Complex Variable

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 & Solution

Correct Answer: Option D

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If f(z) has a pole of order n at z =...

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

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If f(z) has a pole of order n at z =...

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

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