386 F. B. }Viley and G. A. Bliss 
After treating the three remaining terms of [6] in the same 
manner, we have 
J ’Z nr 
f ( 2 :) dz = [uF (t) —vF (t) ]dtAi I [uF (t) + vF (0 ] dt, 
zoK *^^0 to 
where u and v are functions of cp (t) and \f/ {t) as expressed above. 
Furthermore we know that 
[uFit)-vFil)]dt = ^(T) 
and 
[uF{t) + vF{t)]di='^iT) -AFto) 
])rovided that $ (t) and (t) are anti-derivates of the functions 
uF (i) ~ vF (t), uF d) + vF it), 
respectively. But if we write 
F {z) = U ix, y) + i V (x, y), 
we have, according to the proof of the first theorem of this 
section, 
dt/ dF dF dF 
^ U, - = V, = — V, = u, 
dx dx dy ' dy 
and hence 
i \uF it) - vF it) ]dt=V iX, Y) - U ixo, yo), 
{\vF it)FuFit)]dt=ViXY) -Vix^yd). 
Jt, 
From this result and [7] it follows that 
dz = FiZ)~Fizo). 
Our next step is to show that the integral along C, where C 
is any rectifiable curve, is equal to the integral along K, where 
K is a suitably chosen polygon inscribed in C, also joining Zq and 
Z. We accomplish this by showing that the inequality 
