Page 92 - Calculus with Complex Numbers

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2 C o m p I ex f n c ti o n s

8. To lind the maximum of Isin zlon the disc IzI s 1 we use the infinite form of
the triangle inequality which states that
X X
jg zn s jg Ial
1 1
for any sequence of complex numbers (zn)nz1 .
ln particular,

I z 13 Iz 15 IzI7
:% I zl + 31 + 51 + 71 + ' ' '
1 1 1
:i 1 + + + + . . .
V
W
V
== sirzA 1

for Iz I 1qL 1.

Also for z = i we have Isin i I= Ii sirth 1 I= sinh 1.

Therefore Isin zlis maximum on Iz Is 1 at z = i with maximum value
equal to sinh 1.
lf z = .x + i y then
z + 1
= 2
z + 4
isequivalentto
1.x + iy + 1 I2 = 4Iv + iy + 41 2

.
(x + 1)2 + ,2 = 4((.x + 4)2 + y2)
.x 2 + 2 .x + 1 + ),2 = 4.(.x2 + 8.x + 16 + y2)
O = 3w2 + 3y2 + 30 .x + 63

2 = Iz + 51
which is the equation of a circle centre -5 radius 2.