Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

$ f(z) = \frac{z^2 - 2z + 4}{z^2 - 1} $

For poles z$^2$ - 1 = 0; (z-1)(z+1) = 0

Therefore, z = 1,-1 are the two poles of which z = 1 is the only pole which lies inside.

Residue of f(z) at z = 1 is:

$ \lim_{z \to 1} (z-1) \frac{z^2 - 2z + 4}{(z-1)(z+1)} = \frac{1-2+4}{1+1} = \frac{3}{2} \\ \therefore \int_c \frac{z^2 - 2z + 4}{z^2 - 1} \,\, dz = 2 \pi i \frac{3}{2} = 3 \pi i $