336 UNIVERSITY OF VIRGINIA PUBLICATIONS 



the theorem fails for the reasons above. In fact the theorem for an ordi- 

 nary closed boundary (not a contour) on and within which f{z) is regular 

 furnishes a root oi f{z) for each pole oi j'{z)/f{z) inside the boimdary, and 

 if there be n roots the integral is 2ivin. Cauchy's integral must therefore 

 be abandoned in the studj' of the function within the contour when based 

 on the contour integral. 



2. The demonstration of the following theorem is so simple that it is 

 almost incredible to believe it new, but as one does not find it in the classic 

 treatises on analysis it is thought worth while to give this note. 



Theorem. Every monogenic function f{z) has a root inside of any simple 

 closed contour along which \f{z) I is constant and .within which f{z) is regular. 



Let the equation of the contour C be 



1/(2) 1^ = V? -{- v^ = M^ = constant. 



Consider the surface f = it° + v^. This function of x and y is finite, con- 

 tinuous and unlimitedly differentiable at all points inside C. It therefore 

 has an upper or a lower boundary at some point x,y in this region which 

 definite maximum or minimum value f must attain since it is continuous. 

 At this point therefore 



On squaring and adding there must result at x, y 



or what is the same thing 



fiz) . f'iz) = 0. 



At x,y either the function f(z) is zero, or both f{z) and f'(z) are zero and 

 the theorem is proved, unless /(3)=t=0 and f'{z) = 0. At the point x, y must 

 occur a true maximum or minimum of f . 



3. We will now show that it is not possible for f to have a maximum or 

 minimum value where f{z) =(= and 



p{z) =0, r = 1, 2, . . ., n - 1, 



m 



