I'm starting with Complex-Analysis and I'm struggling a bit with some exercises. I have $\Omega = \lbrace z\in C : Re(z)&gt;0 \rbrace$ and $f: \bar{\Omega} \rightarrow C$ continuous and holomorphic in $\Omega$ with $\lim_{z \to \infty}\vert f(z)\vert =0$. What I have to prove is that $sup \lbrace \vert f(z)\vert : z\in \partial\Omega \rbrace = sup \lbrace \vert f(z)\vert : z\in \bar{\Omega} \rbrace$. I've tried with the Maximum Modulus Principle but I get nowhere with it, would be great if someone could just give me some tips to know how to start.
120910
Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)&gt;0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$. I thought about taking the bounded set $H\cap D(0,R), R&gt;0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$.
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...
36023
I have to solve following problem     If $\Re (f)$ is bounded above or below for a function $f$ holomorphic on $\mathbb{C}$ then $f$ is constant.   My attempt: If there is $M$ such that $\Re(f) \le M$, then $\|e^{f}\|=e^{\Re(f)}\le e^{M}$. From Liouville's theorem the entire function $e^f$ is constant, that is $0=(e^f)'=f' e^f$. This means $f' =0$, so $f$ is constant. If $\Re(f)$ is bounded below, we consider $e^{-f}$ and proceed the same way.  Am I correct? Is there a solution using maximum modulus principle?
39255
I know $R^\omega$ is the set of functions from $\omega$ to $R$. I would think $R^\infty$ as the limit of $R^n$, but isn't that $R^\omega$? The seem to be used differently, but I can't tell exactly how.
86184
Let $G\subseteq\mathbb{C}$ be a domain and $f\colon G\to\mathbb{C}$ holomorphic. Show:     $$ f'=0\mbox{ in }G\Rightarrow f\mbox{ is constant} $$      Cauchy-Riemann: $$ f'(z)=f_x(z)=u_x(x,y)+i v_x(x,y), z=x+iy=0\\ \Leftrightarrow u_x(x,y)=0~\wedge v_x(x,y)=0 $$  and $$ f'(z)=-if_y(z)=-iu_y(x,y)+v_y(x,y)=0\\\Leftrightarrow u_y(x,y)=0~\wedge~v_y(x,y)=0 $$  That means: If one derivates u and v to x, one gets 0, and therfore the x-part of these functions must be constant. The ssame for the y-part of u and v. So u and v are constant and so is then f.  Is that okay?
114815
Let $\Omega$ be a domain in $\mathbb{C}$ and let $\{f_n\}_{n \in \mathbb{N}}$ be a sequence of injective functions that converge in $O(\Omega)$ to $f$ . Show that $f$ is either injective or a constant function.   How does the conclusion change if, instead of a domain, we allow $\Omega$ to be an arbitrary open set ?  I know that $f$ is holomorphic as an almost uniform limit. But I dont know how to proceed.
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.
122566
Let $f$ be holomorphic non-constant on $D=\left\{ 0&lt;\left|z\right|&lt;10\right\}$ . Given that for all $n\in\mathbb{N}$: $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ prove that $f$ has an essential singularity at $0$. Find an example of $f$ satisfying the condition.   My idea was to assume for contradiction the singularity is not essential, which means that either the limit $z\to0$ of $f$ exists or of $1/f$, and from $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ we get that the limit of $f$ exists and is $0$.  I then define $h(z)=f(z)$ for $z\neq0$ and $h(0)=0$, which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I'm not sure about it's properties.  Any ways of solving this which avoids using the gamma function?
84033
Let $\Omega$ be the right half plane excluding the imgainary axis and $f\in H(\Omega)$ such that $|f(z)|&lt;1$ for all $z\in\Omega$. If there exists $\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$ such that $$\lim_{r\rightarrow\infty}\frac{\log|f(re^{i\alpha})|}{r}=-\infty$$ prove that $f=0$.  The hint is define $g_n(z)=f(z)e^{nz}$, then by previous exericise $|g_n|&lt;1$ for all $z\in\Omega$.
99990
Suppose $\Omega$ is a region,$f\in H(\Omega)$, and $f\not\equiv 0$.If for every positive integer n,there is $f_n \in H(\Omega)$,$(f_n)^n=f$,prove that there is $g\in H(\Omega),f=e^g$.  My conjecture :  $$g=lim  (nf_n - n)$$  But I don't know how to deal with it.  I will appreciate your help.
826
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
91704
This is an exercise from Remmert's Theory of Complex functions.  Let $D\subset \mathbb{C}$ be a domain and $f:D\rightarrow \mathbb{C}$ a real-differentiable function. Assume that the following limit exists:  $ \mathrm{lim}_{h\rightarrow 0} \left|  \frac{f(c+h) - f(c)}{h}   \right|.$  Show that either $f$ or $\overline{f}$ is complex-differentiable.  I've tried showing that $\frac{\partial f}{\partial \overline{z}} = 0$ or $\frac{\partial \overline{f}}{\partial z} = 0$ by using the fact that there exist continuous functions $g$ and $h$ such that in $D$ one can write   $f(z) = f(c) + (z-c)g(z) + (\overline{z} - \overline{c})h(z)$   and that  $g(c)= f_{z}(c)$ and $h(c) = f_{\overline{z}}$ and then plugging this into the limit above. Does this approach works and I just can´t see how to do it? Can someone give a hint or a guideline solution to this?
90731
I am preparing for qualifying exams, and this is a question from the Penn State Qualifying Exam for Fall 2015. It is stated as follows  Let $\epsilon &gt; 0$ and let $f$ be holomorphic (analytic) on the disk $S = \{z \in \mathbb{C} \ | \ |z| &lt; 1 + \epsilon \}$. Suppose that $f(z)$ is real valued whenever $|z| = 1$. Prove that $f$ is constant.  I have tried a few things in showing this to be true, but I keep finding holes in my logic. My largest issue is that the portion of the domain in which $f$ is real valued is not open, and so I'm not able to use many of the theorems I otherwise feel would be helpful (Open Mapping Theorem, Cauchy-Riemann Equations, Maximum Modulus, etc.).   Most of my attacks towards this problem have centered around showing things for the unit disk $\mathbb{D}$ instead of $S$, because I figure if I can show that $f$ is constant on $\mathbb{D}$, then I can use the Identity Theorem to extend it to $S$. However, since I don't know anything about the specific values that $f$ obtains, I can't use Schwarz' Lemma either.  I played with the idea, also, of suggesting that, if we consider the closure of $\mathbb{D}$, then $f(\partial{\mathbb{D}}) \in \mathbb{R}$. Since $\partial\mathbb{D}$ is closed and bounded and $f$ is analytic, it's image should also be bounded. That would put, for some $M \in \mathbb{R}$, $f(\partial{\mathbb{D}}) \in [-M,M]$, which is bounded and obtains a maximum (since $f$ is continuous and real valued here). I know that this set isn't open, but there is a Corollary in my text (Complex Analysis - Freitag) that says the following:  "If $K$ is a compact subset of the domain $D$ and $f:D \rightarrow \mathbb{C}$ is analytic, then the restriction of $f|K$ being a continuous function has a maximal modulus on $K$. By the Maximum Modulus Principle, we can moreover affirm that the maximal modulus value is necessarily taken on the boundary of $\mathbb{D}$."  The proof of that statement follows from the Open Mapping Theorem. Would it be appropriate to use that in this case, then? If $K = \partial \mathbb{D}$, even though it's technically not an open set, can I say that $f$ obtains its maximum on $\mathbb{D}$, state that it's constant, and then extend this to $S$ using the Identity Theorem? My issue here is that the constant itself isn't actually in $\mathbb{D}$,  I appreciate any help for this problem (or any tips for showing that complex functions will be constant, as these types of problems show up often).  Edit: My function $f$ is not necessarily entire, so Liouville's Theorem does not apply
97775
Find all values of $\omega \in \mathbb{C}$ such that $\frac{\omega - \frac1\omega}{2i} = 2.$  So what I have is:  Let $\omega$ be represented as $x + yi$ (by definition of complex numbers)  Then $\frac1\omega = \frac{x-yi}{x^2 + y^2}$ (by properties of complex numbers)  So, $\frac{x+yi - \frac{x-yi}{x^2 + y^2}}{2i} = 2$  Eventually, I get, via expansion and simplification:  $\frac{x(x^2 + y^2 - 1) + yi(x^2 + y^2 + 1)}{2i(x^2 + y^2)}$  Any hints on how I can continue? I feel like I'm close but missing something.
123196
Let $f$ be holomorphic in an open $\Omega \subset \mathbb{C}$ and $\gamma$ a closed curve in $\text{int}(\Omega)$, along which $f$ is never zero.   Are these hypotheses enough to claim $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is an integer? If not, what are the necessary and sufficient conditions for that?  I can prove that's an integer for some particular cases, for example when $\Omega$ is convex and  $\gamma$ is a circle. But I've seen people claming this in many other contexts, like when $\Omega$ is an annulus around $0$ and $\gamma $ is an arbitrary curve inside it. This isn't obvious to me at all.  Thanks!
114990
Let $0&lt;\delta&lt;r$ and $R&gt;0$, and $a\in\mathbb C$. I want to show that there exists $C&gt;0$ such that for all holomorphic functions $f\colon B(a,r)\to B(0,R)$, we have $$ \vert f(z)-f(w)\vert\leq C\vert z-w\vert $$ for $z,w\in B(a,\delta)$.  I'm not sure how to approach this. I was wondering if I could use that $$ \left\vert\frac{f^{(n)}(a)}{n!}\right\vert\leq\frac{\Vert f\Vert_{C(a,\rho)}}{\rho^n} $$ for some $0&lt;\rho&lt;r$. I also have an expression from biholomorphic maps from $B(0,1)$ to itself, but I'm not sure if I could use that here. Any hints?
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.