Let $f$ be biholomorphic (i.e. holomorphic and injective) on the unit disk with $f(0)=0$. Show that there is a biholomorphic $g$ on the unit disk such that $g(z)^2=f(z^2)$. Show that this function satisfies also $g(-z)=-g(z)$.     My attempt:  Consider the power series expansion  $$f(z)=f'(0)z+\cdots$$ so $$f(z^2)=f'(0)z^2+\cdots=z^2(f'(0)+\cdots)$$ Therefore, we have  $$ g(z)=z\cdot(f'(0)+\cdots)^{1/2}$$ defined on a neighborhood of $0$.   But how to extend the function to the whole unit disk? (Schwarz reflection principal?) And how to make the function to be an odd function?  There seems to be a similar question , but one important condition in that question does not hold here. So both ways do not work.
122942
Let $\Delta=\{z\in \mathbb C: |z| &lt; 1\}$. Let $f: \Delta \rightarrow \mathbb C$ be a one-to-one analytic function fixing the origin. Prove that there is a one-to-one analytic function $g: \Delta \rightarrow \mathbb C$ such that $[g(z)]^2=f(z^2)$. Further, show that such a function is odd.  The only idea which comes to my mind is the following (not sure whether it is the right direction though). Let $h: \Delta \rightarrow \Delta \rightarrow \mathbb C,\ z\mapsto z^2\mapsto f(z^2)$. Note that $h'(0)=0,\ h''(z)=2f'(z^2)+4z^3f''(z^2)$; since $f$ is one-to-one and analytic, its derivative never vanishes, so $h''(0)\ne 0$. So $h$ is an analytic function of $\Delta$ with a zero of order two at the origin. Then there is an analytic function $g$ defined in an open disc about the origin with a simple zero at the origin such that $[g(z)]^2=f(z^2)$.   This is the only thing I came up with, but it seems to be a little bit irrelevant since I don't see any way of extending this $g$ to the whole $\Delta$, and also the injectivity and oddness of such a $g$ are unclear.
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!
105348
Question: Suppose that functions $f$ and $g$ are analytic on the disc $A = \{z : |z| &lt; 2\}$ and that neither $f(z)$ nor $g(z)$ is ever $0$ for $z \in A$.  $\frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$ for $n=1,2, ... ,$  show that there is a constant $c$ such that $f(z) = cg(z)$ for all $z \in A$.  Comments: This is one of the more theoretical problems from an old complex analysis exam. I am not sure which theorems I am supposed to use here. Am I supposed to take the integrals of both sides over the same curve and use the root-pole counting theorem? All input very much appreciated, so far everyone here has been of great help.
115495
I am studying for a qualifier exam in complex analysis and right now I'm solving questions from old exams. I am trying to prove the following:     Prove that if $f$ and $g$ are entire functions such that $f(z)^2 + g(z)^2 = 1$ for all $z \in \mathbb{C}$, then there exists an entire function $h$ such that $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$.     My Attempt  The approach that occurred to me is the following. Since $f(z)^2 + g(z)^2 = 1$ then we have $(f(z) + ig(z))(f(z) - ig(z)) =  1$. Then each factor is nonvanishing everywhere in $\mathbb{C}$ and thus by the "holomorphic logarithm theorem" we know that since $\mathbb{C}$ is simply connected, there exists a holomorphic function $H:\mathbb{C} \to \mathbb{C}$ such that  $$e^{H(z)} = f(z) + ig(z)$$  and then we can write $\exp(H(z)) = \exp\left(i\dfrac{H(z)}{i} \right) = \exp(ih(z))$,    where  $h(z) := \dfrac{H(z)}{i}$.  Thus so far we have an entire function $h(z)$ that satisfies  $$e^{ih(z)} = f(z) + ig(z)$$  On the other hand, we also know that $e^{iz} = \cos{z} + i \sin{z}$ for any $z \in \mathbb{C}$, thus we see that   $$e^{ih(z)} = \cos{(h(z))} + i \sin{(h(z))} = f(z) + ig(z)$$  Thus at this point I would like to conclude somehow that we must have $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$, but I can't see how and if this is possible.    My questions   Is the approach I have outlined a correct way to proceed, and if so how can I finish my argument? If my argument does not work, how can this be proved?   Thanks for any help.
18635
Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.  Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|&lt;1$ and some $n,m\ge 0 $?
96136
I've never really had to think about this problem before, and to be honest complex analysis isn't my strongest suit, so when I suddenly needed to know where $z^z$ is holomorphic, I didn't know where to begin to start proving my hunches. My guess is that it's holomorphic away from $0$.  More generally, given two entire functions $f$ and $g$, where is $f(z)^{g(z)}$ holomorphic? Again, my hunch is that it is away from the zeroes of $f$. This isn't a time sensitive question, so any help you can provide would be appreciated, from proofs to thoughts to hints!
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?
5224
I recently came across the result that if $g$ is holomorphic in a neighbourhood of a point $z_0$ and  $$\sum_{n=0}^{\infty }{g^{(n)}(z_0)} $$ converges then $g $ is an entire function. Now, my doubt is if the above series is absolutely convergent then can we say something more about $g$, like say $g$ is a polynomial? I could not find any non-polynomial  entire function( I tried with functions like $e^z$, $\sin z $, et al), whose successive derivatives converge absolutely as series. Any help please?
11603
Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be a holomophic function in the unit disc. Show that if $$\sum_{n=2}^{\infty} n\vert a_n \vert &lt;\vert a_1\vert$$ and $a_1\not = 0$, then $f$ is injective.   I am able to show that $f'$ never vanishes by showing that $\vert f'-a_1\vert &lt;\vert a_1\vert$, but that doesn't imply that $f$ is injective. For example, $e^z$ is not inejective but its derivative never vanishes. What is required to prove that $f$ is injective? How do you show $f$ is injective on any set?
78557
If $g(w)$ and $f(z)$ are holomorphic functions, show that $g(f(z))$ is also holomorphic.   From the assumptions we have that for some $A,B$,  $$\lim_{h\rightarrow 0}\frac{g(f(z)+h)-g(f(z))}{h}=A,$$  $$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}=B.$$  We want to prove that there exists $C$ with  $$\lim_{h\rightarrow 0}\frac{g(f(z+h))-g(f(z))}{h}=C.$$  The second equation means, by definition of the limit, that for any $\varepsilon &gt; 0$, there exists $\delta$ such that $0&lt;|h|&lt;\delta$ implies $$\left|\frac{f(z+h)-f(z)}{h}-B\right|&lt;\varepsilon.$$  I don't know how to get to $\dfrac{g(f(z+h))-g(f(z))}{h}$ from here.
90763
Let $z_0$ be a point in an open set $U \subset \mathbb{C}$.  Let $f,g:U \to \Delta(0,1)$ where $g$ is a bijection such that $g(z_0)=0$ and $g'(z_0)\neq 0$. Suppose that $f(z_0)=0$, then $|f'(z_0)|\leq |g'(z_0)|$  I am not even sure how to start this.  I was assuming Schwarz lemma.
118626
I am interested in finding explicit formulae for (better yet characterizing) conformal functions from various domains onto the open unit disc $\mathbb{D}\subset\mathbb{C}$, and in understanding the key ideas necessary to establish such functions.  Specifically, what can $f$ look like when $f:G\to\mathbb{D}$ is conformal and  (1) $G=\{x+iy~|~x,y&gt;0\}$ is the open first quadrant.  (2) $G=\{x+iy~|~x&gt;0,~0&lt;y&lt;1\}$ is an open horizontal strip in the first quadrant.  (3) $G=\{z\in\mathbb{C}~|~\frac{1}{2}&lt;|z|&lt;1\}$ is an annulus.  (4) $G=\mathbb{D}\cap\{|z-\frac{1}{2}|&gt;\frac{1}{2}\}$ is something else (torus?).
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.
99254
We know that if $f(z)$ and $g(z)$ are entire functions such that $g(z)\ne0$ and $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then by Liouville's theorem $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  Now my question is this that similar to above argument, if $f(z)$ and $g(z)$ are Meromorphic functions such $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then I want to show $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  I am thinking in this way that because because poles and zeros of Meromorphic functions are isolated , by the Riemann's theorem on removable singularities and By using analytic continuation to eliminate removable singularities also we can have $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  Is this true way?
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 $f(z)=z \ \forall\ z\in\mathbb{C}$
105364
$\textbf{Problem.}$ Suppose $f$ is holomorphic on the unit disk $\mathbb{D}$. Show there are points $a_n\in \mathbb{D}$, $a\in \partial \mathbb{D}$, and $b\in \mathbb{C}$ such that $a_n\to a$ and $f(a_n)\to b$, as $n\to \infty.$   $\textbf{Attempt at a Solution:}$ First, if $f$ has an essential singularity at some point $a\in \partial\mathbb{D}$, I think the statement follows easily.  To contradiction suppose that the statement is false. That is, suppose that $f(a_n)\to \infty$ for all sequences $\{a_n\}$ in $\mathbb{D}$ where $a_n\to a$, $a\in \partial\mathbb{D}.$  Then $$\frac{1}{f(z)}\to 0, \hspace{1cm} \mbox{as} \hspace{1cm}|z|\to 1.$$  $\textbf{ Case 1}$: If $f$ is nowhere vanishing on $\mathbb{D}$, then $1/f(z)$ would be holomorphic in $\mathbb{D}$ and can be extended continuously on $\bar{\mathbb{D}}$. Then, by the maximum principle this would imply that $1/f(z)\equiv 0$ on all of $\mathbb{D}$. But this contradicts the fact that $f$ is holomorphic (non-constant) in $\mathbb{D}.$  $\textbf{Case 2}$: Suppose $f$ has a finite number of zeros in $\mathbb{D}$. That is, let $\{b_1,...,b_n\}$ be points of $\mathbb{D}$ such that $f(b_i)=0.$   Let $r=\min \{|z-b_i| \forall z\in \partial\mathbb{D}\}$, and consider the annulus $$A=\{z\in\mathbb{C}: r&lt;r_1&lt;|z|&lt;1\}$$  Then, $1/f(z)$ is holomorphic on $A$, and can be continuously extended to $\bar A$. (Note that, $1/f(z)$ is already holomorphic for $|z|=r_1$.  Let $$g(z)=\frac{1}{f(z)}\frac{1}{f(r_1/z)}.$$ Then $g(z)=0$ for all $z\in \partial \bar{A}$. By the maximum principle it follows that $g(z)\equiv 0$ on $A$, and by the identity theorem it follows that $1/f(z)\equiv 0$ on $A$. But this, again, contradicts the fact that $f$ is holomorphic on $\mathbb{D}$ and thus on $A$ as well.   $\textbf{Case 3:}$ This is the case where i'm not so sure what i'm doing is correct.   Suppose, now, that $f$ has an infinite number of zeros in $\mathbb{D}$. That is, let $E=\{b_i\in \mathbb{D}: g(b_i)=0\}$. Then, $E$ cannot have an accumulation point in $\mathbb{D}.$  My idea was to take neighborhoods, $\mathbb{B}_{r_i}(b_i)$, and consider $$g(z)=\frac{1}{f(z)}\prod_{w\in A_w}\frac{1}{f(z/w)},$$ where $A_w=\{w\in \mathbb{D}: |w-b_i|=r_i\},$ for all $z\in \overline{\mathbb{D}\setminus\bigcup \mathbb{B}_{r_i}(b_i)}.$  Then, I would argue similarly as in Case 2.  There is probably a much shorter method.  I'd appreciate any thoughts or suggestions!!
104157
If $f(z)$ is analytic in the unit disk and $f(0)=0$, show that $$f(z)+f(z^2)+\cdots+ f(z^n)+\cdots$$ converges locally uniformly to an analytic function in the unit disk.  I am thinking to apply Weierstrass $M$ test for convergence or since all the $f(z^n)$ are analytic in the unit disk, so each of them can be expressed as the taylor series. Then rewriting the series again might work. But I am not sure. Any suggestions!