If f(z) has a pole of order n at z = a, then Residue of function f(z) at a is

A. $${\text{Res f}}\left( {\text{a}} \right) = \frac{1}{{\left( {\text{n}} \right)!}}{\left\{ {\frac{{{{\text{d}}^{{\text{n}} - 1}}}}{{{\text{d}}{{\text{z}}^{{\text{n}} - 1}}}}\left( {{{\left( {{\text{z}} - {\text{a}}} \right)}^{{\text{n}} - 1}}{\text{f}}\left( {\text{z}} \right)} \right)} \right\}_{{\text{z}} = {\text{a}}}}$$

B. $${\text{Res f}}\left( {\text{a}} \right) = \frac{1}{{\left( {{\text{n}} - 1} \right)!}}{\left\{ {\frac{{{{\text{d}}^{{\text{n}} - 1}}}}{{{\text{d}}{{\text{z}}^{{\text{n}} - 1}}}}\left( {{{\left( {{\text{z}} - {\text{a}}} \right)}^{{\text{n}} - 1}}{\text{f}}\left( {\text{z}} \right)} \right)} \right\}_{{\text{z}} = {\text{a}}}}$$

C. $${\text{Res f}}\left( {\text{a}} \right) = \frac{1}{{\left( {\text{n}} \right)!}}{\left\{ {\frac{{{{\text{d}}^{{\text{n}} - 1}}}}{{{\text{d}}{{\text{z}}^{{\text{n}} - 1}}}}\left( {{{\left( {{\text{z}} - {\text{a}}} \right)}^{\text{n}}}{\text{f}}\left( {\text{z}} \right)} \right)} \right\}_{{\text{z}} = {\text{a}}}}$$

D. $${\text{Res f}}\left( {\text{a}} \right) = \frac{1}{{\left( {{\text{n}} - 1} \right)!}}{\left\{ {\frac{{{{\text{d}}^{{\text{n}} - 1}}}}{{{\text{d}}{{\text{z}}^{{\text{n}} - 1}}}}\left( {{{\left( {{\text{z}} - {\text{a}}} \right)}^{\text{n}}}{\text{f}}\left( {\text{z}} \right)} \right)} \right\}_{{\text{z}} = {\text{a}}}}$$

Answer: Option D