I came across a theorem used to calculate residues of rational functions that states that if f and g are analytic functions at $z_k$ and $g'(z_k)$ isn't 0, then the residue of $f(z)/g(z)$ at $z_k$ is $f(z_k)/g'(z_k)$. I want to prove this using Laurent series (taking out the coefficient of $(z-z_k)^-1$) but I can't get around the algebra of actually dividing the two Laurent series. Can someone guide me through this?
123828
Let be $f,g$ functions analytic in $z_0$, with $z_0$ a zero of order one of $g$ and $f(z_{0})\neq 0$. Show that  $$ \operatorname{Res}\Bigl(\frac{f}{g},z_0\Bigr)=\frac{f(z_{0})}{g'(z_{0})} $$ My attemp... If $z_{0}$ is a zero of order one of $g$, then $g(z)=(z-z_{0})h(z)$, with $h(z)$ analytic and $h(z_{0})\neq 0$, then by analyticity of $g$ and $h$: $g'(z)=h(z)+(z-z_{0})h'(z)$. So,  $$ \frac{f(z_{0})}{g'(z_{0})}=\frac{f(z_{0})}{h(z_{0})+0}=\lim_{z\to z_{0}}{(z-z_{0})\frac{f(z)}{g(z)}} $$  But in the ultimate point I'm not sure
52164
Let the degree of the polynomials $P(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n$   $a_n\neq0$ and $Q(z)=b_0+b_1z+b_2z^2+\cdots+b_mz^m$ $b_m\neq 0$ be such that $m\geq n+2$. Show that   if all the zeros of $Q(z)$ are interior to a simple closed contour C,   then $$\int_C\frac{P(z)}{Q(z)} \, dz=0$$   I am lost on how to show it $P(z)$ and $Q(z)$ are polynomials, then the ratio is analytic in all the points except where $Q(z)=0$. If $\frac{P(z)}{Q(z)}$ be analytical in all the points on the contour or inside, so I could tell straight that the integral is zero, but that is not the case.
116826
Find the residue at $z=-2$ for  $$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$  I know that:  $$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$  Let $z \to -1 - z$ to get:   $$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$  therefore we divide by the other part to get:  $$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$  I have to somehow get the coefficient of $\frac{1}{z+2}$ because I want to evaluate the residue of   $g(z)$ at $z=-2$  The problem is I cant ever get a factor of $\frac{1}{z+2}$ what should I do?
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.
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.
81136
Let $f:\mathbb{C}\to\mathbb{C}$ be a holomorphic function such there exists a $c\in\mathbb C$ with $|f(z)|\leq c\cdot |z|^n$ for all $z\in\mathbb C$. Show that there exists an $a\in\mathbb C$ with $f(z)=a\cdot z^n$.   Proof.  My general idea is to show that the taylor series only has one non-zero coefficient:  $\displaystyle |a_k|\leq\frac{k!}{2\pi}\oint_{|w|=r}\frac{|f(w)|}{|w|^{k+1}}dw\leq \frac{k!}{2\pi}\oint_{|w|=r}\frac{c\cdot|w|^n}{|w|^{k+1}}dw\leq \\\displaystyle\leq \frac{c\cdot k!}{2\pi}\oint_{|w|=r}|w|^{n-k-1}dw$  Now using the standard estimate since $\oint=2\pi r$:  $\displaystyle \leq c\cdot k!\cdot r^{n-k}\longrightarrow\begin{cases}0,\ k&gt;n\\c\cdot n!,\ k=n\\ \infty,\ k&lt;n\end{cases}$ as $r\to\infty$  In conlusion the taylor series for now looks like $f(z)=\sum_{k=0}^{n-1}a_k z^k+ a_n z^n$ with $|a_n|\leq n!\cdot c$. What do I do with the remaining summands?  Edit: I would say that for $k&lt;n$ we have $|z|^k&gt;|z|^n$ for $z\in D_1(0)$. So since the estimate has to hold for all $z\in \mathbb C$ we can't have that either.
613
I'm doing a bit of self study, but I'm uncomfortable with a certain idea. I want to show that $f(z)$ is analytic if and only if $\overline{f(\bar{z})}$ is analytic, and by analytic I mean differentiable at each point. Here $f$ is a complex valued function.  What I do is write $f(z)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real functions of two variables. Then $\overline{f(\bar{z})}=u(x,-y)-iv(x,-y)$.  These two forms look very similar, in the sense the one function being differentiable should immediately imply the other is differentiable, since the only thing really changing might be a $-$ sign popping out due to the chain rule.   How can I more rigorously express that $f(z)$ is analytic iff $\overline{f(\bar{z})}$ using this? Many thanks.
90354
$$\frac{1}{z-\sin z}$$ $$\frac{1}{z(1-\frac{\sin z}{z})}$$  the laurent's transform is also used to simplify the function.  the residue is found a $\frac{2*36*2}{5}$ but the actual answer is given as $\frac{3}{10}$ the residue is given as The solution is incorrect and also the pole is of the order 3.
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.
69869
What is the general form of a rational function which has absolute value $1$ on the circle $|z|=1$? In particular, how are the zeros and poles related to each other?   So, write $R(z)=\dfrac{P(z)}{Q(z)}$, where $P,Q$ are polynomials in $z$. The condition specifies that $|R(z)|=1$ for all $z$ such that $|z|=1$. In other words, $|P(z)|=|Q(z)|$ for all $z$ such that $|z|=1$. What can we say about $P$ and $Q$?
123447
Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that   $$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$$  I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$  $$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$$ I put  $$g(z)=f(z)-\frac{c}{z-z_0}$$  Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $$g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$$ The power series of $g$ has radius of convergence $&gt;1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $$\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$  the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?
79633
I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
53428
Find the Laurent Series for $f(z) = \cot(\pi z)$ at $z= 0$.  I just need hints; I am not sure how to do this at all. I am very new to this.
93591
Working on a problem in Gamelin's book.  "Show that if $f(z)$ is a continuous function on a domain $D$ such that $f(z)^N$ is analytic on $D$ for some integer $N$, then $f(z)$ is analytic on $D$."  He starts a hint with the statement "Show that the zeros of $f(z)$ are isolated." This makes me think. I know that if $f$ is analytic and nonconstant, then the zeros are isolated. Is there some sort of converse to this theorem? I.e., for example, if continuous and zeros are isolated then analytic?  His next hint is: "At a zero of $f(z)$, write $f(z)^N=(z-z_0)^mh(z)$ where $h(z_0)\ne 0$ and show that $N$ divides $m$." However, the book has covered nothing on winding numbers thus far. Also, how will the fact that $N$ divides $m$ help to show that $f(z)$ is analytic?
108112
We begin by assuming that a non-constant complex polynomial $$p(z)=a_n z^n+a_{n-1}z^{n-1}+...+a_1 z+a_0,$$ where $a_0 \neq 0$, doesn't vanish in $\mathbb{C}$. Therefore $1/p$ is also an analytical function everywhere. Note that $(p(z)-a_0)/z$ is a polynomial and therefore, due to Cauchy's theorem, the following integral must be zero.  $$\int_{S_r^1}\frac{(p(z)-a_0)/z}{p(z)}dz=\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz=0.$$  On the other hand  $$\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz$$  $$=\int_{0}^{2\pi}\frac{a_n(re^{i\theta})^{n-1}+a_{n-1}(re^{i\theta})^{n-2}+...+a_1}{a_n(re^{\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}(ire^{i\theta})d\theta$$  $$=\int_{0}^{2\pi}i d\theta - \int_{0}^{2\pi}\frac{a_0 i}{a_n(re^{i\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}d\theta$$ $$=2\pi i -\int_{0}^{2\pi}\frac{a_0i}{p(re^{i\theta})}d\theta.$$  Let $r \rightarrow \infty$ and the integral approaches value $2\pi i$ - a contradiction. Therefore $1/p$ can't be entire and $p$ must vanish somewhere in the complex plane.  EDIT: I have changed this post and deleted the idea that the integral $\int_{\partial D_r}\frac{(p(z)-a_0)/z}{p(z)}dz=0$ could be evaluated with the fundamental theorem of calculus. Now, as it stands, this proof is essentially the same as the one that Mr. José Carlos Santos points out in his comment.
83471
I am trying to understand this proof of Rouché's theorem, but I am missing the logic of the last and most crucial step. Here are the assumptions:     Suppose that $ f $ and $ g $ are analytic inside and on a regular   closed curve $ \gamma $ and that $ |f(z)| \gt |g(z)| $ for all $ z &gt; \in \gamma $. Then $$ \mathcal{Z}(f + g) = \mathcal{Z}(f) \text{ inside &gt; } \gamma $$   The proof goes as follows:  First note that $ f \neq 0$ on $ \gamma $ since otherwise $ |g| \lt 0 $ which doesn't make any sense. So, we can write the following:  $$ \mathcal{Z}(f + g) = \frac{1}{2\pi i} \int_{\gamma}\frac{(f+g)'}{(f+g)} = \frac{1}{2\pi i} \int_{\gamma} \frac{(f(1 + \frac{g}{f}))'}{f(1 + \frac{g}{f})} = \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} + \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} $$  We know $ f $ is analytic, so $ \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f}  = \mathcal{Z}(f) $ by the argument principle. What I don't know is, why is $ \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} = 0$?
84192
Let $f$ be a rational function mapping to $\mathbb{C}$, $$f(z)=\frac{P(z)}{Q(z)}$$ with $\deg P\leq \deg Q -2$. I want to show that the sum over all residues is zero.  What am I asked to show? I think: The only singularities are in the zeros of $Q$. So if $N_Q$ is the set of zeros of $Q$ I have to show $$\sum_{q\in N_Q} Res(f,q)=0,$$haven't I? Since there are max. $\deg Q$ many zeros of $Q$, the sum is finite.  In order to show the statement - how can I use the $\deg$ condition?