From the theory of Laurent expansions, it is known that there are constants $a_n$ such that, for 1 < |z| < 4, $\frac{1}{z^2 - 5z + 4} = \sum_{n = - \infty}^{n = \infty} a_nz$
Find $a_{10}$ and $a_{−10}$ by the method of your choice.
So I got $\frac{1}{z^2 - 5z + 4} = \frac{1}{(z - 1)(z - 4)} = \frac{1}{-3(z-1)} + \frac{1}{3(z-4)}$. Then $\frac{1}{-3(z-1)} = \frac{-1}{3z}(\frac{1}{1 - \frac{1}{z}}) = \frac{-1}{3z}\sum_{n = 0}^\infty (\frac{1}{z})^n$ and $\frac{1}{3(z-4)} = \frac{-1}{12}(\frac{1}{1 -\frac{z}{4}}) = \frac{-1}{12}\sum_{n = 0}^\infty (\frac{z}{4})^n$. But I couldn't get $\frac{-1}{3z}\sum_{n = 0}^\infty (\frac{1}{z})^n + \frac{-1}{12}\sum_{n = 0}^\infty (\frac{z}{4})^n$ to be of the form $\sum_{n = - \infty}^{n = \infty} a_nz$.
My next Idea was since $\frac{1}{(z - 1)(z - 4)} = (-3\frac{1}{-3(z-1)})(3\frac{1}{3(z-4)})$ was to play with $(-3\frac{-1}{3z}\sum_{n = 0}^\infty (\frac{1}{z})^n)(3\frac{-1}{12}\sum_{n = 0}^\infty (\frac{z}{4})^n) = \sum_{n = 0}^\infty \frac{-1}{4^{n + 1}z}$ so now I have something of the form $\sum_{n = - \infty}^{n = \infty} a_nz^{-1}$ instead of $\sum_{n = - \infty}^{n = \infty} a_nz$.
Last idea was to play with $a_n = \frac{1}{2\pi i} \int_r \frac{f(z)}{z^{n + 1}}dz = \frac{1}{2\pi i} \int_{1 < |z| < 4} \frac{1}{z^{n + 1}(z -1)(z - 4)}dz$. Here the only singularity in $1 < |z| < 4$ is $z = 0$ and its of order $n +1$ but I feel like there should be a more simple way to do the problem then working with a pole of order $11$ for $a_{10}$ and I don't know what to do for $a_{-10}$ since there is no pole.
Your first idea is correct. You got that, when $1<|z|<4$,\begin{align}\frac1{z^2-5z+4}&=-\frac1{3z}\sum_{n=0}^\infty\left(\frac1z\right)^n-\frac1{12}\sum_{n=0}^\infty\left(\frac z4\right)^n\\&=-\frac13\sum_{n=-\infty}^{-1}z^n-\frac1{12}\sum_{n=0}^\infty\frac1{4^n}z^n.\end{align}So,$$a_n=\begin{cases}-\frac13&\text{ if }n<0\\-\frac1{12\times4^n}&\text{ if }n\geqslant0.\end{cases}$$