I need to prove that if $f$ is an entire function (holomorphic in all the complex space) and $f = f'$ then there is a complex constant $K$ such that $f(z)  = Ke^z$.  Moreover, I need to show that $e^z = \sum\limits_{n \ge 0} \frac{z^n}{n!}$.  Well I am not so sure what to do here. I see that $f-f' = $ constant, but I don't think it helps. I can clearly show that if $f(z) = Ke^z$ it satisfies the conditions, but I am not sure how to prove that $f$ is necessarily such a function.  Help would be appreciated.
120428
How is the function $e^z$ the only function that is entire and has the property $f'(z) = f(z)$ and that $f(0) = 1$ but how would you exactly prove it?  where $e^z = e^x(\cos y + i\sin y)$
48580
Let $f$ be an entire function such that $\mid f(z) \mid \leq A \mid z \mid$ for all $z$, where $A$ is a fixed number. Show that $f(z)=a_1z$, where $a_1$ is a complex constant.
90080
$f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$  What I think is since $f$ do not have any zero for some bounded domain, I can define a branch of logarithm $(\log f)$ on that domain which will gives my desired result $f =e^{\log f}$. I don't know if I am doing it right? If this is right I don't know how do I argue $(\log f)$ is entire. Hint please.
3814
The problem as stated in R. Remmert's Theory of Complex Functions p. 62., is this:  Let $f = u + iv$ be holomorphic in the region $G \subseteq \mathbb{C}$ and satisfy $u = h \circ v$ for some differentiable function $h : \mathbb{R} \to \mathbb{R}$. Show that $f$ is constant.  I've attacked this computationally in each way I can think of. I've tried to show that $\frac{\partial f}{\partial z}$ is zero; played with the Laplace equation, knowing that $h \circ v$ and $v$ must be hamonically conjugate, etc. I even considered spinning an argument based on the maximum principle.  It seems obvious, but I have not succeeded in a proof yet. Perhaps I am missing a concept?
70748
Suppose $h(z)$ is a complex function. I have noticed that $h(z) = f(z)\cdot g(z)$, where $f(z)$ and $g(z)$ are non-holomorphic. Can $h(z)$ be holomorphic?   Can a similar statement be made, if $f(z)$ is holomorphic, but $g(z)$ is not?
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!!
18801
Suppose $f(z)$ is holomorphic in $|z|&lt; 1$ and $\operatorname{Re}f(z) &gt; 0$ in this region. Furthermore, suppose $f(0) = a &gt; 0$. Then why does $|f'(0)| \leq 2a$?  The hint I have is to consider $g(z) = \frac{f(z) - a}{f(z) + a}$ and then show $|g'(0)| \leq 1$. If I do have $|g'(0)| \leq 1$, then the result follows as $|f'(0)| = 2a|g'(0)| \leq 2a$, but I have no idea on how to show $|g'(0)| \leq 1$.
121878
Let $f$ be a meromorphic function on $\mathbb{C}$ such that $|f(z)|\geq |z|$ at each $z$, where $f$ is holomorphic. Then, which of the following is/are true.   The hypothesis are contradictory, so no such $f$ exists. Such an $f$ is entire. There is a unique $f$ satisfying the given conditions. There is an $A\in \mathbb{C}$ with $|A|\geq 1$ such that $f(z)=Az$ for each $z\in \mathbb{C}$   I think $1$ and $3$ are false but not sure. And what about $2$ and $4$? I am unable to find suitable examples. Please help me here.
92203
Function $f(z)$ is holomorphic at least within the unit disk.   If for all arguments $|z|&lt;1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.  How should I tackle this problem? Can it be generalized if for $|z|&lt;R$ we have $|f(z)| \le M$ ?    So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
83190
Let $f$ be an entire function s.t $f(\mathbb{C}) \cap B_R(z_0) = \varnothing$ for some $z_0$ and some $R$. Then $f$ is constant.  I guess since the image of the whole plane isn't dense then $f$ doesn't have an essential singularity at infinity. Now I have to exclude it has a polar singularity at infinity..
40760
As I attempted to solve for a function that is the $k$-th derivative of itself where $k$ is some natural number, and generalise the result, I realised it gets very hard to prove anything. For $k= 1$, apart from trivial corner cases, it's the exponential function with any constant multiplied.  For $k=2$, along with the exponential function and it's reciprocal, it works for their sum along with any mixture by constants multiplied...  For any $k$, I guessed it should be of the form $e^{x w}$ where w is the $k$-th root of unity. And we can add all these $k$ terms for each root with constants multiplied.  While it is easy to verify it's satisfying the condition, it is beyond my scope to prove this is the only family of functions that satisfy it.  I am hoping for some interesting math at work here, and want to know if this is an already established fact. Also, how do I prove anything here?
29355
I need to prove that if $f,g$ are analytic in the unit disk, and $|f|^2+|g|^2=1$ for all $z$ in the unit disk, then $f,g$ are constant.  This is an exercise question so it should not be very hard, but I don't know where to start. Any hint is appreciated.
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?
48119
I want to prove that if $f$ is an analytic function for which $\Im f$ is constant, this implies that $f$ itself is constant.  So to start off, it's not given that the function is entire or anything, ruling out Liouville. Rather, I'm guessing the maximum modulus theorem will prove useful here (?).  If I were to show using that the real part of $f$ was constant, proving the statement would be easy (just take the absolute value and investigate $e^f$), but in this case I'm none the wiser from that. Should I use the same parametrization $e^f$ and apply the Cauchy-Riemann equations somewhere? Any suggestions welcome.
116696
Let $f$ be an entire function such that $\int_0^{2\pi} |f(r e^{it})| dt\leq r^{\frac{41}{3}}$ for all $r&gt;0$. Prove that $f$ is constant.  I thought of using Liouville's theorem, combined with Cauchy's estimates. To do so, I tried to use Cauchy's integral formula, but it is not possible to bound the term $|w-z|$ which appears below in the integral.  Anyone can give me a hint? Thanks!
87244
If $f:\mathbb{D}\rightarrow\mathbb{D}$ is holomorphic then prove that   $$\frac{|f(0)|-|z|}{1 + |f(0)||z|} \leq|f(z)| \leq\frac{|f(0)| + |z|}{1 - |f(0)||z|} $$   I have been wracking my brain for hours and am out of ideas.  The hypothesis seems to suggest an application of the Schwarz lemma but I cannot find a suitable function - I've tried functions of the form $$\frac{f(z) - f(0)}{\mbox{scale factor}}$$ and none have worked.  I also tried working backwards to obtain an inequality and got $$-|z|\leq|f(z)||1\pm f(0)z| - |f(0)|\leq |z|\;\;\;\mbox{(*) }$$ If I could prove $$|f(z)[1\pm f(0)z] - f(0)|\leq 1$$ then I could obtain (*), but with so few conditions on $f$ I have no idea how to proceed.
11977
I'm having trouble proving the following:  Let $f$ be analytic on $D(0,R)$ and suppose there is a real constant A for which $\operatorname{Re}f(z) &lt; A$ for all $z \in D(0,R)$. If $f(0) = 0$, show that $|f(z)| \leq \frac{2A|z|}{R-|z|}$.   I was told to first start by showing that $|\frac{f(z)}{2A-f(z)}| \leq 1$ and also use the following result, which I was able to prove:  Let $f$ be analytic on $D(0,R)$ and let $f(0)=0$. If $|f(z)| \leq M$ for all $z \in D(0,R)$, then $|f(z)| \leq M|z|/R$ for all $z\in D(0,R)$.  Thanks in advance!