To solve the integral $\frac{1}{2\pi i} \oint_{\lvert z \rvert = 2} \frac{2z}{z^2 - 1} \, dz$, we can use the residue theorem.

The integrand is $\frac{2z}{z^2 - 1}$.

First, we factorize the denominator: $z^2 - 1 = (z - 1)(z + 1)$

The integrand can be rewritten as: $\frac{2z}{(z - 1)(z + 1)}$

The singularities (poles) of the integrand are at $z = 1$ and $z = -1$. Both poles are inside the contour $|z| = 2$.

To apply the residue theorem, we need to find the residues at these poles.

1. Residue at $z = 1$:

$\text{Res}\left(\frac{2z}{(z-1)(z+1)}, z=1\right) = \lim_{z \to 1} (z - 1) \frac{2z}{(z - 1)(z + 1)} = \lim_{z \to 1} \frac{2z}{z + 1} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1$

2. Residue at $z = -1$:

$\text{Res}\left(\frac{2z}{(z-1)(z+1)}, z=-1\right) = \lim_{z \to -1} (z + 1) \frac{2z}{(z - 1)(z + 1)} = \lim_{z \to -1} \frac{2z}{z - 1} = \frac{2 \cdot (-1)}{-1 - 1} = \frac{-2}{-2} = 1$

By the residue theorem, the integral is $2\pi i$ times the sum of the residues inside the contour.

So the sum of the residues is $1 + 1 = 2$.

Therefore: $\oint_{\lvert z \rvert = 2} \frac{2z}{z^2 - 1} \, dz = 2\pi i \cdot 2 = 4\pi i$

Finally: $\frac{1}{2\pi i} \oint_{\lvert z \rvert = 2} \frac{2z}{z^2 - 1} \, dz = \frac{1}{2\pi i} \cdot 4\pi i = 2$

So the value of the integral is $2$.

Would you like more details or have any questions?

Here are 5 questions you might consider asking:

1. How does the residue theorem work?
2. Can you explain the process of finding residues?
3. What are the conditions for using the residue theorem?
4. How do you factorize polynomials in complex analysis?
5. Can you provide another example of a contour integral using the residue theorem?

Tip: When solving complex integrals, always check the location of singularities relative to the contour.