244 CHAPTER 6 • COMPLEX INTEGRATION
We sometimes state the maximum modulus principle in the following form.- EXAMPLE 6.26 Let f (z) = az + b. H we set our domain D to be D 1 (0),
then f is continuous on the closed region D1 (0) = {z: lzl ~ l }. Prove that
max If (z)I = lal + lbl
1 •1:5 1
and that this value is assumed by f at a point z 0 = ei^9 o on the boundary of
D 1 (O).
Solution From the triangle inequality and the fact that lzl ~ 1 in D1 (0), it
follows that
If (z)I = laz + bl ~ lazl + lbl ~ la l + lbl.
If we choose zo = ei^90 , where 80 E arg b - arg a, then
arg(azo) = arga+ argzo
= arga + (argb- arga)
= argb,(6-58)so the vectors azo and b lie on the same ray through the origin. This is the
requirement for the Inequality (6-58) to be an equality (see Exercise 21, Section