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2

Just insert the identity $$ (\mathrm{i} t)^{-k}=\frac{1}{\Gamma(k)}\int_0^\infty \omega^{k-1}e^{-\mathrm{i}\omega t}d\omega\ , $$ in the RHS, with $k=-\alpha$. This gives $$ \frac{1}{2\pi\Gamma(-\alpha)} \int_0^\infty d\omega\ \omega^{-\alpha-1} \int_{-\infty}^{\infty} f(y)dy\int_{-\infty}^{\infty}dt\ e^{it(x-\omega-y)}\ , $$ and you can recognise the ...


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Your map might not be a map from the unit disc to itself. To prove your original statement, consider the map $w \mapsto \frac{z_0-w}{1-\bar{z_0}w}$ which is a biholomorphism on the disc sending the point $z_0$ to $0$.


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Consider $f(z) = \sin(1/(1-z^2)) \cdot \sin(1/(1+z^2)).$


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Consider the map $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $$f(x,y) := (u(x,y),v(x,y)) = (x,0).$$ This is differentiable since it's a polynomial. However it does not satisfy the C-R equations because for all $x,y \in \mathbb{R}$, $$\frac{du}{dx} = 1 \neq 0 = \frac{dv}{dy}.$$ It therefore is not analytic when considered as a map from the complex ...


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Hint: $|(1-w)/(1+w)|<1$ iff $w$ is in the open right half plane. This follows from basic properties of linear fractional transformations.


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If I have understood your ideas correctly: Setting $g(z)=z^2-z+2 = (z-\frac12)^2+\frac74$ we may compare with $f(z)=(z-1/2)^2$ for the contour $|z|=2$ where $|g(z)-f(z)| =\frac74< \frac94=(\frac32)^2\leq ||z|-\frac12|^2\leq |f(z)|$. For $|z|=1$ not so clear if you don't want to consider $f$ of more or less the same form as $g$.


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Well, you have $f(0+yi)=f(x+0i)=0$ for every $x$ and $y$. Thus the partial derivatives are both zero at the origin, and C-R is trivially satisfied. On the other hand, $$df_{0}=\lim_{|z| \to 0}\frac{\sqrt{|xy|}}{\sqrt{x^{2}+y^{2}}}$$ does not exist, because along the line $x=y$, the directional derivative (the limit along that line) is $\frac{|x|}{\sqrt{2}|x|}...


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Note that each one of the three balls is relatively open AND CLOSED in $K$, being the intersection of a closed ball of $\mathbb{C}$ with $K$. For instance, $B_1 = \overline{B_1} \cap K$, where the closure is meant in $\mathbb{C}$. So you have found a subset of $K$ (namely, $B_1$) which is relatively closed and open in $K$, hence $K$ is disconnected. Your ...


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user147263 Jun 4 '14 at 22:21 gave an answer. Let D be domain in w-plane defined by -\pi/4< arg w <5 \pi/4 and let G=D +(-1) and f conformal mapping of the unit disk onto G such that f(0)=0. Since G is non-convex / non-starlike (wrt 0) domain, f is not starlike.



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