Page 76 - Calculus with Complex Numbers
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7.4 Boundedness of cot.z
From Section 2.8 we have
2 cos2 x + sinhz y
I cot zl = = .
2 COS Z
sin z sin2 x + sinhz y ,
.
where z = .x + iy.
For y lixed we have
cos2 x + sinhz y 1 + sinhz y coshz y
.
s = = ()0th
zy
sin2 x + sinhz y sinhz y sinhz y ,
.
which shows that I cot z Is coth zr/2 = 1.09033 141 1 . . . for all z eE the horizontal
sides of yx for all N k: 1.
For .x lixed = (N + 1/2):v we have
cos2 x + sinhz y sinhz y
.
= s 1,
sin2 x + sinhz y 1 + sinhz y
.
which shows that I cot z I s 1 for all z eEthe vertical sides of yx for all N k: 1.
Hence we have Icot zls M = tl0th zr/2 for all z eEyx for all N k: 1.
W e can now show
cot z
2 dz --> 0
(JN . !
:
as N --> co. For z e yx we have IzIk: (N + 1/2)> > Nn', and the length of
yx is 8(.V + 1/2)> :é 9Nn' for N k: 4. Therefore by the estimate lemma (4.4)
we have
7.5 U se of cosec az
Having shown that
we can obtain the sum of
