1758 CHAPTER 55. COMPLEX MAPPINGS
Now from 55.5.26 this is dominated by
exp(
π exp2(|H (0)|+24C ln
(1
1−θ
)))= exp
(π exp(2 |H (0)|)exp
(48C ln
(1
1−θ
)))(55.5.27)
Consider exp(2 |H (0)|). I want to obtain an inequality for this which involves β . Thisis where I will use the convention about the logarithms discussed above. From 55.5.25,
2 |H (0)|= 2
∣∣∣∣∣log
(√log( f (0))
2πi−√
log( f (0))2πi
−1
)∣∣∣∣∣
≤ 2
(ln
∣∣∣∣∣√
log( f (0))2πi
−√
log( f (0))2πi
−1
∣∣∣∣∣)2
+π2
1/2
≤ 2
∣∣∣∣∣ln(∣∣∣∣∣√
log( f (0))2πi
∣∣∣∣∣+∣∣∣∣∣√
log( f (0))2πi
−1
∣∣∣∣∣)∣∣∣∣∣
2
+π2
1/2
≤ 2
∣∣∣∣∣ln(∣∣∣∣∣√
log( f (0))2πi
∣∣∣∣∣+∣∣∣∣∣√
log( f (0))2πi
−1
∣∣∣∣∣)∣∣∣∣∣+2π
≤ ln(
2(∣∣∣∣ log( f (0))
2πi
∣∣∣∣+ ∣∣∣∣ log( f (0))2πi
−1∣∣∣∣))+2π
= ln((∣∣∣∣ log( f (0))
πi
∣∣∣∣+ ∣∣∣∣ log( f (0))πi
−2∣∣∣∣))+2π (55.5.28)
Consider∣∣∣ log( f (0))
πi
∣∣∣log( f (0))
πi=− ln | f (0)|
πi+
arg( f (0))π
and so
∣∣∣∣ log( f (0))πi
∣∣∣∣ =
(∣∣∣∣ ln | f (0)|π
∣∣∣∣2 +(arg( f (0))π
)2)1/2
≤
(∣∣∣∣ lnβ
π
∣∣∣∣2 +(π
π
)2)1/2
=
(∣∣∣∣ lnβ
π
∣∣∣∣2 +1
)1/2
.