CCM322: Complex Analysis
Assignment 6
You may assume any holomorphic function has continuous derivative f 0 .
Notation: as usual, D(a, R) denotes the open disc centred at a 2 C of radius
R > 0, r (a) denotes the circle centred at a 2 C of radius r > 0, whilst H(⌦)
denotes the set of holomorphic functions on an open subset ⌦ of C.
[1] (i) Let f 2 D(a, R)) and choose r 2 (0, R). Using the CIF for a disc
(Propn 8.3, p.41 of the notes), show that
Z 2⇡
1
f (a) =
f (a + rei✓ ) d✓.
2⇡i 0
(ii) Suppose that
|f (z)|  |f (a)|
Show that
Z
8 z 2 D(a, R).
2⇡
0
(Hint: use the estimate |
|f (a)|
Rb
a
|
|f (a + rei✓ )| d✓ = 0.
Rb
a
| | for a path
: [a, b] ! C.)
(iii) Explain why, therefore, |f | must be constant on D(a, R).
(iv) Deduce that f is constant on D(a, R).
(Hint: With f = u+iv, compute @/@x and @/@y of |f |2 , then use the Cauchy
Riemann equations to deduce u(x, y) and v(x, y) are constant.)
(v) Let ⌦ be a connected open subset of C and let f 2 H(⌦). Deduce
that if |f | achieves its maximum M 2 [0, 1) at some point z0 2 ⌦ then f is
constant on ⌦.
(Hint: Let UM be the subset of points of ⌦ at which |f | is equal to M , use
(iv) to see that UM is open.)
(vi) Deduce that if ⇤ is a bounded open subset of C and f is holomorphic
on ⇤ and continuous on its closure ⇤, then |f | attains its maximum on the
boundary @⇤ := ⇤\⇤ of ⇤. (For example, if ⇤ is the unit disc, then @⇤ is the
unit circle, can you see where |f | has its maximum when f (z) = ez ?)
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(v) Infer that if f is holomorphic on ⇤ and non-constant, then |f | attains
its minimum either at a zero of f or on the boundary.
For the remaining points, you may assume the fact proved in lectures that a
function is holomorphic i↵ it has a power series expansion around each point
of its domain (i.e. f is C-analytic).
[2] Prove Liouville’s Theorem (see §8.3 on p.54 of the course notes– printed
and on Keats).
[3] Prove the fundamental theorem of algebra (see §8.3 on p.54 of the course
notes– printed and on Keats).
[4] Show that If f is an entire function (holomorphic on all of C) and
z 2 f (z) ! 2 as z ! 1 then f (z) = 2z 2 + az + b where a and b are some
constants.
[5] Let f be an entire function satisfying f (2z) = 2f (z) for all z 2 C. Show
that f is a polynomial of degree at most 1 with f (0) = 0.
Hint: show f (0) = lim f (1/2n ) = f (1)/2n = 0, by continuity of f , so f (z) =
zg(z) with g holomorphic and g(2z) = g(z).
[6] In each of the following cases, determine whether the statement is true
or false. Give a brief explanation or a counterexample as appropriate.
(a) There is an entire function f (z) such that |f (z)|  |z| for all z 2 C,
f ( 1) = f (1) = 1 and f (0) = 0?
(b) There exists a nonzero polynomial P (z) of degree 3 such that the equation
2
d
d2
(P (z) ez ) = 0 has six distinct solutions z 2 C (here dz
2 denotes the second
dz 2
derivative).
Hint: You may assume the results on p.54 of the course notes– printed and
on Keats
(c) A continuous function defined on an open set ⌦ is holomorphic on ⌦ if
and only if it has a primitive (a primitive for f is a holomorphic function F
with F 0 = f ).
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(d) If f is an entire function such that f 0 (z) = 2z f (z) then f (z) = c ez ,
where c is a constant.
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