonable proof since I know very little nothing about functional analysis. Can anyone provide a (counter)example or prove that such function does not exist?
49561
Is it true that  $ f(\bar{z})=\overline {f(z)}$, Where z is complex?  I think it holds when $f(z)$ is holomorphic since we have $f(z)=p(x,y)+iq(x,y)=p(z,0)+iq(z,0)$ Any help...
31626
Suppose that $f$ is holomorphic on $\mathbb C$ and that $\Re(f(z))\ge 0$ for all $z$. Show that $f$ is constant. [Hint: consider $e^{−f(z)}$.]   My thoughts: If $\Re(f(z))\ge 0 $ holds, then $e^{−f(z)}$ is a bounded holomorphic function (do I need to prove this or is it obvious?) So then by Liouville's Theorem $e^{−f(z)}$ is constant.   But then I'm not sure how to rigorously go from this $\exp(−f(z))$ back to $f(z)$.   Could anyone help me piece this together please?  Thanks