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**Spec** **1)** Find the Taylor series around the point 2i and specify the radius of convergence:

(a) $\displaystyle f(z)=\frac{1}{z}$

(b) $\displaystyle f(z)=log(z)$

(a): $\displaystyle f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $\displaystyle c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $\displaystyle z_0 = 2i$

So the Taylor series is: $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$

My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

**2) **Find the first three nonvanishing terms for the Maclaurin series of: $\displaystyle f(z)=\frac{(1-z)^{1/2}}{1+z^2}$

**3)** Determine the Laurent series for $\displaystyle \frac{1}{z+z^2}$, $\displaystyle |z+1/2|> 1/2$

I've solved this for different areas, but I'm having a little trouble with this one.