That's it. I don't really know what to do. I've tried to used analytic continuation with the function $g(z)=\frac{1}{1+z}$ but of course, such a function is not analytic in $\mathbb{C}$ so i can't use the theorem. And i can't find the contradiction
124104
Does there exist a function $f$ which is holomorphic on $B_0(2)$ (open disc of radius 2 in the complex plane) such that $f(1/n)=1/(n+1) \forall n \in \mathbb{N}$? At the moment I'm thinking not but a proof is seeming elusive. Any hints would be appreciated.
50538
Title says it all I looked for a proof on this site but couldn't find one.  Prove if $f$ is analytic on $D$ and $g$ is analytic on $\Omega$ containing the range of $f$ show $g(f(z))$ is analytic.  The statement seems obvious but I can't seem to be able to prove it.
108215
Studying for an exit exam and it's been years since I work with any series expansions. Here's a past problem:  Show that $$\sum_{n=1}^{\infty} \frac{z^n}{1+z^{2n}}$$ converges to an analytic function on the set $A=\{z \in \mathbb{C}: |z|&lt;1 \}$.  I'm looking to use Taylor's theorem for this problem but how would I compute the radius of convergence? Do I have to do a comparison test?
5755
Let $f(z)$ be analytic on $\mathbb{D}$ = {${z\in\mathbb{C}:|z-1|&lt;1}$} such that $f(1) = 1$, if $f(z) = f(z^2)$ for all $z\in\mathbb{D}$, then which of the following statements are correct?  1) $f(z) = [f(z)]^2$  2) $f(\frac{z}{2}) = \frac{1}{2}f(z)$  3) $f(z^3) = [f(z)]^3$  4) $f'(1) = 0$  I can't understand how to do solving this problem. please anyone give me some hints.
97979
Is there an analytic function $f$ on $B(0,1)\subset\mathbb{C}$ such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? I know the following doesn't work:  Let $g(z)=\exp(-1/z)$. Then, $f=g$ on a sequence with a limit point in $B(0,1)$ and so $f=g$ on $B(0,1)$. Since $g$ is not $\mathbb{C}$-differentiable at $0$, neither is $f$ and so such a function cannot exist.  This is not the solution because you cannot use the identity principle with a non analytic function like $\exp(-1/z)$ is not analytic at $z=0$. Any help using the identity principle another way?
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?
13616
I've been asked by some schoolmates why we have $$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$ I couldn't say much besides that the $\Gamma$ function, analytic continuation of the factorial, is defined with an integral involving $e$. Then I also know that actually $$ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$ Is there a reason for these facts?  P.S. I added the tag "intuition", please remove it if you think it is not pertinent.
31277
Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance.
111707
Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.   Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.   I am not answering the question and inviting all users to answer it.  It is a nice problem.   I will answer it within a week, but I am letting everyone try.
66397
Use the Cauchy-Riemann equations to show that the function $$g(z) = \sin (\bar z)$$ is not analytic at any point of $\mathbb{C}$.  Here's as far as I got -  $$\sin \left(\frac{\bar z}{1}.\frac{z}{z}\right) =\sin \left(\frac{|z|^2}{z}\right) =\sin \left(\frac{x^2 + y^2}{x+iy}\right)$$  I can't see how to separate the real and imaginary parts so that I can apply the Cauchy-Riemann equations.
51351
So the problem states that if $f(z)$ is entire, and satisfies the relation $f(z+i) = f(z)$ and $f(z+1) = f(z)$, show that $f(z)$ is constant. So I was thinking that since any point in $\mathbb{C}$ can be written as $\alpha * 1 + \beta * i $ we can say that $f(z + z_0) = f(z) $ in which case it is constant, but I'm having trouble breaking down the steps and using the fact that $f$ is entire, which makes me feel like I'm missing something. What should I review to figure this out?
8783
I have the following problem:     Find the constants $c_n$ so that  $$ \frac{1}{\sin^2 z } = \sum_{n= -\infty} ^ {+\infty} \frac{c_n}{(z-\pi n)^2} $$ and the series converges uniformly on every bounded set after dropping finitely many terms. Justify all your claims. Hint: Use Liouville's theorem to prove the equality.     Let $c_n = 1 $ for all $n$.   Let $\displaystyle f(z) := \frac{1}{\sin^2 z}$ and $\displaystyle g(z) := \sum_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $.   Begin by showing that $h$ is an analytic function which converges uniformly on compact subsets of $\mathbb{C} \backslash \mathbb{Z}$.   Suppose $K$ is a compact set which contains no integers. Define  $$ \delta_n = \inf_{z \in K} |\pi-z/n| =\frac{1}{n} \inf_{z \in K} |\pi n-z|, $$ where the infimum exists because $K$ is compact. Also, compactness implies boundedness. Thus as $n \to \pm \infty$, we have $\delta_n \to \pi$.   Therefore, for sufficiently large $n$, we have $\delta_n &gt; 2$, and  $$ \frac{1}{|z \pm n | ^2} \leq \frac{1}{\delta_n ^2 n^2} &lt; \frac{1}{4n^2}. $$ By the Weierstrass M-test, $g$ converges absolutely uniformly on $K$. Since each term is analytic on $K$, we conclude that the series converges to an analytic function on $\mathbb{C} \backslash \mathbb{Z}$.   Clearly the only poles of $g(z)$ are at $\pi n$ for each integer $n$, with corresponding principal part $\frac{1}{(z-\pi n)^2}$.   For each integer $n$, we have $\sin^2 (\pi n) = \frac{d}{dz} \sin^2(\pi n) = 0$ and $\frac{d^2}{dz^z} \sin^2 (\pi n) \neq 0$, so $f(z) = 1/ \sin^2 (z)$ also has a pole of order two at each integer multiple of $\pi$, and no other poles. Furthermore, the principal part of $f(z)$ is, using the Laurent formulas and contour integration, equal to $\displaystyle \frac{1}{(z-n\pi)^2}$.   Thus $h(z) := f(z)-g(z)$ has removable singularities at the points $n\pi$.   Note that both $f$ and $g$ are periodic with period $\pi$. That is,  $$ f(z) = f(z +\pi) \quad \text{ and } \quad g(z) = g(z+\pi) \quad \text{ for all } z \in \mathbb{C}\backslash\mathbb{Z}. $$ Thus, since $h$ is bounded on the square $\{ z: |\operatorname{Re} z| &lt; \pi, |\operatorname{Im} z |&lt; \pi \}$ and periodic, we may conclude that $h$ is bounded on the set $\{ z: |\operatorname{Im} z |&lt; \pi \}$. To show boundedness on the entire plane, we show that it holds on the vertical strip (with center removed)  $$ S = \{ z: 0 \leq \operatorname{Re} z \leq \pi, | \operatorname{Im} z | \geq \pi \}. $$  For $z$ in $S$, \begin{align*}  \sum_{n= -\infty} ^ {+ \infty} |z-\pi n | ^{-2}  &amp;= \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(x-\pi n)^2 + y^2}   &amp;\leq \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi (n-1))^2 +y^2}   \end{align*} \begin{align*} &amp; = \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2}   &amp;&lt; 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2}   &amp;\leq 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +\pi ^2}.  \end{align*}  Thus $g$ is bounded on the set. It is easy to show that $f$ is also bounded on $S$. Thus the difference $f-g$ is also bounded. By the periodicity of $h$, the function is bounded on the entire plane, and is constant by Liouville's theorem.   I'm wondering if my reasoning so far is valid; and if so, how to show that the relevant constant is in fact zero.   Thanks.
48836
I'm currently self-studying &quot;Theory of Functions of a Complex Variable&quot; by A. I. Markushevich and now I've encountered several difficult problems in the book from the start. This is just one of them. Prove that every complex number $z$ where $|z|=1$ but $z\not=-1$ can be represented as $$z=\frac {1+it}{1-it}$$ where $t \in \mathbb{R}$. I've tried writing $$z=e^{i(t+\frac{\pi}{2}-\frac{\pi}{2})}$$ $$=\cos(t+\frac{\pi}{2}-\frac{\pi}{2})+i\sin(t+\frac{\pi}{2}-\frac{\pi}{2})$$ and then reducing by the addition formula but it doesn't seem to go anywhere. If someone would give me a hint to start, not the full answer, I would appreciate it very much.
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?
8597
Let $f(z)$ be an entire function such that $$|f(z)|&lt;\frac{1}{|\text{Im}(z)|},\qquad z\in\Bbb C-\Bbb R.$$ The question asked me to prove that $f(z)=0$. At least looking at it, it really seems to have an application of Liouville's theorem lurking around somewhere, but I haven't found it.   My thoughts first led me to think about doing this by contradiction and using Picard's little theorem. So, I've considered a strip containing the real axis (say of width $2$ for simplicity). This will imply the complement maps into the unit disk, which implies that this strip has to map almost everywhere else on the complex plane. Now, on the imaginary axis, I know that $f(z)$ will vanish as $z\to\infty$, and this seems like it might be useful. I'm sort of under the impression $f(z)$ might have an essential singularity at $\infty$, which would mean $f(1/z)$ has an essential singularity at $0$. Either way, it of course can't have a pole at $\infty$ because of $f(z)$ vanishing on the imaginary axis, and if it is a removable singularity, it must be $0$, which still gives the solution. Therefore, I think it must have something to do with $f(z)$ having an essential singularity at $\infty$.   I was hoping to combine this with the above inequality and deduce a contradiction, but I haven't thought of one yet.   Any hints or suggestions? I'm studying for a prelim. exam, so this isn't homework.  Update: So, in the spirit of searching methods using the essential singularity at $z=0$, we have $$\left|f\left(\frac{1}{z}\right)\right|\leq\frac{1}{\left|\text{Im}\left(\frac{1}{z}\right)\right|}=\frac{|z|^2}{|y|}.$$ Thus, for $|y|&gt;1$, we have $$\left|f\left(\frac{1}{z}\right)\right|&lt;|y|\left|f\left(\frac{1}{z}\right)\right|&lt;|z|^2.$$ Thus, if I can show that $|f(1/z)|$ is a polynomial, we'll know it can't have an essential singularity at the origin, concluding my proof.
67562
Let g be an entire function such that   $2g(z)=g(\frac{z}{2}) + g(\frac{z+1}{2})$ holds $\forall z\in \mathbb{C}$   the I want to prove that $ g $ is constant.  Hint: Apply maximum modulus principle on $cl(B_r (0))=${$ |z| \leq r $}$ $  also   if $g$ has no zeros and  $cg(z)=g(\frac{z}{2}) g(\frac{z+1}{2})$  holds $\forall z\in \mathbb{C}$ and for some $c\in \mathbb{C}$ prove that $g(z)=ce^{\frac{-b}{2}}e^{bz}$ for some $b\in \mathbb{C}$   I have no idea how to prove this statement . Any Hints leading to solutions  are appreciable
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.