What Is a Complex Polynomial Differentiable Only on the Unit Circle?

jpb1980
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Find a (complex) polynomial function f of x and y that is differentiable at the origin, with
df/dz = 1 at the point z=0, and differentiable at all points on the unit circle x^2 + y^2=1, but is not differentiable at any other point in the complex plane. (Bruce Palka, Page 101)

I think we use the Cauchy Riemann relations. I am having a hard time with this one.



I have tried (1+x^2 +y^2) + (1 + x^4 + 2 x^2 y^2 + y^4)*i but that did not work.
 
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Remember that a real-differentiable function f on a subset of the plane is complex-differentiable exactly where \partial f/\partial\overline{z} = 0. This means that the set of differentiability you are given, the union of the origin and the unit circle, should be the zero set of \partial f/\partial\overline{z}. Try to construct a polynomial g with this zero set, and then see if you can make f with \partial f/\partial\overline{z} = g.
 
ystael said:
Remember that a real-differentiable function f on a subset of the plane is complex-differentiable exactly where \partial f/\partial\overline{z} = 0. This means that the set of differentiability you are given, the union of the origin and the unit circle, should be the zero set of \partial f/\partial\overline{z}. Try to construct a polynomial g with this zero set, and then see if you can make f with \partial f/\partial\overline{z} = g.

I am aware of that identity, but I do not see how that will help me here. What you said makes perfect sense, but it does not help me construct the polynomial. I do know that, with a0 + a1*z + a2*z^2 +..., we would certainly have a1 =1. But otherwise, I'm stumped.
 
To get an idea for what the degree and the shape should look like, try to construct a real polynomial in one variable with zeroes at -1, 0, 1 that you can use as an inspiration. Hint: the obvious degree is cubic, but that's likely the wrong inspiration.
 
Nevermind. I have figured out all my other problems. This one is not meant to be. I dare say that only a few mathematician in the world could solve this one. No one online knows, lol. Anyhow, about your hint, (x-1)(x+1)*x = x^3-x. But I have already tried x-x^3 = y+y^3. It does not work. Even if I did (x^2-1)(x^2+1)*x^2, It would be of no use.

But hey, if all else fails, I have stumped some really smart people, lol.
 
In any event, thanks for trying. Your time is appreciated. At least you had the courtesy and dignity to tackle this problem. BTW, I'm not in grad school yet. I'm trying to solve every problem out of Palka's Complex Analysis book. There is no grade involved here. I leave here in sorrow not because I will lost grade points, but because there is something I cannot know.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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