Lecture 15: Simple Connectivity and tan−1
Hart Smith
Department of Mathematics
University of Washington, Seattle
Math 428, Winter 2017
Theorem
If E ⊂ C is open and connected, and f (z) is analytic
Z on E, then
f (z) = F 0 (z) for an analytic F on E if and only if
f (z) dz = 0
γ
for every closed path γ contained in E.
Z
Proof. (⇒) If f = F 0 then f (z) dz = 0 by Fund. Thm. Calc.
γ
Z
(⇐) If f (z) dz = 0 all closed γ, we fix z0 ∈ E, c0 ∈ C, and let
γ
Z
z
f (w) dw
F (z) = c0 +
z0
Z
z
denotes integration along any path from z0 to z.
where
z0
By the proof of Theorem 2.6.1, F 0 (z) = f (z).
Definition
We say a connected open set E ⊂ C is simply connected if, for
every closed path γ in E, we have indγ (z) = 0 for all z ∈
/ E.
• That is: a closed path in E can’t wind around any z ∈
/ E.
• By Cauchy’s Theorem, if E is simply connected then for all
Z
closed γ in E, all analytic functions f on E,
f (z) dz = 0
γ
Corollary
If E ⊂ C is simply connected, and f is analytic on E, then f has
an anti-derivative on E : f (z) = F 0 (z) for some analytic F on E,
and F is determined on E up to a constant.
If we fix z0 ∈ E, the general anti-derivative F of f is given by
Z z
F (z) = c0 +
f (w) dw ,
c0 ∈ C.
z0
Example: E = C \ (−∞, 0] is simply connected.
Proof. We need verify indγ (z) = 0 for all closed paths γ in E,
for all z ∈ C \ E, that is, for z ∈ (−∞, 0].
Key fact: entire half-line (−∞, 0] lies in unbounded component
of C \ {γ}, since it’s connected (and unbounded), therefore
indγ (z) = 0 on the entire set z ∈ (−∞, 0].
• f (z) =
1
has anti-derivative on C \ (−∞, 0], F (z) = log z.
z
• Principal branch of log z is the unique anti-derivative of
C \ (−∞, 0] that vanishes at z = 1.
1
on
z
Example of interest for tan−1 (w) :
5
4
3
2
i
1
E = C \ [i, +i∞) ∪ [−i, −i∞)
0
is simply connected.
-i
-1
-2
-3
-4
-5
-1
-0.8
-0.6
-0.4
0
0.2
• There exists an analytic
function
g(w)
on-0.2E such
that
g(0) = 0,
g 0 (w) =
1
1 + w2
• By last lecture, tan(g(w)) = w, so g(w) = tan−1 (w).
0.4
0.6
0.8
An explicit formula for tan−1 (w)
tan z = −i
e iz − e −iz
e 2iz − 1
=
−i
e iz + e −iz
e 2iz + 1
Let u = e 2iz . Then tan z = w is equivalent to
u−1
= iw
u+1
tan z = w
⇔
e
tan
which gives
2iz
−1
i −w
=
i +w
⇔
u =
i −w
i +w
1
i −w
z =
log
2i
i +w
1
i −w
(w) =
log
2i
i +w
• If w ∈
/ {i, −i}, there exists (many) choices for tan−1 (w).
• Values of log differ by 2πk i ; values of tan−1 (w) differ by k π.
Principal branch of tan−1 (w) is:
−1
tan
1
i −w
(w) =
log
with principal branch of log .
2i
i +w
n
o
i −w
• This is defined on the set w :
∈
/ (−∞, 0] , w 6= ±i ,
i +w
and is the value of z : tan z = w and − 21 π < Re(z) < 12 π.
• The “cut” set is the set of w : u =
u−1
• Write w = i
. Then
u+1
i −w
∈ (−∞, 0].
i +w
(
u ∈ (−∞, −1) :
u ∈ (−1, 0) :
w ∈ (i, +i ∞)
w ∈ (−i, −i ∞)
Principal branch of tan−1 (w) maps C \ [i, +i∞) ∪ [−i, −i∞)
to the vertical strip − π2 < Re(z) < π2 .
Complex plane representation of z → tan z
10
8
6
4
2
!
:
2
0
:
2
-2
-4
-6
z
-8
-10
-10
-5
w =0 tan z
5