<s>[INST] Radius of convergence of a complex function with a Taylor Series expansion 
The function
  $$f(z)=\frac{1}{1+i-\sqrt{2}z}$$
  has a Taylor series expansion around $z_0=0$. What is its radius of convergence?

So far I have computed the singularity point to be $z = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ So therefore We take the distance from $z_0$ to this singular point is $1$. Please can someone confirm this for me.
 [/INST] One also get at the radius of convergence like this:
Set
$\alpha = \dfrac{1 + i}{\sqrt{2}}; \tag{1}$
then
$\vert \alpha \vert = 1 \tag{2}$
and so
$\vert \alpha^{-1} \vert = 1 \tag{3}$
as well; we write
$f(z) = \dfrac{1}{1 + i - \sqrt{2}z} = \dfrac{1}{\sqrt{2}} \dfrac{1}{\alpha - z} = \dfrac{1}{\sqrt{2} \alpha} \dfrac{1}{1 - \alpha^{-1} z}; \tag{4}$
for $\vert z \vert < 1$ we have 
$\vert \alpha^{-1} z \vert = \vert \alpha^{-1} \vert \vert z \vert < 1; \tag{5}$
thus the series
$\dfrac{1}{1 - \alpha^{-1} z} = 1 + \alpha^{-1}z + \alpha^{-2}z^2 + \ldots = \sum_0^\infty (\alpha^{-1} z)^n \tag{6}$
converges.  Allowing $\vert z \vert = 1$, we may take $z = \alpha$ and not only is 
$\dfrac{1}{(1 - \alpha^{-1} z)} = \dfrac{1}{(1 - \alpha^{-1} \alpha)} = \dfrac{1}{1 - 1} \tag{7}$
undefined, but
$\sum_0^\infty (\alpha^{-1} z)^n = \sum_0^\infty (\alpha^{-1} \alpha)^n = \sum_0^\infty 1^n = \infty; \tag{8}$
the series clearly explodes for $z = \alpha$.  Since this series converges for $\vert z \vert < 1$ but not for $z = \alpha$, the radius of convergence is clearly $1$.  The same holds for the series of $f(z)$, by (4).
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