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## Main Question or Discussion Point

the Residue theorem states that :

[tex]\oint {f(z)dz} [/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [/tex]

and the summation is taken for all the poles of f(z) enclosed by the counter at which the integration is performed .

now i have read somewhere that

[tex]\oint \frac{f(z)dz}{z^{n+1}}[/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [tex]a^{n}[/tex]

[tex]\oint {f(z)dz} [/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [/tex]

and the summation is taken for all the poles of f(z) enclosed by the counter at which the integration is performed .

now i have read somewhere that

[tex]\oint \frac{f(z)dz}{z^{n+1}}[/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [tex]a^{n}[/tex]

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