课程名称︰复分析
课程性质︰数学系大三必修
课程教师︰蔡忠润
开课学院：理学院
开课系所︰数学系
考试日期（年月日）︰2015/11/12
考试时限（分钟）：150
试题 :
[Total: 34 points]
(1) [8 points] True/False questions, no justifications needed.
￣￣￣￣￣￣￣￣￣
︿
(a) Let € = € ∪ {∞}. There exists a rational function q(z) such that
-1 -1 ︿ ︿
#{q (0)} = 2 and #{q (∞)} = 0 (regard q(z) as a map from € to €,
and count zeros with multiplicity).
(b) There exists an analytic function f(z) defined on {z ∈ €| |z| < 2}
(k)
such that f (0) = 1 for any k ≧ 0.
1
(c) Consider the function f(z) = cosh─. There exist positive constants ε
z
and M so that |f(z)| > M for any 0 < |z| < ε.
(d) Let D = {z ∈ €| |z| < 1}. Suppose that f: D → D is analytic with f(0)
= 0. Then |f(z)| ≦ |z| for any z ∈ D.
(e) There does not exist an entire function whose value on the positive real
∞ -t z-1
axis is ∫ e t dt.
0
(f) These does not exist an entire function whose value on n ∈ |N is
(n-1)!.
(g) These exists a meromorphic function whose pole is exactly n ∈ |N with
j
n exp(2 )
the singular part Σ ────.
j=1 j
(z-n)
(h) Let B be the set of entire functions with f(0) = 1 and |f(z)| ≦ 999
when |z| = 100. For any n ∈ |N, there always exists f(z) ∈ B which has
exactly n zeros in {z ∈ €| |z| < 50}.
3/4
(2) [5 points] Suppose that f(z) is an entire function obeying |f(z)| ≦ n
when |z| = n ∈ |N. Prove that f(z) is a constant function. (Hint: There are
(k)
many ways to do it. You can show that f (0) = 0 for any k ∈ |N, or f(z )
1
= f(z ) for any z , z ∈ €, etc.)
2 1 2
2015 304 12 4 3
(3) [5 points] How many roots does the equation z + 2z + 11z + z + z
2
+ z + z + 1 = 0 have in {z ∈ €| |z| < 1}? Justify your answer.
(4) [6 points] Evaluate the integral
ax
∞ e
∫ ─── dx for 0 < a < 1.
-∞ x
1 + e
z
(Hint: e has a period of 2πi.)
(5) [6 = 2 + 4 points] Consider the function
2πz
e - 1
f(z) =─────.
z
(a) Show that z = 0 is a removable singularity, and determine f(0).
(b) Construct the Weierstrass product development of f(z), and find its
genus. You shall briefly explain the convergence of your expression. You
can simply invoke the theorem of Weierstrass, or derive the estimate by
hand.
(Hint: You may need the following formula:
2πz
2πe 1 1 1
──── = π + ─+ Σ (─── + ─).)
2πz z n≠0 z - in in
e - 1
(6) [4 points] Let f(z) be a non-constant entire function of finite order. Prove
that the image of f(z) can miss at most one value in €. (Hint: You may
assume f(z) misses α and β. What can you say about the entire function
f(z) - α? Could it be possible that f(z) - α never equals to β - α?)