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I was preparing for my exam of complex analysis and i have few questions from previous one that sounded tricky to me.

1)
  i) Is it possible for $f:\mathbb{C}\to\mathbb{C}$ to be differentiable only in one point?
  ii) Analytic only in one point?

My guessing is that first (i) one is true, but cant explain why. Perhaps need an example.

I highly doubt about second (ii) one because for function to be analytic you have to find a domain were function would be differentiable.

2)
Write down a function $f:\mathbb{C}\to\mathbb{C}$ that is differentiable in two points only.

I need an example too.

Next question kind'a beat me out.

3)
Make an example of analytic $f:\mathbb{C}\to\mathbb{C}$ function that maps circle $|z-1| = 1$ to circle $|z|=2$ and line $\text{Im } z=0$ to line $\text{Re } z=0$.

My guess is you have to find a fraction function for circle and then use it with composition of $e^{\varphi\theta}$ which rotates the coords. But that's only my guess.

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    I think you should have an equal sign (instead of a $<$) in "Make an example of analytic $f:\mathbb{C}\to\mathbb{C}$ function that maps circle $\left|z-1\right|<1$ to circle $\left|z\right|=2$." Note that non-constant analytic mappings are $\textit{open mappings}$.2011-06-15
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    maybe there was a mistake..2011-06-15
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    Hint: if a function is analytic at $z\in\mathbb{C}$, then there is a disk $U\ni z$ such that $f(w)=\Sigma_{n=0}^{\infty} a_nw^n$ ($a_n\in \mathbb{C}$ for all $n\geq 1$) for all $w\in U$. In particular, $f$ is also analytic in $U$.2011-06-15
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    I can be wrong, but for (3) I think $z\mapsto i(\frac{z}{2}-i)$ might do it. @Amitesh Could you tell me why this is incorrect (Re: open mappings)2011-06-15
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    Dear yayu, the disk $D=\{z\in\mathbb{C}:\left|z-1\right|<1\}$ is an open set in the complex plane. The open mapping theorem states (in particular) that if $f:\mathbb{C}\to\mathbb{C}$ is a non-constant analytic mapping, then it is an open mapping. Hence the image of the disk $D$ under any non-constant analytic mapping is an open set. However, $\{z\in\mathbb{C}:\left|z\right|=2\}$ is not an open set. Q.E.D.2011-06-15
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    @yayu: You're close but it's not quite right. Note that Amitesh's comment addressed an earlier version of the question. As it is now, the question makes sense.2011-06-15
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    @Theo what is the mistake?2011-06-16
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    @yayu: your map maps the circle $|z-1|=1$ to the circle $|z-(1 + \frac{i}{2})|=\frac{1}{2}$ (first you shrink by a factor $1/2$, then move by $-i$ then rotate by $\pi/2$ around zero). Actually, I misread the question (I mixed up the real and the imaginary axes). There is no such map as the OP asks for, as the imaginary axis is tangent to the circle $|z-1|=1$ while the real axis intersects the circle $|z| = 2$. What I had in mind was $z \mapsto 2i(z-1)$ and this maps the real axis to the imaginary axis and the circle of radius $1$ around $1$ to the circle of radius $2$ around $0$.2011-06-16

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