Let $f(z)=2z^2-1$.Then what is the maximum value of  $|f(z)|$ on the unit disc  $D=\{z\in C:|z|\le1\} $ equals   $2$ $3$ $1$ $3$ more than minimum value  This question can have more than 1 answer? I dont know how to proceed
121454
I am thinking about the following problem:     Let $f(z)=2z^2-1.$Then what is  the maximum value of $|f(z)|$ on the unit disc $D=\{z \in \mathbb C:|z|\leq 1\}$ ?   I guess i have to use the maximum modulus principle.But  i also notice that $|f(z)|=|2z^2-1|\leq 2|z|^2+1\leq 2+1=3.$  So, is $3$  the maximum value of $|f(z)|?$ Can someone point me in the right direction? Thanks in advance for your time.
94842
I am trying to prove that:  Let $f$ be analytic in the closed unit disc with $f(z)\leq 1$ for $|z|\leq1$.  For a given $0&lt;r&lt;1$ and $|z_1|,|z_2|\leq r$, $z_1\neq z_2$ then,   $$\left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|\leq \frac{1}{(1-r)^2}.$$  My attempt is: part a  of the question asks to prove $$\left|\sum_k^\infty a_nz^n\right|\leq \frac{ |z^k|}{1-|z|}.$$   So, \begin{eqnarray*}  \left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|&amp;=&amp; \frac{|\sum_1^\infty a_nz_2^n-\sum_1^\infty a_nz_1^n|}{|z_2-z_1|}\\  &amp;\leq&amp; \frac{\frac{ |z_1 |}{1-|z_1|}+\frac{ |z _2|}{1-|z_2|}}{|z_2-z_1|}\\  &amp;\leq&amp;\frac{1}{(1-r)^2} \frac{   |z_2| ({1-|z_1|})+  |z _1| (1-|z_2|) }{|z_2-z_1|}\\ \end{eqnarray*} But could not prove that the last term is controlled by 1.
60820
Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you.
61181
How can one show there is no holomorphic function $f$ on the open unit disc $\mathbb{D}$ such that it extends continuously to $\frac{1}{z}$ on $\partial\mathbb{D}$? I mean $f$ takes value $\frac{1}{z}$ when $|z|=1$ and continuous here.  Thank you.
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!
120974
Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.  I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have no prove.
34284
Let $f(z)$ be a holomorphic function on a disk of radius $&gt;1$ about the origin, and suppose that $|f(z)|\leq 1$ for $z$ with $|z|=1$ and $\text{Re}(z)&gt;0$ and $|f(z)|\leq 2$ for $z$ with $|z|=1$ and $\text{Re}(z)\leq 0$.  How does one show the bound $|f(0)|\leq \sqrt{2}$?
57591
Suppose $f: \overline{\mathbb{D}} \to \mathbb{C}$ is continuous, analytic in $\mathbb{D}$ and satisfies $|f(z)|&lt;1$ for $|z|=1$. Find the number of solutions to the equation $f(z) = z^n$ where $n$ is a positive integer. ($\mathbb{D}$ is the unit disk).  Some of my own attempts: Using the Maximum/Minimum Principle, we can split up the the problem in $2$ parts: Suppose $f$ is constant. It then holds that $z^n = c$ has exactly $n$ zeros because of the fundamental theorem of algebra.  Suppose $f$ is not constant. It holds that $f$ has at least one zero  in $\mathbb{D}$. It of course also holds that $z^n$ has a zero of multiplicity $n$ at the origin. I think I'm supposed to hit with stuff like Rouché's theorem, the Argument Principle or Schwartz lemma, but nothing seems to fit.   Thanks in advance.
106227
Suppose that  $f(z)$ is analytic and $f(z)\neq0$  on the disc $|z|&lt;1$ .  Show that for $0&lt;r&lt;1$,  $$ \exp\left[\frac{1}{2\pi}\int_{0}^{2\pi}{\log}|f(re^{i\theta})|\,d\theta\right] = |f(0)| $$  Any other theorems needed to solve this problem? I don't have any idea just with "Gauss's MVT"  Thank you in advanced.
229
I am completely stuck in the following problem:  let f be holomorphic function on the unit disk $\{z:|z|\lt 1\}$ in the complex plane.which of the following is/are necessarily true?  (a) if for each positive integer $n$ we have $f(\frac{1}{n})=\frac{1}{n^2}$,then $f(z)=z^2$ on the unit disc.  (b) if for each positive integer $n$ we have $f(1-\frac{1}{n})=(1-\frac{1}{n})^2$,then $f(z)=z^2$ on the unit disc.  (c) $f$ cannot satisfy $f(\frac{1}{n})=\frac{(-1)^n}{n}$ for each positive integer $n$  (d) $f$ cannot satisfy $f(\frac{1}{n})=\frac{1}{n+1}$ for each positive integer n   what I thought in this problem is the following:  in the first part,by identity theorem we can say that $f(z)=z^2,\forall$z in the unit disc.  but,in the second part ,if we choose $g(z)=z^2$ then,$f(1-\frac{1}{n})=g(1-\frac{1}{n})$ but,the limit point ,i.e, $1$ is not in the unit disc.so,in this case,we cannot use identity theorem to conclude whether $f(z)=z^2$ on the unit disk or not .so,what procedure can we follow in this case???also,explain the options (c) and (d).  please answer in detail..  thanks a lot...
4597
Greets  This is exercise 15.d chapter 3 of Stein &amp; Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:  Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon&gt;0$ $g(D_{\epsilon&gt;0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.  I would like to know an answer with the the maximum modulus principle.  Thanks
42133
Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?   I've been attempting this with Lagrange multipliers in a few different ways. However, the resulting system of equations with two lagrangians has so many variables that it becomes very complicated. Can someone show how this is to be done manually?  I also attempted turning it into two unknowns by replacing $z$ with $x+y$. However, this also led nowhere.
66962
So i have to prove if $f(z_1)=f(z_2)$ then $z_1=z_2$ .  I tried substituting $z=x+yi$ i also tried proving their differences in absolute value are zero.None of this work .Im sure im missing a property or something that will make this very easy. I also tried making the first part of the eq belonging is $R$  so the  second part of the eq must have imaginary part  zero. I just dont get to the result i want.
27340
Suppose that $f$ is analytic on a domain $D$, which contains a simple closed curve $\gamma$ and the inside of $\gamma$. If $|f|$ is constant on $\gamma$, then I want to prove that either $f$ is constant or $f$ has a zero inside $\gamma$  Here is my take: if $f$ is constant, i dont see a reason why $|f|$ wouldnt be constant :)  if $f$ is not constant, then the max/min modulus principle applies ... meaning $|f|$ can not have any local max/min on D now i am lost at this point ...
21389
Is the following statement true?  Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ D $.  P.S. I think I got a solution. But I was wondering if this problem has a trivial solution.  Here is my solution, since nobody has posted a solution yet:-  Suppose, $ \gamma $ is the boundary of $ D $(assuming $ \gamma(t)=e^{2\pi it} $). Then $ f\circ \gamma $ is a simple continuously differentiable curve. So by Jordan curve theorem, $ f\circ \gamma $ divides $ \mathbb C $ into 2 path connected regions $ A_1, A_2 $ such that if $ a\in A_1 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=2\pi i$ and if $ a\in A_2 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=0$. So, $ f(z)=a $ for some $ a\in A_1,z\in D $ implies $ \int_{\gamma} \frac {f'(z)}{f(z)-a} dz=2\pi i $. So $ f(z)=a $ has only one solution in $ D $. Similarly if $ a\in A_2 $, then $\int_{\gamma} \frac {f'(z)}{f(z)-a}=0 $, so $ f(z)=a $ has no solution in $ D $. If $ f(z)=a $ for some $ z\in D, a\in f\circ \gamma $, then by open mapping theorem, $ f(z)=a' $ for some $ z\in D, a'\in A_2 $, which is not possible. So $ f $ is one-one in $ D $.
18552
$f$ is an analytical function on the unit disk $\mathbb{D}$ for which there is a $C &gt; 0$ such that  $\forall z \in \mathbb{D} \colon |f(z)| \leq \dfrac{C}{1-|z|}$.  I know have to prove that    $\forall z \in \mathbb{D} \colon |f'(z)| \leq \dfrac{4C}{(1-|z|)^2}$.  I tried using the Cauchy integral formula but I can't find a fitting circle to get this result.
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?