Question: Let $f$ be an entire function that satisfies $|f(z)| \leq 1 + |z|$ for all $z \in \mathbb{C}$. Show that $f(z) = az + b$ for fixed complex numbers $a$ and $b$. My attempts: I realise that I need to apply the generalised Cauchy Integral Formula $\displaystyle f^{(n)}(z_0)=\frac{n!}{2\pi i} \int_{\Gamma}\frac{f(z)}{(z-z_0)^{n+1}}\,dz$ on an arbitrarily large circle to show that $f^{(n)}(0)=0$ for all $n \geq 2$. While I know what to after this is proven, I'm unsure how to actually prove this. Any help or guidance would be greatly appreciated!
122300
Let $f$ be an entire function that satisfies $|f(z)| \leq 1 + |z|$ for all $z\in \mathbb{C}$. Show that $f(z) = az +b$ for fixed complex numbers $a,b$.    The hint tells us to try and use Cauchy's Integral Formula on an arbitrary circle.    This is my attempt: Consider an arbitrary large circle with centre $0$ and let $\Upsilon$ be the contour around this circle.    Then $$\int_{\Upsilon} \frac{f(z)}{z^{n+1}} \mathrm{d}z = \frac{2\pi i}{n!}f^{(n)}(0).$$    Note that $$\int_{\Upsilon}\frac{f(z)}{z^{n+1}} \leq \int_{\Upsilon}\left|\frac{f(z)}{z^{n+1}} \right|\leq \int_{\Upsilon} \frac{1}{|z^{n+1}|} + \frac{1}{|z|^n}$$ and the right hand side is $0$ by Cauchy's Integral Formula. I'm not sure if the first equality is valid, and not sure where to go from here.
105860
How would I show that an entire function $f$ with the property $|f(z)| \geq 1$ must be constant?    I'm aware of Liouville's theorem, just not sure how to apply it here.
40890
Let $f$ be holomorphic in the unit disk $D(0,1)$, and $f(0)=0$. Suppose $|\Re f(z)|\leq 1$ for all $|z|&lt;1$. Show that  $$|\Re f(z)|\leq\frac{4}{\pi}\arctan|z|;$$ $$|\Im f(z)|\leq \frac{2}{\pi}\ln\frac{1+|z|}{1-|z|}.$$  What I could do is just to observe that $$\frac{e^{i\frac{\pi}{2}f(z)}-1}{e^{i\frac{\pi}{2}f(z)}+1}:D(0,1)\to D(0,1)$$ with $0$ being fixed. And Schwarz lemma applies. However, I could not get any information on the real and imaginary part of $f$.
38091
I've encountered the following problem in a set of course notes on complex analysis, but I can't seem to solve it:     Prove that if f is a continuous real-valued function with $|f(z)| \leq 1$, then $\left|\int_{|z| = 1} f(z)\,dz\right| \leq 4$.   This is my work so far:     \begin{align*} \left|\int_{|z| = 1} f(z) \,dz \right| &amp;= \left|\int_0^{2\pi}f\left(e^{i\theta}\right)ie^{i\theta}\,d\theta\right| \\ &amp;= \left|\int_0^{2\pi}f\left(e^{i\theta}\right)\cos\theta\,d\theta + i\int_0^{2\pi}f\left(e^{i\theta}\right)\sin\theta\,d\theta\right| \end{align*}   I initially tried to use the triangle inequality at this point and then bound the real integrals, but this gives me an upper bound of $4\pi$. Using the estimation lemma on the initial integral gives me an upper bound of $2\pi$. I can't manage to get it any tighter.  The emphasis on real-valued makes it seem that this is a key point. Any ideas?
14252
Could someone advise me on how to determine all entire functions $f$ such that $|f'(z)| \leq |f(z)|, \forall z\ ?$  Hints will suffice, thank you.
88074
Find all entire functions such that $|f(z)|\leq |\sin(z)|$ for all $z\in\mathbb C$.  Half an hour ago I asked the same question for $|f(z)|\leq |z^2-1|$  but this time I cannot say something like $$|f(z)|\leq M|z|^m$$ so I don't know how to start. The problem is that $$ \sin(z) = \frac{e^{iz}-e^{-iz}}{2i} $$ gets really large if $z$ gets large.
6711
Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic
22336
Let $f(z)$ be entire function. Consider the functions $e^{if(z)}$ and $e^{−if(z)}$ and applying the Maximum Modulus Theorem, show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function.  (We take $f(z)=u(z)+iv(z)$)  I am confused as so far I have $|g(z)|=|e^{if(z)}|=|e^{-v(z)}|$ and then since $f(z)$ is real, $f(z)=u(z)$ and $v(z)=0$ so I assumed it would follow that $|g(z)|=|e^{v(z)}|=1$.  Similarly, $|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1$.  Using Liouville I assumed one could say that both $g(z)$ and $h(z)$ are bounded entire functions, they are constant and so it follows that $v(z)$ is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of $u(z)$ are equal to zero.  It would then follow that $f(z)$ is constant.  I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of $|z|=1$ perhaps?  Any help would be much appreciated!!
122888
This is a problem from my complex analysis textbook.  The hint is to consider $g(z)=(z-iR)^2(z+iR)^2 f(z)$ and to show that $|g(z)| \leq 8R^2$.  This is what i have tried: Consider $Re z \geq 0$, then $|(z-iR)^2 f(z)|\leq \left(\frac{|z-iR|}{Re z}\right)^2\leq 2$   But then I don't know what to do next. Any help I'd appreciate :)
29631
There are three parts of the question, the first two which I proved.   $a.$ Proving $f(z)$ is entire analytic.  $b.$ $f(z+1) = f(z)$ and $f(z+i) = e^{\pi}e^{-2\pi iz}f(z)$  $c.$ Inside the unit square in the 1st quadrant, prove that $f(z)$ has a unique $z_0$ such that $f(z_0)=0$  I need some help regarding the $c.$ part of the question. I don't understand how to apply the argument principle here (which is a hint). It is also given that $f(z)$ does not vanish on the boundary of the square.  What I could do:  $$f(0) = f(1) = 1+2\sum_{n=1}^{\infty}e^{-\pi n^2}$$  and   $$f(i) = f(1+i) = e^{\pi}\big(1+2\sum_{n=1}^{\infty}e^{-\pi n^2}\big)$$  Using the hint given below by a user: I end up with $$\dfrac{1}{2\pi i}log f(z)$$ and the integral evaluates to -1/2 and 1/2 on the two vertical sides and 0 on the horizontal lines. Resulting in the closed integral to 0. Any ideas where I might be going wrong?  Any idea how to proceed?
54970
If the function $f(z)$ is analytic and one valued in $|z-a|&lt;R$, How do we prove that $0&lt;r&lt;R$, $f^{'}(a)$ = $\frac{1}{{\pi}r}\int P(\theta)e^{-i\theta}d\theta$, where $P(\theta)$ is the real part of $f(a+re^{i\theta})$. I am guessing Cauchy's integral formula can be used but that has lead me to $f^{'}(a)$ = $\frac{1}{2{\pi}r}\int f(a+ re^{i\theta})e^{-i\theta}d\theta$. How to proceed from here.
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!
97133
Let $f : \mathbb{C} \to \mathbb{C}$ be analytical. How does one prove that there are finitely many positive $n$ such that $f(1+1/n) = \frac{1}{n+1}$ ? A hint is given: think of the expression $\frac{z-1}{z}$.  I think this is a pretty hard homework assignment. Here's what I tried so far: Let $g(z) = \frac{z-1}{z} $ . Then $g(1 + 1/n) = \frac{1 + 1/n - 1}{ 1 + 1/n} = \frac{1/n}{1+1/n} = \frac{1}{n+1} $. So the given hint is probably indeed useful, but I'm not sure yet how.   Another thing I tried to use to obtain the solution was the given fact that $f$ is analytical. Let $f = g$. Then $f(z) = f(x+iy) = \frac{ x + iy - 1}{x+iy} = \frac{x+iy-1}{x+iy} \cdot \frac{x-iy}{x-iy} = \frac{x^2 - 2 i x y + y^2 + iy}{x^2 + y^2} = 1 + i \frac{y(1-2x)}{x^2 + y^2} $ . Now I have written $f$ in the form $ f = u + i v$, with $u,v$ being real-valued functions. So now we can use the Cauchy-Riemann Integral formulas: $ \frac{d}{dx} 1 = 0 = \frac{d}{dy} \frac{ y(1-2x) }{x^2+y^2} \implies \frac{y(1-2x)}{x^2+y^2} = c $. The other formula yields a similar result.   But now what? I still have no idea how to prove that there are finitely many positive $n$ such that the aforementioned equality holds. Should I use any of the results I already obtained? If so, how? If not, do you know how to prove this?
85733
Let $f$ be an entire function such that $f\circ f$ has no fixed points. Prove that $f$ is a translation   $$z\mapsto f(z)=z+b \qquad (b\neq 0)$$   Firstly, we prove that there exists a constant $c\in \mathbb{C}\backslash \{0,1\}$ such that  $$f(f(z))-z=c(f(z)-z) $$ applying Picard's little theorem. If $c=0$, then $f(f(z))=z$, so $f\circ f$ has a fixed point (absurd). If $c=1$, then $f(f(z))=f(z)$, so $f$ is the identity $f(z)=z$ and of course it has fixed point (absurd). Then,  $$F(z)=\frac{f(f(z))-z}{(f(z)-z)}$$ is an entire function which does not take the values 0 and 1 so, by Picard's little theorem, it must be constant.  Also, I've proved that $f'\circ f$ is a constant function. Let's see this. Differentiating  $$f(f(z))-z=c(f(z)-z) $$ we have $$f'(z)f'(f(z))-1=cf'(z)-c$$ $$f'(z)[f'(f(z))-c]=1-c$$ Again, the entire function  $$G(z)=f'(z)[f'(f(z))-c]$$ does not take the values $0$ and $1$ so, by Picard's Little Theorem, then is constant.   However, I don't know how to prove this problem. Any help would be appreciate.
99702
Let $\Gamma$ denote the positively oriented circle of radius $2$ with center at the origin. Let $f$ be an analytic function on $\{z \in\mathbb{C} : |z|&gt;1 \}$, and let $$\lim_{z\to \infty}f(z)=0.$$  Prove that $$ f(z)=\frac{1}{2\pi i} \int_{\Gamma} \frac{f(\xi)}{z-\xi}d\xi,  $$ for all $z\in\mathbb{C}$ with $|z|&gt;2$.  I have no idea how to even start. Any hint will be appreciated.
76141
Let $D := \{z \in \mathbb{C}: |z| &lt; 1\}$ and $f\colon D \rightarrow \mathbb{C}$ be holomorphic. Suppose $\lvert f(z)\rvert \leq 1$ on $D$, show that $$\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|} \ \forall z \in D.$$  I have tried using Cauchy Integral Formula to $f$ and expanding $f$ at $0$, but I have no idea why $f(0)$ appears in the inequality in the former case. Both answers and hints are appreciated and thanks in advance!
2445
An analytic function $f(z) = f(x+iy)$ in $\mathbb{C}$ cannot have modulus $\frac{A}{\cosh(x)}$ for some constant $A \neq 0$.  Can we do so simply using the Cauchy-Riemann Equations?  I tried working by contradiction:  Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$  We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (\frac{A}{\cosh x})^2$$  I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.   My start point is:  $$u_x = v_y \text{ and } u_y = -v_x$$ along with $$u(x,y)u_x + v(x,y)v_x = -A^2 \frac{sinh x}{(cosh x )^3}$$ and  $$u(x,y)u_y + v(x,y)v_y = 0$$  I would appreciate some hints!