I need to expand the following functions to a Taylor series and find the radius, and I'm not sure how to do so: (already solved similar questions, but stuck with those.)

$f(z) = {\frac{z-1}{3-z}}$ near $z=1$ (I did solve $f(z)={\frac{1}{3-z}}$ near $z=2i$ already, but this is somewhat different)

$f(z) = {\frac{1}{3z+1}}$ near $z=-2$

when $z$ is the complex number.


For the first function you can write $$f(x) = \frac{z-1}{3-z} = \frac{z-1}{2 - (z-1)} = \frac{z-1}{2} \cdot \frac{1}{1 - \frac{z-1}{2}} = \frac{z-1}{2} \cdot \sum_{k=0}^\infty \left( \frac{z-1}{2} \right)^k = \sum_{k=0}^\infty \left( \frac{z-1}{2} \right)^{k+1}.$$

This is a standard trick and is not difficult to adapt to the second function.

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