587 
Then for |z| =e we have a certain number u for which 
1 
SP (ON 
u 
in which we may suppose u > 1. 
By applying the formula (1) to /(z) in the cireles having the 
different points of the circumference of the circle |z| =p as centres 
and @ as radii, we get: 
| F(z) | and | F(2)|-!<ame for |2|=£. 
a 
By applying the formula in the circles with the different points 
of the circumference of the circle | z | = Fas centres and 5 as radii, 
we find: 
| F(z) | and | (2) |-! <Ca(apr)yp = apt) wr? for | z | = 
Going on in this way we find 
v—l1 v—2 y 
pri(zylend|F(z)|-1<a? (Fe teel = 
y 
pl 1 4 
v Ë 
Ed! u la) for Ken 
We have therefore 
| F(z)|and| F(2)|'<e” for lijm 2 
Qo» 
in which 
1 
g == Log (ar 1) . 
3. Log | #(z)| is harmonical in 2, with the exception of O. 
For 0< |z|<o we have therefore, if we pul 2=re%: 
Log | F (2) |= A Logr + & (an cos nO + by sin nO) vr. 
Here me 
Qn ° 
ar de 
An +4, 7° = „je nO . Log | F (re) | dO 
JE 
0 
and 
1 i 
b, mbr = „je nO Log.) E (re), dO et a ne ate 
Ed 
0 
where 0< r<g. r is for the rest arbitrary. 
If we now put r =< there follows. from (2): 
