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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.

  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.

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.

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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dot = point...? – Aryabhata Jun 15 '11 at 22:19
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 $\textit{open mappings}$. – Amitesh Datta Jun 15 '11 at 22:27
@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. – t.b. Jun 15 '11 at 23:30
@Theo what is the mistake? – kuch nahi Jun 16 '11 at 1:24
@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 $0$. – t.b. Jun 16 '11 at 1:42

For 1i) Consider the function $f(z) = |z|^2$. It's easy to see that it's not differentiable at any point other than possibly $(0,0)$ by, say, checking the Cauchy-Riemann equations. A (not so hard) direct calculation shows $f$ really is differentiable at $(0,0)$.

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Would the downvoter care to leave a comment? – Jason DeVito Jun 15 '11 at 23:55
$@$Jason: I upvoted your answer. But question 1 is clearly about subtle distinctions between "differentiable" and "analytic". Maybe an even better answer would explicitly ask the OP which definitions were given in his course! Some people use the terms "complex differentiable" and "analytic" synonymously, but I'm guessing (as you are) that for the latter the function is supposed to be differentiable in an open neighborhood of every point of the domain. Maybe the downvoter has a different definition.... – Pete L. Clark Jul 16 '11 at 2:06

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