# every point on boundary of region of convergence is singular

I am given the following function:

$$f(z)=1+z^2+z^4+z^8+z^{16}+ \cdots$$

and shall show that it is holomorphic in the unit disc, that $f\to\infty$ as $z\to e^{2i\pi/2^n}$, and that every point on the circle $|z|=1$ is singular. I struggle with the last part. It seems intuitive, since we are summing up different points on the circle that seem not to have any structure, but how can I show this?

thank you very much,

-m.p.

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The identity $f(z^2)=f(z)-z^2$ might help. –  Jonas Meyer Nov 10 '11 at 0:26
For an easy idea of some of what is going wrong, check what happens at (1,0) and (-1,0) –  gary Nov 10 '11 at 0:37

Firstly, $f(z)\rightarrow\infty$ as $z\rightarrow 1^-$. So $z=1$ is a singularity.

Since $f(z)=z^2+f(z^2)$ we have that if $z^2\rightarrow 1^-$ then $f(z)\rightarrow\infty$. So $z$ such that $z^2=1$ are singularities.

We may write $f(z)=z^2+z^4+f(z^4)$ so if $z^4\rightarrow 1^-$ then $f(z)\rightarrow\infty$.

Continuing in this way we see that for any $n$, any $z$ such that $z^{2^n}=1$ is a singularity of $f(z)$. Such $z$ are dense in the unit circle.

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that helps. thank you very much. –  Marie. P. Nov 10 '11 at 1:34

The second part helps answer the last part. It implies (with a little tweaking) that every point on the circle is a limit point of points where $f$ blows up.

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thanks for this finishing. –  Marie. P. Nov 10 '11 at 1:33