Verification of Laurent Series calculation I tried to calculate the Laurent series of these functions but I have no way to verify my answers. 
i)
$$
\begin{align}
 f(z)=\frac{e^{z^2 }-1}{z^4}, \mathbb{D}=\mathbb{C} \backslash \{0\}  
\end{align}
$$
We know that $ e^z=\sum_{n=0}^\infty(\frac{1}{n!})z^n $ , hence 
$$
 f(z)=\frac{e^{z^2 }-1}{z^4}=\frac{1}{z^4}(-1+e^{z^2})=\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})(z^2)^n)=\\
=\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})z^{2n})=\frac{1}{z^4}\sum_{n=1}^\infty(\frac{1}{n!})z^{2n}=\\
\sum_{n=1}^\infty(\frac{1}{n!})z^{2n-4}
$$
ii)
$$
\begin{align}
 f(z)=\frac{Logz}{(z-1)^2}, \mathbb{D}=\{ z\in \mathbb{C}:0<|z-1|<1\}  
\end{align}
$$
We know that $Logz=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n$, for $|z-1|<1$ ,hence
$$
 f(z)=\frac{Logz}{(z-1)^2}=\frac{1}{(z-1)^2}\left(- \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n \right)= \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^{n-2}
$$
iii)
$$
\begin{align}
 f(z)=z^6\cos^2{z^{-2}},z_0=0, \mathbb{D}=\mathbb{C} \backslash \{0\}  
\end{align}
$$
We know that $$\cos z=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}$$ therefore
$$\cos(\frac{1}{z})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n}$$ and 
$$\cos(\frac{1}{z^2})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z^2})^{2n}$$ 
We also know that when we a function to the power of 2, its correspondent series will be the derivative of the original series, so:
$$\cos^2(\frac{1}{z^2})=1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n}$$  (actually my result here was slightly diferrent, but wolframalfa returned this)
Finally, $$f(z)=z^6+ \left( 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n} \right)$$
iv)
$$f(z)=\frac{1-2z}{z^2-z},\mathbb{D}=\{ z\in \mathbb{C}:|z-1|>1\} $$
$$f(z)=\frac{1-2z}{z^2-z}=\frac{1-2z}{z(z-1)}=-\frac{1}{z}-\frac{1}{z-1}=-\frac{1}{z}+\frac{1}{1-z}==-\frac{1}{z}+\sum_{n=0}^\infty z^n$$
v)
$$f(z)=\frac{1}{1+z^2} , \mathbb{D}=\{ z\in \mathbb{C}:1<|z-2i|<3\}  $$
$$f(z)=\frac{1}{1+z^2}=\frac{1}{(i-z)(i+z)}=\frac{1}{(2i)}\frac{1}{(i-z)}+\frac{1}{(2i)}\frac{1}{(i+z)}$$
Let $z=2i+u$ so $$\frac{1}{(2i)}\frac{1}{(i-(2i+u))}+\frac{1}{(2i)}\frac{1}{(i+2i+u)}=\frac{1}{(2i)}\frac{1}{(-i-u)}+\frac{1}{(2i)}\frac{1}{(3i+z)}=\\
\frac{1}{(2i)}\frac{1}{u}\frac{-1}{(1-\frac{1}{iu})}+\frac{1}{(2i)}\frac{1}{3i}\frac{1}{(1-\frac{iu}{3})}=\\
\frac{1}{(2i)}\frac{1}{u}\sum_{n=0}^\infty(\frac{1}{iu})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{iu}{3})^n=\\
\frac{1}{(2i)}\frac{1}{(z-2i)}\sum_{n=0}^\infty(\frac{1}{i(z-2i)})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{i(z-2i)}{3})^n
$$
If anyone will take the time to verify my results and point my errors I will be grateful because I have no other means of checking their validity.
Sorry for the long post.
 A: You first computation is correct. The second is mostly correct, only at the end you dropped a minus sign. (It might have been better to absorb it into the sign in the sum and write $(-1)^{n-1}$ there rather than $(-1)^n$.)
In the third,

We also know that when we a function to the power of 2, its correspondent series will be the derivative of the original series

is wrong, consider for example polynomials to see that that generally isn't the case.
The square of $\sum\limits_{n=0}^\infty a_n w^n$ is the Cauchy product of the series with itself,
$$\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k a_{n-k}\right) w^n,$$
and $\sum\limits_{k=0}^n a_k a_{n-k}$ rarely is equal to $(n+1)a_{n+1}$. For the cosine, a trigonometric identity makes the computation of the power series for $\cos^2 w$ easy:
$$\cos (2w) = \cos^2 w - \sin^2 w = 2\cos^2 w - 1,$$
and hence $\cos^2 w = \frac{1}{2}(1+\cos (2w))$. Thus
$$\cos^2 \frac{1}{z^2} = \frac{1}{2}\left(1+\cos \frac{2}{z^2}\right) = \frac{1}{2} + \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{2^{2n}}{z^{4n}} = 1 + \sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}}{(2n)!z^{4n}}.$$
Multiplying that with $z^6$ yields the result.
In part iv), you need to expand the function in powers of $z-1$ since the domain is an annulus with centre $1$.
$$\frac{1-2z}{z^2-z} = \frac{1}{z-1}\left(\frac{1}{z} - 2\right) = (z-1)^{-1}\left(\frac{1}{(z-1)+1}-2\right).$$
Now expand $\frac{1}{(z-1)+1}$ into a geometric series, keeping in mind that $\lvert z-1\rvert > 1$ in the domain.
In the last, you made a sign error in your partial fraction decomposition,
$$\frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)} = \frac{1}{2i}\left(\frac{1}{z-i} - \frac{1}{z+i}\right).$$
From then on, you have once typed $z$ where it ought to have been $u$, but made no real mistake. So to obtain the correct Laurent series, you only need to adjust the signs to match the correct partial fraction decomposition.
