A First Course in Complex Analysis · Problem 36

The key is recognizing that $z^2 - 4 = (z-2)(z+2)$, so the denominator has a repeated factor at $z=2$—your center point. Partial fractions will reveal a pole of order 2 at $z=2$ and a simple pole at $z=-4$. For the Laurent series, you'll expand the $z+2$ term as a power series in $(z-2)$ to handle the singularity structure. Part (b) then becomes a residue calculation: identify the coefficient of $\frac{1}{z-2}$ in your Laurent series and apply the residue theorem.

The key is recognizing that $z^2 - 4 = (z-2)(z+2)$, so the denominator has a repeated factor at $z=2$—your center point. Partial fractions will reveal a pole of order 2 at $z=2$ and a simple pole at $z=-4$. For the Laurent series, you'll expand the $z+2$ term as a power series in $(z-2)$ to handle the singularity structure. Part (b) then becomes a residue calculation: identify the coefficient of $\frac{1}{z-2}$ in your Laurent series and apply the residue theorem.