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4

Hint: Let $g(z) = \frac{1/4 -z}{1-(1/4)z}.$ I assume you know this map is a holomorphic bijection of $D$ onto $D.$ Note $g(0)= 1/4$ and $g(1/4)=0.$ Apply the Schwarz Lemma to $g\circ f \circ g.$


3

Multiply both sides by $e^{-\alpha z}$ and, with a little magic, you can make it happen. I won't spoil things by saying any more.


3

Some advice about studying: do not forget to be as analytical as possible. Why is the theorem this way? Which hypotheses are necessary? What are examples? Counterexamples? Can you reformulate the theorem? Come up with a different proof? Why is this a theorem? What depends on it? What does it follow from? Can you come up with interesting consequences? Find ...


3

There is a mistake in the real part. $$ \frac{q^n - 1}{q - 1} = \frac{e^{in\phi} - 1}{e^{i\phi} - 1} = \frac{e^{i(n-1/2)\phi} - e^{-i\phi/2}}{e^{i\phi/2} - e^{-i\phi/2}} = \frac{- i e^{i(n-1/2)\phi} + i e^{-i\phi/2}} {2\sin{\phi/2}} $$ the real part is $$ \frac{\sin ((n-1/2)\phi) + \sin(\phi/2)} {2\sin{\phi/2}} $$ yielding the right result. However, ...


2

I think the key information is in the sentence Every component of $\mathbb{S}\setminus K_n$ contains a component of $\mathbb{S}\setminus \Omega$. (Geometrically/topologically, this is just saying that "$K_n$ does not have any unnecessary holes".) This way, whenever one needs to prescribe a pole $\alpha$ in a component of $\mathbb{S}\setminus K_{n-1}$ ...


2

If $z\in\mathbb C$, let $x\equiv\operatorname{Re}z$ and $y\equiv\operatorname{Im}z$, so that $z=x+iy$. Then, using Euler's formula, $$\exp(x+iy)=\exp(x)\exp(iy)=\exp(x)[\cos y+i\sin y]=-1+i.$$ Separating the real and imaginary parts, one must have \begin{align*} \exp(x)\cos(y)=&\,-1,\\ \exp(x)\sin(y)=&\,\phantom{-1}1. \end{align*} In particular, ...


2

The rectangle encloses the pole at $z=i$. Hence, the integral is $$2 \pi i \left( \text{Residue of $f(z)$ at }z=i\right)$$ The residue of $f(z)$ at $i$ is given by $\lim_{z \to i} (z-i)f(z)$. We have $f(z) = \dfrac{z^2+i}{z^2+1} \sin(z)$. Hence, $$\lim_{z \to i} (z-i) \cdot \dfrac{z^2+i}{z^2+1} \sin(z) = \lim_{z \to i} \dfrac{z^2+i}{z+i} \sin(z) = ...


2

use the unit circle as the contour, so that we have $z=e^{i\theta}$ with $d\theta = \frac{dz}{iz}$ since $\cos n\theta = \Re z^n$ we require $\Re I$ where: $$ I = \int_{|z|=1} \frac{z^n}{1-a(z+\frac1{z})+a^2}\frac{dz}{iz} \\ $$ $$ = -\frac1{ia}\int_{|z|=1} \frac{z^n}{z^2-(a+\frac1{a})z+1}dz $$ $$ = -\frac1{ia}\int_{|z|=1} \frac{z^n}{(z-a)(z-\frac1{a})}dz ...


1

Let's let $z=e^{i\theta}$ so that $dz=i\,e^{i\theta}d\theta$. This complex plane contour $C$ is the unit circle. Thus, $$\begin{align} \int_{-\pi}^{\pi}\frac{\cos (n\theta)}{1-2a\cos \theta+a^2}d\theta&=\frac12\,\oint_C \frac{z^n+z^{-n}}{1-a(z+z^{-1})+a^2}\frac{dz}{iz}\\\\ &=\frac{i}{2a}\,\oint_C \frac{z^n+z^{-n}}{(z-a)(z-1/a)}dz\\\\ ...


1

Look at the integral $$ I_n = \int_{-\pi}^{\pi} \frac{e^{in\theta}}{(1-ae^{i\theta})(1-ae^{-i\theta})} \, d\theta. $$ It is easy to see the denominator expands to $1-2a\cos{\theta}+a^2$. Since the interval of integration is symmetric, we can add the integral with $\theta \mapsto -\theta$ to $I$, and we find the integrand is $$ ...


1

Induction is fairly straightforward. Let $P_k$ be the statement $$\dfrac1{(1-z)^{k+1}} = \sum_{n \geq 0} \dbinom{n+k}k z^n$$ Consider the base case. Recall that the geometric series $$\dfrac1{1-z} = \sum_{n \geq 0}z^n$$ converges for $\vert z \vert < 1$. This validates the inductive step for $k=0$. Assume that induction is true for $k=m$, i.e., we have ...


1

You can this via Taylor's theorem with the integral form of the remainder; what this actually amounts to is a lot of integration by parts: $$ \frac{1}{(1-z)^{k+1}} - \sum_{n=0}^{m} \binom{n+k}{n} = \int_0^z \frac{(z-t)^{m}}{m!} \frac{1}{(1-t)^{(k+1)+(m+1)}} \frac{(k+m+1)!}{k!} \, dt, $$ which can be shown by induction on $m$. Of course, the factorials on the ...


1

$$\lim\limits_{(x,y)\to (0,0)}\frac{x+iy}{x-iy} $$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+}\frac{r\cos\phi+ir\sin\phi}{r\cos\phi-ir\sin\phi} $$ $$=\lim\limits_{r\to 0^+}\frac{\cos\phi+i\sin\phi}{\cos\phi-i\sin\phi} $$ $$=\lim\limits_{r\to 0^+}\frac{e^{i\phi}}{e^{-i\phi}}= e^{2i\phi} $$ This limit is clearly dependent on $\phi$. Therefore, ...


1

We know $f(0)=0.$ Unless $f\equiv 0,$ there will be a first power series coefficient that is nonzero, say $a_N.$ Then we can write $f(z) = z^Ng(z),$ where $g$ is entire and $g(0) = a_N.$ If $N=1, $ then $|f(1/n)|$ is on the order of $1/n$ as $n\to \infty,$ which contradicts the hypothesis. Thus $N\ge 2,$ giving the conclusion.


1

Consider the holomorphic function $g(z) = e^{-\alpha z} f(z); \tag{1}$ we have $g'(z) = -\alpha e^{-\alpha z} f(z) + e^{-\alpha z} f'(z)$ $= -\alpha e^{-\alpha z} f(z) + e^{-\alpha z} \alpha f(z) = 0; \tag{2}$ thus $g(z) = c, \;\; \text{a constant}; \tag{3}$ and thus from (1), $f(z) = ce^{\alpha z}. \tag{4}$ QED!!! Note Added Saturday 25 April 2015 ...


1

By Leibniz's rule, $$\dfrac{\partial}{\partial z} (g \overline{g}) = \dfrac{\partial g}{\partial z} \overline{g} + g \dfrac{\partial \overline{g}}{\partial z} = g' \overline{g} + 0 = g' \overline{g}$$


1

You can treat the square as a function. Lets say $f(x)=x^2$. Then what you have is: $$\lim_{z\to\infty} f(\frac{az+b}{cz+d})$$ As f is continuous at $\frac{a}{c}$ and $$\lim_{z\to\infty} \frac{az+b}{cz+d}\iff \lim_{z\to\infty} \frac{\frac{az}z+\frac{b}z}{\frac{cz}z+\frac{d}z}=\lim_{z\to\infty} \frac{a+0}{c+0}=\frac ac$$ as you wrote, we have: $$ ...


1

$\bf{My\; Solution::}$ Let $z=x+iy\;,$ Then $|x+iy| = 1\Rightarrow x^2+y^2 =1 $ Now We have To Maximize $$f(z) = \left|z^3-z+2\right| = \left|(x+iy)^3-(x+iy)+2\right|$$ We Get $$f(x,y) = \left|x^3-iy^3+3ix^2y-3xy^2-x-iy+2\right|$$ $$f(x,y)=\left|x(x^2-3y^2)+iy(3x^2-y^2)-x-iy+2\right|\;,$$ Using $x^2+y^2 = 1$ $$f(x,y) = ...


1

i am going to parametrize the unit circle by $z = \cos t + i \sin t.$ we have $$\begin{align}|z^3 - z + 2|^2 &= (\cos 3t - \cos t + 2)^2 +(\sin 3t - \sin t)^2 \\&=\cos^2 3t + \cos^2t+4-2\cos 3t \cos t+4\cos 3t-4\cos t \\ &+\sin^2 3t + \sin^2 t-2\sin 3t \sin t\\ &=6-2\cos 2t+4\cos 3t-4 \cos t\end{align}$$ the critical points of $ |z^3 - z + ...


1

You have $x = \Re(z)$ and $y = \Im(z)$ for all $z\in\mathbb{C}$. These are defined on all of $\mathbb{C}$. Your function is a polynomial in these, so the composition is defined everywhere in $\mathbb{C}$.


1

The fact is indeed a consequence of the CR equations. Note that $$\frac{\partial f}{\partial \bar z}=\frac12 \frac{\partial (u+iv)}{\partial x}+\frac12 \frac{\partial (iu-v)}{\partial y}=\frac12 \left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\,\frac12 \left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)=0$$ ...


1

Let $z(t)=x(t)+iy(t) \neq 0 \ \forall t\ $, then $$ \frac{d}{dt}|z(t)| = \frac{d}{dt} \sqrt{x^2(t)+y^2(t)} = \frac{1}{2} \frac{2[x(t)x'(t)+y(t)y'(t)]}{\sqrt{x^2(t)+y^2(t)}} = \frac{x(t)x'(t)+y(t)y'(t)}{|z(t)|} \ \ \ \ \ \ (1) $$ On the other hand, $$ \frac{z'(t)}{z(t)} = \frac{x'(t)+iy'(t)}{x(t)+iy(t)}=\frac{x(t)x'(t)+y(t)y'(t)+ i [x(t)y´(t)-y(t)x'(t)]} ...


1

$$\frac{d}{dt}|z(t)| = \frac{d}{dt}\sqrt{z\bar{z}} = \frac{1}{2\sqrt{z\bar{z}}}\left(\bar{z}\frac{dz}{dt} + z\frac{d\bar{z}}{dt} \right) = \frac{1}{|z|} \Re \left(\overline{z}\frac{dz}{dt}\right) = |z|\Re \left(z^{-1}\frac{dz}{dt}\right)$$ because $\frac{\bar{z}}{|z|} = |z|z^{-1}$


1

$$e^{x+iy}=-1+i$$ $$e^x\left(\cos(y)+i \sin(y)\right)=-1+i$$ $$\left\{ \begin{array}{l l} e^x\cos(y)=-1 \\ e^x\sin(y)=1 \end{array} \right. $$ $$\left\{ \begin{array}{l l} e^{2x}=2 \\ \tan(y)=-1 \end{array} \right. $$ with condition $\cos(y)<0$ and $\sin(y)>0$ $$\left\{ \begin{array}{l l} x=\frac{1}{2}\ln(2) ...


1

What you are trying to prove is not correct. Let $n=1$, $\varphi = \frac{\pi}{3} $.



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