math185f08-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

MATH 185: COMPLEX ANALYSIS
FALL 2008/09
PROBLEM SET 5 SOLUTIONS
1. Consider the n th Taylor polynomial approximant to exp( z ),
f
n
( z ) := 1 + z +
1
2!
z
2
+ · · · +
1
n !
z
n
.
Show that for all z ∈ C with Re( z ) < 0,
| exp( z ) - f
n
( z ) | ≤ | z |
n +1
for all n ∈ N .
Solution. Let Ω := { z ∈ C | Re( z ) < 0 } . Let z ∈ Ω and let Γ
z
be the line segment from 0
to z . Then by Proposition 3.9 ,
Z
Γ
z
exp( w ) dw = exp( z ) - exp(0) = e
z
- 1 .
By Proposition 3.5 ,
Z
Γ
z
exp( w ) dw ≤ sup
w ∈ Γ
z
| exp( w ) | ×
Z
Γ
z
| dw |
= | z | sup
w ∈ Γ
z
e
Re( w )
≤ | z |
where the last step follows since Γ
z
⊂ Ω and thus Re( w ) ≤ 1. Hence
| e
z
- 1 | ≤ | z | .
Suppose the statement is true for n - 1, ie.
| exp( z ) - f
n - 1
( z ) | ≤ | z |
n
for all z ∈ Ω. Now observe that
Z
Γ
z
exp( w ) - f
n - 1
( w ) dw = exp( z ) - z +
1
2!
z
2
+ · · · +
1
n !
z
n
- exp(0)
= exp( z ) - f
n
( z ) .
By Proposition 3.5 and the induction hypothesis,
Z
Γ
z
exp( w ) - f
n - 1
( w ) dw ≤ sup
w ∈ Γ
z
| exp( w ) - f
n - 1
( w ) | ×
Z
Γ
z
| dw |
≤ | z | sup
w ∈ Γ
z
| w |
n
= | z |
n +1
.
Hence
| exp( z ) - f
n
( z ) | ≤ | z |
n +1
as required. By mathematical induction, this holds for all n .
2. Let f : C → C be an entire function. As usual, we write f ( z ) = u ( x, y ) + iv ( x, y ) for z = x + iy .
(a) Show that if u is positive valued, ie. u ( x, y ) > 0 for all x, y ∈ R , then f is constant.
Date : October 24, 2008 (Version 1.0).
1

(b) Show that if | u ( x, y ) | < | v ( x, y ) | for all x, y ∈ R , then f is constant.
(c) Can we still draw the same conclusions if ‘ < ’ is replaced by ‘ > ’ in (a) and (b)?
Solution. Note that e
x
is a monotone increasing function on R .
• For (a), we choose g ( z ) = e
z
and note that
| e
f ( z )
| = | e
u
e
iv
| = e
u
≤ e
0
= 1 .