The point where (1), ∫z(z-a)(za- 1)= 0 is f(z)=(z? +1)/[z (z-a) (za-1)], ∴ has three unique points, namely z 1=0, z2=a and z3 =1/a.

∴ By residue theorem, RES [f (z), z1] = lim (z → z1) (z-z1) f (z) = lim (z → 0) (z? +1)/[(z-a) (za-1)] =1/a. Similarly, Res[f(z), z2]=(a? + 1)/[a(a？ - 1)]，Res[f(z)，z3]=(a？ + 1)/[a( 1-a？ )]。

(2) Let z = e (i θ) and ∴ d θ = dz/(iz). ∴ cos θ = (z+1/z)/2. ∴ I = (I/2) 丨丨 = 1(z? + 1)dz/{z[z？ + 1-( 1+a？ )z]} .

(3)∵a & gt； 1, ∴ In the field 丨丨 = 1, f(z) has two poles: z 1=0 and z3 = 1/a. ∴ From Cauchy integral theorem, there is an original formula. - 1)]。

For reference.