# Complex singularities of $f$ - Radius of convergence

A example in my textbook explain that $x^2-2x+2=0$ has the solutions $x = 1± i$. The distance between $x=0$ and $x = 1± i$ is $\sqrt{2}$ in the complex plane. So the radius of convergence of the Taylor series for $h(x)=(x^2-2x+2)^{-1}$ is $\sqrt{2}$ around $x=0$. The link Radius of convergence - Complex plane help me to understand the behavior of this argumentation.

According to this link, the problem is we can exploit the singularities of $h$. If a series is a Taylor series of some function $f$, i.e. $f(x)=∑^∞_{n=0}a_n(x−x_0)^n$ where $a_n=\frac{f^{(n)}(x_0)}{n!}$, then the radius of convergence is equal to the distance between $x_0$ and the singularity of f the is closest to $x_0$ in the complex plane, as long as the function $f$ is sufficiently "nice".

Does someone could explain to me rigoriously why this comment is true?

• as long as $f$ is holomorphic, and the proof is with the Cauchy integral formula. if $f$ in holomorphic in the region $|z-a| < r$ : $\int_{|z-a| = r} \frac{f(z)}{z-s} dz = -2 \pi \int_0^{2 \pi} \frac{f(r e^{i\theta})}{1-s e^{-i \theta} / r } d\theta = 2 i \pi f(s)$ (that you can also prove with the Fourier series) expanding $\frac{1}{1-s e^{-i \theta} / r } = \sum_{k=0}^\infty s^k e^{-i k \theta} / r^k$ and inverting $\sum$ and $\int$ gives an analytic series representation of $f(s)$ in the region $|s-a| < r$ – reuns Mar 11 '16 at 11:49
• and that analytic series is in fact the Taylor series at $a$ with a radius of convergence of at least $r$. (the converse is that if $f$ is analytic on $|z-a| < r$ then it has no singularity there) – reuns Mar 11 '16 at 11:51

If $f$ is a holomorphic function on an open set $U \subset \mathbb{C}$ and $z_0 \in U$, as well as $r \gt 0$ so that $D_r(z_0) \subset U$. Then $$f(z) = \sum_{n=0}^{\infty}{c_n(z-z_0)^n}\ \ \ \ \forall z \in D_r(z_0)$$
1.) $\forall n$ $$c_n = \frac{f^{(n)}(z_0)}{n!}$$ 2.) (Cauchy Integral Formula) $\forall n$ and $\forall s \gt 0$ with $\overline{D_s(z_0)} \subset U$ $$c_n = \frac{1}{2 \pi i} \int_{|z-z_0|=s} \frac{f(z)}{(z-z_0)^{n+1}} dz$$
Now the convergence radius $R$ of the power series is then the supremum over the radius $r$ of these discs. Which is, informally speaking, you extend the disc untiil you encounter the first "problem" with holomorphy. And this is the first singularity. Suppose there is a singularity $w_0$ of distance less than $R$ to $z_0$. What would happen is that the numerator in the Cauchy integral blew up and the coefficients would not be defined. So $R$ is the distance of $z_0$ to the closest singularity.
If $f$ is holomorphic ($\Bbb C$-differentiable) in $D(z_0,r)$, $0<\rho<r$ and $z\in D(z_0,\rho)$: $$f(z) = \frac1{2\pi i}\int_{|w-z_0|=\rho}\frac{f(w)}{w-z}\,dw.$$ But the integrand can be written as a series using $$\frac1{w-z} = \frac1{w-z_0 - (z-z_0)} = \frac1{w-z_0}\,\frac1{1 - (z-z_0)/(w-z_0)} = \frac1{w-z_0}\,\sum_{k=0}^\infty\left(\frac{z-z_0}{w-z_0}\right)^k,$$ where the geometric series converges for $|z-z_0|<|w-z_0| = \rho$. Using the uniform convergence: $$f(z) = \frac1{2\pi i}\int_{|w-z_0|=\rho}\left(\sum_{k=0}^\infty f(w){(z-z_0)^k\over(w-z_0)^{k+1}}\right)dw =$$ $$\sum_{k=0}^\infty\left(\frac1{2\pi i}\int_{|w-z_0|=\rho}\frac{f(w)}{(w-z_0)^{k+1}}\,dw\right)(z-z_0)^k.$$ Now, we have that $f$ is a power series (the Taylor series by unicity) at least in $D(z_0,r)$, i.e., the radius of convergence is at least $r$ and if $f$ has some singularity in the border, the radius of convergence is exactly $r$ because can't be greater.