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2

$f(z) = \dfrac{z^2}{z-z_p}$ is holomorphic (except at $z=z_p$). It follows from Cauchy-Riemann's equations that the real and imaginary parts of a holomorphic function are harmonic.


1

The function $f(z)=z^{3/2}$ is not single-valued in the entire plane. We can see this easily by noting that $z=re^{i\theta}=re^{i(\theta+2n\pi)}$. However, $z^{3/2}=(re^{i(\theta+2n\pi)})^{3/2}=(-1)^n(re^{i\theta})^{3/2}$. This means that for each point in the $z$-plane, there are two points mapped by $z^{3/2}$. Now, if one deletes ("cuts") from the ...


0

Hint: if the series converges at a point with $|z|=r_1$, its terms are bounded there. Use that to get an exponentially decreasing uniform bound on $|z| \le r$, where $r < r_1 < R$.


1

If $\phi: B \to \Bbb D$ is a biholomorphic map from $B$ to the unit disk, then $$g := \phi \circ f \circ \phi^{-1}$$ is a biholomorphic map from the unit disk to itself. But we know (via the Schwarz Lemma) how to parameterize all such maps, and given any such map $g$, we can recover the corresponding map $f$ by conjugating $g$ with $\phi^{-1}$.


0

We will use the Schwarz theorem for the second derivatives, which states that $$ \frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x} $$ for a function $f:\mathbb{R}^2\to\mathbb{R}$. Set the Cauchy-Riemann equations $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, ~~~~~(1) \\ \frac{\partial u}{\partial ...


1

The restriction of $P_n$ to $T$ converges uniformly, so $(P_n|_T)$ is uniformly Cauchy on $T$. By the maximum modulus principle, $(P_n)$ is also uniformly Cauchy on $\bar D$, and thus converges uniformly to some (continuous) function $F$ on $\bar D$. On the other hand, Morera's theorem shows that a uniform limit of holomorphic functions is holomorphic, so ...


0

Note that $A=|A|e^{i\alpha}$ and $B\in {\mathbb R}$ are given, and we want to find a $z=re^{i\phi}$ such that your equation holds. In terms of the new variables $(r,\phi)$ this means that $$r^2+|A|r\cos(\alpha+\phi)+B=0\tag{1}$$ should have at least one solution $(r,\phi)$ with $r\geq0$, $\phi\in{\mathbb R}$. The discriminant is ...


2

For (a), look at the Maclaurin series for $f$: $$ \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,z^k. $$ Use the given estimate on $|f^{(k)}(0)|$ together with (for example) the root test to show that the series converges for all $z$. This gives you an analytic continuation of $f$ to all of $\mathbb{C}$. (b) If $f$ is not identically zero, we can write $f(z) = ...


0

(1) : If $|z|^2+Re(Az) =-B$ then $$0 \leq |2z+A|^2$$ $$=4|z|^2 +4Re(Az)+A^2$$ $$=-4B+|A|^2.$$(2) : If $|A|^2-4B \geq 0 $, let $z=(\sqrt { |A|^2-4B} -A)/2$. Then $|z|^2+Re(Az)+B=0.$


0

I believe there might be mixing-up of some ideas. It's not the formula per se the one that needs a definition for the formula. It is the expression $e^x$. This expression, as mentioned by Steven, may be defined in various ways, which all represent the same thing. It is like the expression $e^x$ is some kind of ubiquitous expression, if you may see it that ...


5

The complex exponential function needs some definition. There are several ways one could define it: Power series As a solution to a differential equation As the unique holomorphic extension of the real exponential function to the whole plane Via the limit definition $\lim_{n \to \infty} (1+\frac{z}{n})^n$ Via Euler's formula ... One can take any one of ...


1

This function is bounded and entire, since is is both 1 and i periodic and continuous. It is therefore constant.


0

Continuing your answer: $$ {(\cos{(\alpha +\beta)})}^{2}\leq 1\\ 4B\leq |A^2|{(\cos{(\alpha+\beta)})}^{2}\leq |A^2| \\ |A^2|\geq 4B $$


1

For real $z$, say, $z := x + 0 i$, we have $|\cosh z|^2 = \cosh^2 x$, which is unbounded, but substituting this real quantity in your book's formula gives $\cos^2 x$, which is bounded, and so the book's formula is wrong. Your derivation gives the correct formula.


4

The identity you quote as being the textbook's answer - $|\cosh(x+i y)|^2 = \cos(x)^2 + \sinh(y)^2$ - is false in the case $y=0$ for all $x \not = 0$. Your identity and proof are correct. It is the case that $|\cosh(x+i y)|^2 = \sinh(x)^2 + \cos(y)^2$.


1

If $f:D \to D$ is holomorphic and $f(D) \subset D,$ then $|f''(0)| \le 2$ by Cauchy's estimates. We don't need $f(0)= f'(0)$ for this. We do need $f(0)= f'(0)$ for the estimate $|f(z)|\le |z|^2,$ but this is independent of first part; it follows from the proof idea in the Schwarz Lemma as @A.G. suggested in a comment.


0

$p$ is trivially increasing on $\mathbb{R}^+$, hence its real zeros are negative. However, $x\leq -1$ implies $p(x)\geq 11$, hence the real roots lie in $(-1,0)$. Since: $$ p(z)=(z^6+9z^4+1)+(z+1)(z^2-z+3)$$ there are no real roots at all. Since $1+1+2+4<9$, by Rouché's theorem there are exactly four roots of $p(z)$ inside the unit disk, that obviously ...


1

Concerning the first part: You want to prove that there is no real root to the polynomial $p(z)=z^6+9z^4+z^3+2z+4$. Now, the idea is to group terms to prove that $p(z)>0$ for all real $z$. First of all, you can get away that $z^3$ term using $z^6+z^3+\frac{1}{4} \ge 0$ which is equivalent to $(z^3+\frac{1}{2})^2 \ge 0$ which holds for all real $z$. ...


2

To show the point $a$ is a removable singularity of $f/g$ suffices to prove that $$\lim_{z\to a}(z-a) \dfrac{f}{g}(z) = 0.$$ In other words, given $\epsilon > 0$ there is $\delta > 0$ such that if $|z-a|<\delta$ then $$\left|(z-a)\dfrac{f(z)}{g(z)}\right| < \epsilon.$$ Constructing $\delta$ $$\left|(z-a)\dfrac{f(z)}{g(z)}\right| = ...


0

It is clear from the equation that $f(a)=0$ ... let in a considerably small ndb of $a$ $f(z) = (z-a)^n h_1(z)$ and $g(z)=(z-a)^m h_2(z)$... where $h_1,h_2$ are non vanishing holomorphic on that nbd... then $f/g (z) = (z-a)^{n-m} h(z)$ in the deleted nbd of a... now if $n-m <0$ then consider a small nbd inside the previous choosen nbd where $h$ is bounded( ...


0

Your inequality implies that $f/g$ is locally bounded. So indeed any singularity (except maybe the one at infinity) is removable. Concerning the second part of your question: Claim If $|h(z)|\le 1 + |z|$, then $h(z) = \lambda z + \mu$ with $|\lambda|<1$, $|\mu|<1$. Proof Define $k(z) = h(z) - h(0)$. Then it is easy to see that $|k(z)| \le ...


4

No, in fact, any discrete subset $I\subset\mathbb{C}$ must be countable. For any $i\in I$, discreteness of $I$ says that we can find an open set $U_i\subset \mathbb{C}$ such that $U_i\cap I=\{i\}$. Inside $U_i$, we can find an open disk $V_i$ containing $i$ such that the real and imaginary parts of the center of $V_i$ as well as the radius of $V_i$ are ...


1

If you are looking to simplify the angles, then you can use $\frac 13\arctan(\frac{11}{2})=\arctan \frac 12$ You can find this by using $$\tan3\theta=\frac{3t-t^3}{1-3t^2}=\frac{11}{2},$$ where $t=\tan\theta$ $$\Rightarrow2t^3-33t^2-6t+11=0$$ $$\Rightarrow(2t-1)(t^2-16t-11)=0$$ from which the acute angled solution for $\theta$ which is less than ...


1

$|w|=1\Rightarrow |az+b|=|cz+d|\Rightarrow|a||z+\frac ba|=|c||z+\frac dc|$ This is a straight line iff $|a|=|c|$, namely the perpendicular bisector of $-\frac ba$ and $-\frac dc$. When $|a|\neq|c|$, the locus of $z$ is a Circle of Apollonius.


1

The domain of $w=f(z)=z^2$ is not restricted to $(\frac{-\pi}{2},\frac{\pi}{2})$ to obtain an inverse. Rather, the interval $[0,\pi ]$ (or $[\pi ,2\pi ]$ is obtained as the image of a bijection from the slit $w$ plane. This map is then a branch of the inverse. The procedure is as follows: it's no harder to look at the more general case so let ...


2

Let $$\begin{align}p(z)&=a_nz^n+a_{n-1}z^{n-1}+\ldots +a_1z+a_0 \\&=a_nz^n\cdot\left(1+\frac{a_{n-1}}{a_nz}+\ldots + \frac{a_{1}}{a_nz^{n-1}}+\frac{a_{0}}{a_nz^{n}}\right)\end{align}$$ The expression in parentheses tends to $1$ as $|z|\to\infty$. Especially, there exists $r_0$ such that its absolute value is between ...


0

Suppose $p(z) = a_nz^n + \cdots.$ Then $|p(z)| \sim |a_n||z|^n$ as $z \to \infty.$ So given $\epsilon > 0,$ we will have $$(1-\epsilon)|a_n|R^n < |p(Re^{it})| < (1+\epsilon)|a_n|R^n$$ for large $R.$ Not sure if you are looking for something more precise.


0

By parity, $$ I = \int_{0}^{+\infty}\frac{x\sin(ax)}{1+x^4}\,dx = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{x\sin(ax)}{1+x^4}\,dx = \frac{1}{2}\text{Im}\int_{\mathbb{R}}\frac{z e^{aiz}}{1+z^4}\,dz $$ and the poles of $\frac{z}{1+z^4}$ with a positive imaginary part occur at $z=\frac{\pm 1+i}{\sqrt{2}}$, hence by setting $f(z)=\frac{z e^{aiz}}{1+z^4}$ and ...


1

It is easiest to show that this is path connected. Step 1: Prove that every point in the solution set is connected to a point where $|x|=1$. Namely, if $xy=1$, then consider multiples $x/\lambda$ and $\lambda y$, but varying $\lambda$ between $1$ and $|x|$, you can continuously change $xy=1$ into a solution where $|x|=1$. Step 2: Observe that when ...


5

The set $S=\{\,(x,y)\in\mathbb C:xy=1\,\}$ is even path-connected: Given $(x_1,y_1),(x_2,y_2)\in S$, we have $x_1,x_2\in\mathbb C^\times$, which is path connected. If $\gamma\colon [0,1]\to \mathbb C^\times$ is a curve with $\gamma(0)=x_1$, $\gamma(1)=x_2$, then $\tilde\gamma\colon[0,1]\to S$, $t\mapsto (\gamma(t),1/\gamma(t))$ is a path in $S$ from ...


0

We have $F=f\cdot\overline{f}$. Hence,$$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}\cdot\overline{f}+f\cdot\frac{\partial\overline{f}}{\partial x}.$$Since we have$$\frac{\partial\overline{f}}{\partial x}=\overline{\left(\frac{\partial f}{\partial x}\right)},$$we deduce$$\frac{\partial F}{\partial x}=2\mathrm{Re}\left(\frac{\partial ...


1

You are mixing up complex differentiability and real differentiability.


1

Just look at $f(x)=\sin(\frac{\pi}{x})$ and $g(x)=0$, they agree on all points $x_n=\frac{1}{n}$ but are not the same...


0

From the article Identity theorem: [G]iven functions f and g holomorphic on a connected open set D [...] Specifically, if two holomorphic functions f and g on a domain D agree on a set S which has an accumulation point c in D then f = g on all of D So the accumulation point must be part of the domain. For example you cannot apply your argument for two ...


0

Take $f$ defined on the real line by $f(x) = e^{-1/x}$ for $x>0$ and $f(x)=0$ if $x\leq 0$. The function $f$ is continuous, and is even infinitely differentiable, on the real line, but is not analytic. Indeed, you prove that $f^{(n)}(0)=0$ for all $n$, which shows that $f$ that cannot be analytic on any interval containt $0$ in its interior, as it would ...


7

Your second point is where it breaks. The antiderivative exists iif the integral from one point to another is independent from the path that is taken which is not guaranteed by continuity.


1

Note : See that here $u$ is NOT harmonic. So there does not exist any analytic function $f=u+iv$ , where , $u=\frac{x}{x^2-y^2}$.


0

Let the center of the circle be the origin $0$. Let $p$ be the complex number representing $P$. I will represent the vertex $A_i$ by $\displaystyle 2(e^{\frac{2\pi}{11}})^i = 2\omega^i$ for $i=1,2,3,...,11$. I will give an outline of the proof. $\displaystyle \sum_{i=1}^{11} |p - 2\omega^i|^2 = \sum_{i=1}^{11} (|p|^2 + |2\omega^i|^2 -4p\omega^i) = 11|p|^2 ...


0

Hints: You can represent points of the plane as complex numbers, say $p$ for $P$ and $a_j$ for $A_j$. Also, you are free to choose the origin (=zero) to be the center of the circle. Then we suddenly know $|a_j|=2$ and $|p|=3$ where absolute value of a complex number is distance from origin. For any complex number $z$, we have $|z|^2=z\bar z$ where $\bar z$ ...


0

It is fairly hard to visualize the symmetry, because as was written in the other two answers the objects in the Complex case are in general four-dimensional. The symmetry can be displayed however, by dropping one dimension. To augment the two answers given and help you visualize it, here is the case for the two maps: $\exp$, the principal branch of $\log$ ...


1

Call $A$ and $B$ the two open sets. Assume that a biholomorphism $\phi:A\to B$ exists and call $\psi:B\to A$ its inverse. $\phi$ and $\psi$ extend to holomorphic maps $\overline{\phi},\overline{\psi}:D\to D$ (why?) and $\overline{\phi}\circ\overline{\psi}=\overline{\psi}\circ\overline{\phi}=id$ (why?). So $\overline{\phi}:D\to D$ is an automorphism of the ...


0

Integrate first one $$\int 3-2y\ dy = 3y-y^2 + g(x) = v$$ $$\frac{dv}{dx}=g'(x)$$ compare it to $\frac{du}{dx}$ from above $$\frac{du}{dx}=g'(x)=-2x$$ $$g(x)=\int-2xdx=x^2 +c $$ $$v(x,y)=3y-y^2+x^2 +c$$ $$f(x+iy)=f(i)=2i \quad=> \quad x=0, y=1$$ $$f(i)=i(3-1+c)=2i\quad=>\quad c=0$$ Finally $$v(x,y)=3y-y^2+x^2$$ $$f(x,y)=x(3−2y)+i(3y-y^2+x^2)$$ As for ...


1

In the first place: Don't flip the plot, because it mixes up the names of the variables. When studying $f$ and $f^{-1}$ at the same time you see the graph of $f^{-1}$ in the original $f$-plot when you tilt your head $90^\circ$ sideways. For functions $f:\>{\mathbb C}\to{\mathbb C}$ and their inverses $f^{-1}$ there is an analogous mischievous symmetry; ...


0

$\newcommand{\Cpx}{\mathbf{C}}$If $g:\Cpx \to \Cpx$ is a function, you can reflect the graph $w = g(z)$ across the (complex) line $w = z$ to get $z = g(w)$, or $w = g^{-1}(z)$. (Of course, $g^{-1}$ is generally multiple-valued, particularly if $g$ is entire.) Oh right, there's that small obstacle of living in a three-dimensional universe, where the graph of ...


2

If not, $1 - f(z)$ never takes a value in $(-\infty, 0]$, and so there is an analytic branch $g$ of $\sqrt{1 - f(z)}$ with real parts always positive. What can you say about $\exp(-g)$?


1

Hint: We have $f(z) = n/z + h(z)$ in $D\setminus \{0\},$ with $h$ analytic in $D.$ We want this equal to $(z^na(z))'/z^na(z)$ for some $a$ analytic and nonzero in $D.$ See where this leads you.


9

Yes, $\cos(x)^x$ is not hard to "explore". I don't know if it's been the subject of any published work, other than OEIS sequence A215347. It satisfies, for example, the differential equation $$ y y'' = (y')^2 - (2 \tan(x) + x \sec^2(x)) y^2$$ and it has the Maclaurin series $$ y = 1 - \dfrac{x^3}{2} - \dfrac{x^5}{12} + \dfrac{x^6}{8} - \dfrac{x^7}{45} + ...


1

The straightforward argument works too. That the power series for $f$ and $g$ at a point $a$ both converges absolutely within at least the smallest of two radii of convergence, so their Cauchy product converges to $fg$.


5

Whatever proof you know for $(fg)' = f'g+fg'$ from elementary calculus will work nearly verbatim in this situation.


4

Here is an easy way to see it, using one of the definitions. A function is analytic if its derivative with respect $\overline w$ is zero. We have $$ \frac{\partial fg}{\partial \overline w}(w_0)=f(w_0)\frac{\partial g}{\partial \overline w}(w_0)+g(w_0)\frac{\partial f}{\partial \overline w}(w_0)=0, $$ because $f$ and $g$ are analytic at $w=w_0$. Following ...



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