How does
$$\frac{1}{b^2i}\oint\frac{1}{z}\frac{dz}{\lbrack k + \frac{1}{2i}\left( z - \frac{1}{z}\right) \rbrack^2}$$
become
$$\frac{1}{b^2i}\oint\frac{1}{z}\frac{dz}{\lbrack \frac{1}{2iz} \left( z^2 + 2kiz - 1\right) \rbrack^2}$$
Is the book wrong or am I missing something?
Notice that $$k+\frac{1}{2i}(z-\frac{1}{z})=\frac{1}{2i}(2ki+z-\frac{1}{z})=\frac{1}{2i}(\frac{2kiz}{z}+\frac{z^2}{z}-\frac{1}{z})=\frac{1}{2iz}(z^2+2kzi-1)$$