Given the series $\displaystyle \sum _{n=-\infty}^{+ \infty} \frac{z^n}{10^{|n|}}$, calculate its region of convergence and the function it represents.

Attempt: $\displaystyle \sum _{n=-\infty}^{\infty} \frac{z^n}{10^{|n|}}=2 \sum _{n=0}^{+ \infty} \frac{z^n}{10^n}-1=2\left (\frac{1}{1-\frac{z}{10}} \right ) -1$ for $\displaystyle |z|<10$. Which is equal to $\displaystyle \frac{10+z}{10-z}=f(z)$. The radius of convergence is 10, so the region is a disk (without the border) centered at 0 with radius 10. Is that good?

2)Is $\displaystyle \sum _{n=-\infty}^{\infty} z^n$ a Laurent series of a function? Why?

Attempt: No, it isn't. Because $\displaystyle \sum _{n=-\infty}^{\infty} z^n=\sum _{n=0}^{+\infty} z^n+ \left ( \sum _{n=-\infty}^{0} z^n \right ) -1$ which would be equal to $\displaystyle \frac{1}{1-z}+\frac{z}{z-1}-1$. The problem is that $\displaystyle \sum _{n=0}^{+\infty} z^n=\frac{1}{1-z}$ for $\displaystyle |z|<1$ and $\displaystyle \sum _{n=-\infty}^{0} z^n = \frac{z}{z-1}$ for $\displaystyle |z|>1$. Since the modulus cannot be greater and lesser than 1 at a time, the series doesn't converge to any function. (Yeah I know, I didn't prove it well. How would you solve the exercise?)