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 1. $f(z) = z ^{-\frac{1}{2}}$ can be expanded as a Laurent series convergent in the annulus $0 < |z| < \infty$. It has a simple pole at $z = 0$.
 2. A function $f(z)$ which is analytic on the annulus $r < |z| < R$ can be represented by a Laurent series centered at $0$, which converges for all $z$ in the annulus.
 3. A function $f(z)$ which is analytic on a domain $D$ can always be expanded in a power series which converges on the smallest disk containing the domain $D$.
 4. A function $f(z)$, analytic in a domain $D$ which contains the annulus $1 < |z| < 4$ can always be expanded in a convergent power series (with no negative exponents) centered at $2i$ and converging in the disk $|z - 2i| < 1$.
 5. If a Laurent series converges in an annulus $r < |z| < R$ then differentiating the Laurent series term by term produces a new Laurent series which converges on an annulus $\hat r < |\hat z| < \hat R$ but the new annulus might be smaller than the original annulus $($i.e $r < \hat r$ and $\hat R < R )$.