I really don't know how to start with this.  Can you please help me with this? Suppose $\displaystyle{f: \overline{B_1(0)} \rightarrow \overline{B_1(0)}}$ is continuous and $f$ is analytic in $\displaystyle{B_1(0)}$.  Prove that $f(z)$ has a fixed point in $\displaystyle{\overline{B_1(0)}}$. How to do this?  Please. I would appreciate any help.  Thanks a lot.
122605
Let $f$ be holomorphic on $\overline{N(0,1)}$. Suppose $|f(z)| \leq 1$ for every $|z| = 1$. Without using fixed point theorem, show that there exists $z \in \overline{N(0,1)}$ such that $f(z) = z$.  My attempt:  Using Rouché's Theorem, I managed to show that if $|f(z)| &lt; 1$ for every $|z| = 1$, then there is a unique $z$ such that $f(z) = z$.  However, I am not sure how I should proceed if $|f(z)| = 1$ for some $|z| = 1$.
107154
if $f:[a,b]\longrightarrow [a,b]$ is continuous and $\alpha , \beta \in \mathbf{R}^+$ then  there is some $c\in [a,b]$, $$f(c)=\dfrac {\alpha f(a)+\beta f(b)}{\alpha+\beta}.$$ Should I use Intermediate Value Theorem?
62345
I came across this question whilst doing some research into complex analysis, and I just can't see what to do!  Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\overline{z})}$ is holomorphic, whilst $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.  I know that holomorphic means that the function is differentiable everywhere, and I need to apply the Cauchy-Riemann equations somehow, but I'm not sure how to approach this.
95079
Suppose that $f(z)$ is analytic in the unit disk $\Delta:\,|z|&lt;1$. Then $f(z)$ has a Taylor series $\sum=\sum a_nz^n$ in the unit disk. One may assume that $\sum$ has $R=1$ as its radius of convergence for convenience.  Question 1 (local version):  If $f(z)$   is continous at some point, say $z=1$, on the unit circle $S^1$, then its Taylor series $\sum$  is  covergent at $z=1$?   Question 2 (global version):  Suppose that  $f(z)$   is continous on the closed unit disk $\overline\Delta$.  Is the  Taylor series $\sum$ covergent at every poin of $S^1$?   Of course, if the answer to the local version is "yes", then so is the answer to the global version.  EDIT. It seems that  Question 1 obviously has a negative answer. So I modify it into a new question.  Question 3 (New local version):  Suppose that  $f(z)$   is continous at some point, say $z=1$, on the unit circle $S^1$ and the coefficients of $\sum$ satisfies $a_n\to 0$.  Is the Taylor series $\sum$  covergent at $z=1$?   Note. If $f$ is analytic at $z=1$, then the answer to Question 3 is yes. This is Fatou's Theorem.
75614
Here is the question: Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be an analytic function.  Suppose that there exists $\epsilon&gt;0$ such that for every $\theta$, $|\theta|&lt;\epsilon$, and $\lim_{z\rightarrow e^{i\theta}}f(z)=0$.  Prove that $f$ is identically $0$ on $\mathbb{D}$. My thoughts:  Well, honestly, I am quite stumped here.  I was thinking I could let $f$ be analytic in some other region (can I even do this?), say the lower half plane and suppose $\lim_{z\rightarrow x}f(z)=0$ whenever $-1\leq x\leq1$, by assumption.  Maybe then I could show that $f$ is identically $0$ on the bigger region, and then consider $f$ on the intersection of those two regions, and then ultimately making my way towards using the uniqueness theorem to show what I want... but I am quite stuck on how to do that... any thoughts, suggestions, etc. are greatly appreciated!  Thank you.
82525
Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.   I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.  Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.  Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$  We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$  How can we obtain this from the above relations?
76619
Let $f$ be continuous at $0$. Suppose $\displaystyle \lim _{x\rightarrow 0} \frac{f(2x)-f(x)}{x} =L$. Prove that $f'(0)=L$.   My Work:  $\displaystyle \Bigg|\frac{f(x)-f(0)}{x}-L\Bigg|=\Bigg|\frac{f(x)-f(2x)+f(2x)-f(0)}{x}-L\Bigg|$  $\displaystyle \leq \Bigg|\frac{f(x)-f(2x)}{x}-L\Bigg|+\Bigg|\frac{f(2x)-f(0)}{x}\Bigg|$. Since $f$ is continuous at $0, |f(2x)-f(0)|$ goes to $0$ as $x$ goes to $0$. Now I want to get rid of $|x|$ of the second term. How can I do it? Can somebody please help me?
22753
Prove that if one of $\displaystyle\lim_{x \rightarrow 0 }{f(x)}$ and $\displaystyle\lim_{x \rightarrow 0}{f(x^3)}$ exists, then the other one also exists. Can anyone guide me on this ? I have no idea on how to start .
75169
Suppose $f$ is an analytic function on a domain $D$. Then I need to show that $\overline {f(\bar z)}$ is also analytic.  Here is what I did -   Suppose $f(z) = u(x,y) + iv(x,y)$ where $u$ and $v$ are real functions of $x$ and $y$ and $z = x + iy$. Now $f(\bar z) =  u(x,-y) + iv(x,-y) $ and then $\overline {f(\bar z)} = u(x,-y) - iv(x,-y) $.   To show that a function is analytic, I need to verify that it satisfy Cauchy-Riemann Equations. I differentiated $\overline {f(\bar z)}$ and checked that it actually satisfies these equations. But here is my doubt - Being analytic means that the function if complex differentiable. Now here to check for analyticity, I am differentiating my function without proving that it's actually analytic. So is this the right way to do it?
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
110092
I need help to prove these theorems: 1)If $\mathcal f $ is continuous on $\mathbb C $  and $\mathcal f(z) = f(2z)  \forall z \in \mathbb C $ then $\mathcal f $ is constant .  2)If $\mathcal f $  is entire and $\mathcal f(2z)=2f(z)$ then $ \exists $ $\lambda $ such that $\mathcal f(z)=\lambda\mathcal z $.
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.
23367
As a part of self study, I am trying to prove the following statement:  Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.   Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
90936
Let $f$ be an analytic function on $\mathbb{C}$ which takes values in $\mathbb{C}\backslash(-\infty,0]$, i.e. takes values in the complement of the nonpositive part of the real axis. Show that $f$ is constant.  My first observation is that there is a branch of the square root defined on $\mathbb{C}\backslash(-\infty,0]$ which takes positive real values along the positive part of the real axis. But I am not sure what do from thereon out. Any help would be appreciated!
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!!
15812
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $\gamma$, then $\int_\gamma\overline{f(z)}f'(z)dz$ is purely imaginary.  I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $\int_a^b\overline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
51130
Let $f:U\rightarrow U$  be analytic, where $U \subsetneqq \mathbb{C}$ is a simply connected domain. Suppose that $f$ has at least two fixed points. Prove that $f$ is the indentity map.  I am able to prove the theorem in the case where $U = \mathbb{D}$ is the unit disc (the proof uses Schwarz lemma, and is discussed ).  The proof however skips the (rigorous) details of how the Riemann Mapping Theorem allows us to reduce from $U$ to $\mathbb{D}$. Any details are appreciated.