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Guo Dunqing replied:

(4) in (e z)/(z? - 1),

(e^z)/(z? - 1)=(e^z)/[(z- 1)(z+ 1)],

(e z) in | z- 1 | < δ and | z+ 1 | < δ,

∴z 1= 1 and z2 =- 1 are isolated singularities and both are first-order poles.

Residual definition and residual calculation rules 1:

Residual Res[f(z), z0] = z→ z0lim [(z-z0) f (z)].

When z0= 1, Res[f(z),1] = z→1lim [(z-1) (e z)/(z? - 1)]

=(e^z)/(z+ 1)

= e/2;

When z0 =- 1, Res[f(z),-1] = z →-1lim [(z+1) (e z)/(z? - 1)]

=(e^z)/(z- 1)

=- 1/2e;

∴ Residual Res[f(z), z0]=Res[f(z), 1,-1]

= e/2- 1/2e;

(7) In (z n)/(z- 1) n, n is a positive integer,

Z = 1 is an isolated singularity and an N-order pole.

Residual Res[f(z), z0] = z→ z0lim [(z-z0) f (z)], z0= 1,

∴ Residual Res[f(z),1] = z→1lim [(z-1) (z n)/(z-1) n]

=(z^n)/[(z- 1)^(n- 1)]