Expansion of the function  \(\rm f\left( z \right) = \frac{1}{{z + 3}}\) about the point z =1 will be

\(\rm \frac{1}{3}\left[ {1 - \frac{3}{z-1} + {{\left( {\frac{3}{z-1}} \right)}^2} - {{\left( {\frac{3}{z-1}} \right)}^3} \cdot \cdot \cdot \cdot \cdot \cdot } \right]\)

\(\rm \frac{1}{4}\left[ {1 - \left( {\frac{{z - 1}}{4}} \right) + {{\left( {\frac{{z - 1}}{4}} \right)}^2} - {{\left( {\frac{{z - 1}}{4}} \right)}^3} \cdot \cdot \cdot \cdot \cdot \cdot } \right]\)

\(\rm \frac{1}{3}\left[ {1 - \frac{z-1}{3} + {{\left( {\frac{z-1}{3}} \right)}^2} - {{\left( {\frac{z-1}{3}} \right)}^4} \cdot \cdot \cdot \cdot \cdot \cdot } \right]\)

\(\rm \frac{1}{4}\left[ {1 - \frac{4}{z-1} + {{\left( {\frac{4}{z-1}} \right)}^2} - {{\left( {\frac{4}{z-1}} \right)}^3} \cdot \cdot \cdot \cdot \cdot \cdot } \right]\)