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3

Suppose you have a global holomorphic section $s$ of a line bundle $L$ on a compact, connected Riemann surface $M$. $s$, being nowhere zero, gives a holomorphic trivialization of the line bundle; i.e., $L \cong M\times \Bbb C$. Holomorphic sections of the trivial bundle are holomorphic functions, and the only holomorphic functions on such an $M$ are constant ...


1

Hint: compare the real part of $z_n$ to $|z_n|$.


1

Three points $z_1,z_2,z_3$ are vertices of an equilateral triangle $$\iff \frac{z_3-z_1}{z_2-z_1}=\cos(\pm 60^\circ)+i\sin(\pm 60^\circ)=\frac{1\pm\sqrt 3i}{2}$$ $$\iff 2z_3-z_1-z_2=\pm\sqrt 3i(z_2-z_1)$$ $$\iff (2z_3-z_1-z_2)^2=(\pm\sqrt 3i(z_2-z_1))^2$$ $$\iff z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1.$$


1

Your proof is correct. To make it more rigorous you should add the following: a. Specify the direction of the contour of your integrals. b. It does not hurt to explain the necessity that $a\in (-1,3)$. c. Most important, in order to define $z^a$ you need to pick the branch of the logarithm which is defined in: $$ \Omega=\mathbb C\smallsetminus\{it: t\le ...


0

There's no need for $a$ to be $0$. Your mistake is that your $C$'s don't really have to be constant: integrating a function of $x$ and $y$ with respect to $x$, say, the "constant of integration" can be a function of $y$. Hint: the real part of an analytic function is a harmonic function.


2

In several complex variables there is a standard term domain of holomorphy. Hardly anyone ever writes "domain of holomorphicity". However, outside of "domain of" combination, both holomorphicity and holomorphy are about equally used. My feeling is that people using complex analysis in relation to Banach spaces, Lie groups, etc are more likely to write ...


1

Let $w=z-1$ so $$\frac{2z+3}{1+z}=\frac{2w+5}{w+2}=2+\frac{1}{w+2}=2+\frac12.\frac{1}{1+\frac{w}{2}}=2+\frac12\sum_{n=0}^\infty(-1)^n\Big(\frac{w}{2}\Big)^n$$ and this series with convergence redius 2 expand as $$\frac52-\frac{w}{4}+\frac{w^2}{8}-\frac{w^3}{16}+\cdots$$


2

How can he just say that $\gamma$ is a union of two circles? That's how $\gamma$ is chosen. That choice is made because it is convenient, of course. One could take wobbly paths, but that would not be as nice for the expansion of $\frac{1}{w-z}$ into a geometric series. To apply Cauchy's integral formula, you need a contour in $\Omega$ that winds once ...


0

If $\tan(a+ib)=x+iy, \tan(a-ib)=x-iy$ $2x=\tan(a+ib)+\tan(a-ib)=\dfrac{\sin(a+ib+a-ib)}{\cos(a+ib)\cos(a-ib)}=\dfrac{\sin2a}{\cos^2a-\sin^2(ib)}$ $2iy=\tan(a+ib)-\tan(a-ib)=\dfrac{\sin[a+ib-(a-ib)]}{\cos(a+ib)\cos(a-ib)}=\dfrac{\sin2ib}{\cos^2a-\sin^2(ib)}$ Now, $\sin(ib)=\dfrac{e^{i(ib)}-e^{-i(ib)}}{2i}=-i\dfrac{e^{-b}-e^b}2=i\sinh(b)$


2

Use $\sin(a) \cos(b) = \frac{1}{2} \sin(b+a) - \frac{1}{2} \sin(b-a)$ and multiply your sum with $\sin\left(x/2\right)$. $$ \sum_{m=0}^{n} \sin\left(\frac{x}{2}\right) \cos(m x) = \frac{1}{2} \sum_{m=0}^n \left\{\sin \left(\left(m+1-\frac{1}{2}\right)x \right) - \sin\left( \left(m-\frac{1}{2}\right)x \right) \right\} $$ The sum telescopes, i.e. ...


0

$\textbf{Hint:}$Use De Moivre's formula to compute $z^{n+1}$. $\textbf{Edit:}$ The other way to compute this sum is writing $\cos x$ as: $$\displaystyle \cos x=\frac{e^{ix}+e^{-ix}}{2}$$ In my opinion it's the easiest way. You simply get two geometric series: $$\sum_{k=0}^{n}\cos kx=\sum_{k=0}^{n} ...


1

Hint: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}=-i\sinh ix$$ $$\cos x=\frac{e^{ix}+e^{-ix}}{2}=\cosh ix$$ Replace $x$ by $ix$ and obtain $\sin ix, \cos ix$ interms of $\sinh x$ and $\cosh x.$ $$\sin(x+iy)=?$$ $$\cos(x+iy)=?$$


1

My guess is that you mean you want to consider each contour as a set of points in $\mathbb C$ and take the union of these two sets. For example $$ \{e^{it} \; :\; t \in [0, 2 \pi]\} \cup \{1 + e^{it}\; : t \in [0, 2 \pi]\}$$


0

The condition implies that $$ {f(z) \over z^2} $$ is bounded, hence it has a removable singularity at $0$ and therefore is constant?


0

Consider $f(z)=a_1z+a_2z^2$ and $f(iz)=ia_1z-a_2z^2$. Using triangle inequality: $$|f(z)+f(iz)| \leq |f(z)|+|f(iz)|$$ So: $$|f(z)+f(iz)| \leq |f(z)|+|f(iz)| \leq k|z|^2+k|iz|^2=2k|z|^2$$ But: $$|f(z)+f(iz)|=|a_1||z||i+1|$$ So: $$|a_1||z||1+i| \leq 2k|z|^2$$ Now you see (for example consider small real $z$), that the inequality is satisfies if only if ...


0

Hint: near 0 what term of $f(z)=a_1z+a_2z^2$ is "in charge"? EDIT: another idea. $$|f(z)|\leq k|z|^2$$ implies (Riemann's theorem) that $$z\longmapsto\frac{a_1z+a_2z^2}{z^2}=\frac{a_1}z+a_2$$ has a removable singularity at $z=0$.


0

Note that provided $ad - bc \neq 0$, which is enough to guarantee it is bijective on the Riemann sphere, any linear fractional transformation $$z \mapsto \frac{az + b}{cz + d}$$ that maps $U$ onto $U$ must also map the boundary $\partial U = \mathbb{R} \cup \{\infty\}$ of $U$ (also in the Riemann sphere) onto itself. So, up to multiplication by a common ...


1

Your formula for the analytic self-map of the U.H.P. isn't correct, your formula is for self-maps of the unit disc. Note that if $a$ is not real, then either $a$ or $\bar{a}$ is in the U.H.P., and it is sent to $0$ or infinity, respectively; if $a$ is real, $f_3$ is just a rotation around $0$, which except for the trivial rotation isn't a self-map either.) ...


2

Note that the only singularities of $$ \frac1{1-z^2}=\frac12\left(\frac1{1-z}+\frac1{1+z}\right)\tag{1} $$ are at $z=1$ and $z=-1$. If $\gamma$ is any closed path that does not circle these singularities, then by Cauchy's Integral Theorem $$ \int_\gamma\frac1{1-z^2}\mathrm{d}z=0\tag{2} $$ No path in $G$ can circle either singularity. Since the difference of ...


1

Expanding the comment: $$\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right),$$ hence $$\int_\Gamma \frac{1}{(z-a)(z-b)}\,dz = \frac{1}{a-b}\left(\int_\Gamma\frac{dz}{z-a} - \int_\Gamma \frac{dz}{z-b}\right) = \frac{2\pi i}{a-b} \left(n(\Gamma,a) - n(\Gamma,b)\right).$$ Since $a$ and $b$ belong to the same component of the ...


0

The sum can be written as the integral: $$\int_0^{\infty} \frac{x}{e^x-1} dx $$ This integral can be evaluated using a rectangular contour from 0 to $\infty$ to $\infty + \pi i$ to $ 0$ .


1

$$ - \frac{\pi i}{2n} e^{\frac{\pi i}{2n}} \frac{1-e^{\pi i}}{1-e^{\frac{\pi i}{n}}}=-\frac{\pi i}{2n}e^{\frac{\pi i}{2n}}\frac2{1-e^{\frac{\pi i}n}}=-\frac{\pi i}n\frac1{e^{-\frac{\pi i}{2n}}-e^{\frac{\pi i}{2n}}}=-\frac{\pi}{2n}\frac{2i}{e^{-\frac{\pi i}{2n}}-e^{\frac{\pi}{2n}}}=$$ $$=\frac\pi{2n}\frac1{\sin\frac\pi{2n}}$$ Using the definition $$\sin ...


1

The usual sine and cosine are the solutions to the differential equation $f''(x)=-f(x)$ that satisfy $\sin(0)=0$, $\sin'(0)=1$ and $\cos(0)=1$, $\cos'(0)=0$. The hyperbolic functions are solutions to $f''(x)=f(x)$ with the same boundary conditions. These characterizations work with either $\mathbb R$ or $\mathbb C$ as the domain.


5

I think the nicest way to 'picture' this intuitively without having started your course is the 'amplitwist' concept coined by Tristan Needham (I'd recommend his book, Visual Complex Analysis, if this idea interests you). Essentially, to quote from his exposition: Analytic mappings are precisely those whose local effect is an amplitwist: all the ...


1

By assumption, $f$ maps the unit disk into the set $$A = \left\{w\in\mathbb{C} : \operatorname{Re} w > \lvert \operatorname{Im} w\rvert\right\} = \left\{ w \in \mathbb{C} : - \tfrac{\pi}{4} < \arg w < \tfrac{\pi}{4}\right\}.$$ Hence $g(z) = f(z)^2$ maps $\mathbb{D}$ into the right half-plane. And $g(0) = f(0)^2 = 1^2 = 1$. So $$h(z) = ...


1

Like N3buchadnezzar just said the residues are given by $$\mathrm{Res}\left(\frac{f(z)}{g(z)},z_k\right) = \frac{f(z_k)}{g'(z_k)}$$ In your case the algebra involved in the calculation may lead to many errors if you consider the residues as you listed them. I suggest you to write the singularities of $\frac{1}{z^4+1}$ as \begin{align*}z_1 &= ...


3

What really helped me to understand how special holomorphic functions are is that, if $U \subset \Bbb C$ is some region (open and connected, e. g. $U = \Bbb C$), then every holomorphic function $f\colon U \to \Bbb C$ is uniquely determined by its values on any subset of $U$ that contains a limit point. In particular, any holomorphic function $f \colon \Bbb ...


0

As $z$ tends to $i$, $z-i$ becomes as little as you please, so bound the other term. Let's say that $\quad|z-i|<\delta<1$ It implies that $\quad|1+z+2i|=|1+3i+z-i|\le|1+3i|+|z-i|<4+\delta<5$ So you have $$|(z - i)||(1 + (z+2i))| < 5\delta$$


1

Try to put $w=z-i$, so that $w \to 0$. Then $z^2=(w+i)^2=w^2+2iw-1$ and $(1+i)z= (1+i)(w+i)=w+i+iw-1$. Hence $$ z^2+(1+i)z+2=w^2+2iw-1+w+i+iw-1+2=w^2+(3i+1)w+i. $$ Now, pick $\epsilon>0$ and choose $\delta<\min \{\sqrt{\frac{\epsilon}{2}},\frac{1}{2}\frac{\epsilon}{|3i+1|}\}$ and deduce that $$ \left| z^2+(1+i)z+2 -i \right| = \left| w^2+(3i+1)w+i -i ...


2

Thank you user179549. Let $H$ denote the right plane minus y axis and define $g:H \to D(0,1)$ by $g(z) = \dfrac{z-1}{z+1}.$ It follows that $|g(f(z))| < 1, \forall z \in D(0,1), $ $g \circ f$ is holomorphic in $D(0,1)$ and $g(f(0)) =0.$ By Schwarz lemma, $|g'(f(0))||f'(0)|\leq 1,$ so $|f'(0)| \leq 2.$


1

Hint. Find a Möbius transformation $g$ that takes the right half-plane to the unit disk and $1$ to $0$. Then apply Schwarz's Lemma to $g \circ f$.


2

It is not entire: it has branch points at the zeros of $\cosh(9z/(2\pi))$.


2

As you mentioned, $(1-z)^i = \exp(i \log(1-z))$. However, $\log$ is a multi-valued function, and different branches of $\log$ will give you different branches of $(1-z)^i$. I'll assume we're using the principal branch of the logarithm, i.e. the one with imaginary part in $(-\pi, \pi]$. For any complex $z$, $\log(1-z)$ is thus in the horizontal strip with ...


0

Assume that $$ f(z)=\sum_{n=-\infty}^\infty a_nz^n. $$ If $a_{k}\ne 0$, for some $k>1$, then $$ a_{-k}=\frac{1}{2\pi i}\int_{\lvert z\rvert=r}z^{k-1}f(z)\,dz, $$ and hence $$ \lvert a_{-k}\rvert\le r^{k} \int_{0}^{2\pi}\lvert\, f(r\mathrm{e}^{i\vartheta})\rvert\,d\vartheta $$ and thus $$ \frac{\lvert a_{-k}\rvert}{r^k}\le \int_{0}^{2\pi}\lvert\, ...


0

Suppose $f(z)$ has a pole of order $n \ge 2$ at the origin. Then is it not the case that $g(z) = z^n f(z) \tag{1}$ for some $g(z) \in H(\Delta)$? It is indeed; see this widipedia entry. Furthermore, $g(0) \ne 0$, whence $\vert g(0) \vert > 0$. Thus there is a real $\gamma > 0$ with $\vert g(0) \vert > \gamma$, and since $g(z)$, hence $\vert ...


0

From the previous post we know that $$|\int_{|z|=1}fdz|\leq \int_{|z|=1}|f||dz|$$ The right hand is exactly the integral $$\int_{|z|=1}|f|\cdot |z|dz=\int_0^{2\pi} r|f(re^{i\theta}|d\theta$$, where $r=1$. So $\int_{|z|=1}zf(z)$ is finite. Assume the integral is zero, if not u can subtract the constant on both side. By Morera Thorem, $zf(z)$ is holmorphic on ...


0

Hint for (1) with the additional hypotesis of $D$ connected: the image set $$f(D)\subset\{\cdots,-2\pi,-\pi,0,\pi,2\pi,\cdots\}.$$ But $f(D)$ is connected (why?)...


15

For convenience let me set $k = 2n$. We can represent the binomial coefficient by $$ \binom{x}{2n} = \frac{1}{(2n)!}\prod_{j=0}^{2n-1} (x-j), $$ so to make this an even polynomial we will replace $x$ by $y+n-\tfrac{1}{2}$. This will make the roots of the equation $$ \binom{y+n-\tfrac{1}{2}}{2n} = c \tag{0} $$ symmetric about the real and imaginary axes ...


1

Do you know that $f$ can be continuously extended to a homeomorphism from the closed disk to the closed square? What would happen at the points being mapped to the vertices of the square if an analytic extension existed?


1

As in Hans Lundmark's answer, your reasoning for point 1 is correct. For point 2. you're on the right track: now gather up the real and imaginary parts to get: $$\begin{array}{lcl}|\sin z| &=& \frac{1}{2} |-2\,\sinh y\,\cos x + 2\,i\,\cosh y\,\sin x|\\&=&\sqrt{(\sinh y)^2\,(\cos x)^2 + (\cosh y)^2\,(\sin x)^2}\\&=&\sqrt{(\sinh ...


1

(1) is fine. For (2), use $e^{ix}=\cos x + i \sin x$ to separate real and imaginary parts. (3) follows from the expression in (2), since $\sinh y$ is unbounded.


1

If $\Gamma$ is the circle (positively oriented), and the $N$ zeros are counted by multiplicity, $$\int_0^{2\pi} R e^{i\theta} \dfrac{f'(R e^{i\theta})}{f(R e^{i\theta})} \; d\theta = \dfrac{1}{i} \oint_{\Gamma} \dfrac{f'(z)}{f(z)} dz = 2 \pi N$$ which says that the average value of $\text{Re}(z f'(z)/f(z))$ over the circle is $N$ (and the average value ...


0

I'm sure this proof could be tightened a bit, but it works. Consider every subset $S$ of $\mathbb{P}$, the prime numbers, and the corresponding set $$G(S)=\left\{\prod_{p\in S}p^{f(p)}:f:S\rightarrow\mathbb{N}>0\right\}$$ that is, the set of $n$ such that $S$ is the set of primes dividing it. We can show your the sum is now equal to ...


2

Let $a\in G$. We need to show that $h$ is differentiable at $a$. As $G$ is open there exists a disc $D(a,r)\subset G$. Now as $f$ does not vanish in $G$, and hence in $D(a,r)$, we define the function $$ F(z)=\int_{[a,z]}\frac{f'}{f}. $$ Then $F$ is holomorphic in $D(a,r)$ and the product $\exp(-F(z))f(z)$ is constant in $D(a,r)$, as $$ ...


2

Hint: try with branch of logarithmic function.


1

You are correct that this can also be done using the open mapping theorem. Here is a sketch. Since $f$ is not constant, its complex derivative is nonzero somewhere. In particular, viewing $f$ as a smooth map $\mathbb{R}^2 \to \mathbb{R}^2$, there is a point $p \in \mathbb{R}^2$ where $Df_p$ has full rank (=2). Note that $g : \mathbb{R}^2 \to \mathbb{R}$ ...


4

The given equality implies that $\operatorname{Im} f(z)$ is bounded. Apply Liouville's theorem to $\exp(i f(z))$ to see that it is constant. It follows that $f$ is also constant.


2

A sequence of functions $\{g_N\}$ is increasing if for every $x$, $\{g_N(x)\}$ is an increasing sequence of numbers. Note that this is completely different from saying that the functions $g_N$ are increasing functions; neither of two properties implies the other. One property describes what happens when $N$ changes, the other describes what happens when $x$ ...


0

Most of the time, you can make a similar statement when $f$ is holomorphic while $g$ is not, but you have to modify it a bit to rule out silly cases like $f(z)=z^2$ and $g(z)=\frac1z$ when $z\neq0$ and $g(z)=0$. In this case, $h(z)=z$ is holomorphic, but in reasonable cases, $h$ will not be holomorphic. Specifically, When $f$ is holomorphic and $g$ is not ...


5

Yes, consider $\bar z \cdot \dfrac 1{\bar z} = 1$. However, if $f$ is holomorphic and $h$ is holomorphic, then $g=h/f$ is holomorphic wherever $f\ne 0$.



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