A FEW EXAMPLES IN THE THEORY OF FUNCTIONS 239 



center and being entirely within the region of regularity of f(z). In the 

 given example the point a is within the circle of radius i and having the 

 origin as a center. We have 



-z)(z-a) w+1 



To integrate this, write . _ w _ , n+1 in partial fraction form: 



(i-z)(z-a)»+ I_ i-z (z-a) (z-a) 2 "*"' ' *" t "(z-a)"+ I " W 



The integrals of these parts along the circle c all disappear with the 



exception of j — ^ dz. It is therefore sufficient to determine the 



coefficient B n+1 . For this purpose the following method of recurrence 

 may be used. We have 



4i_ + A + _^ a+ ... + -^-. (5 ) 



(1— z)(z— a) n x—z z—a (z—a) 2 (z—a) H 



Multiplying (5) by ^--- we get an expression which must be indentical 



z a 



with (4); i. e., coefficients of equal powers are equal. From this we find 



A n yln + i , "n + t 



(i—z)(z—a) i—z z—a ' 



A n 



or A n =A n+I (z-a)+Bn +l (i-z) , and A n+t = B n+1 ; 4 W+I = — 

 To determine A t assume the case of w=o, so that (4) becomes 



1 A 1+ B t 



(1— z)(z— a) 1— z z—a 



Here A =i: hence A, = . By recurrence A 2 = -, rr , A = 



° ' * 1— a J (1—a) 2 ' 3 



(l _ g)3 , • • • , A n+1 =B„ +1 = (i _ a - )n+l . The value of the required 

 integral is therefore 



. _ »/ f dz _ n! 



™ a > ~^i) c (z-a)(i-a) n + l ~ (i-a) n+ * ' 



