Complex Analysis Cheat Sheet
Guide
•
Unless explicitly stated, G is a domain (open and connected).
• Hol (G)
is the set of holomorphic functions on G, Mer (G) and Har (G) are the sets of mero-
morphic (holomorphic, save for countable isolated singularities) and harmonic functions on
G
.
• C (G)
is the set of continuous functions on G.
Theorems
Topology
1. The set S is closed ⇐⇒ S
c
is open
2. Let G ∈ C be an open set. The following de nitions for connectivity are equivalent:
(a) G cannot be decomposed into two disjoint open sets: if X ⊆ G is open and X \ G is open,
then either X = G or X = ∅.
(b) Let a, b ∈ G. So there exitsts a polygonal curve that starts at a and ends at b.
(c) For each locally constant f : G → C, f is necessarily globally constant.
(d) Every continuous f : G → R satis es the intermediate value property: ∃α, β.f (α) =
s, f (β) = t ⇒ [s, t] ⊆
Img (f)
3. γ : [a, b] → C is cont. and does not vanish. Then ∃ψ : [a, b] → C such that e
ψ(t)
= γ (t)
. It is
uniqe up to ±2πi.
4. ind
γ
(z
0
) =
1
2πi
´
γ
dz
z−z
0
5. γ ⊂ G is closed and piecewise C
1
, f ∈ Hol (γ), them
´
γ
f
0
f
dz = 2πi ·
ind
f ◦γ
(0)
6. f ∈ Hol (G) ; γ
0
, γ
1
: [α, β] → G
. Then γ
0
∼
G
γ
1
⇒
´
γ
0
f =
´
γ
1
f
7. Jordan: if γ is simple, it divides C into two connected components.
8. If G is simply connected:
(a) γ is closed, f ∈ Hol (G). Then
´
γ
f = 0
(b) f ∈ Hol (G) has a primitive, F (z) =
´
z→z
0
f
(c) γ is closed, z
0
/
∈ G
, then ind
γ
(z
0
) = 0
(d) f ∈ Hol (G) and does not vanish, then it has a branch of log f,
√
f
there.
9. f ∈ Hol (G) , γ ⊂ G is closed. Then: G is simply connected ⇐⇒ C\G is connected ⇐⇒
∀z
0
/
∈ G,
is closed them ind
γ
(z
0
) = 0 ⇐⇒
´
γ
f = 0 ⇐⇒
if f does not vanish on G there
exists log f ∈ Hol (G).
10. G
1
, G
2
⊂ C, f : G
1
→ G
2
is continuous, 1-1 and onto with a continuous f
−1
, then G
1
is simply
connected ⇐⇒ G
2
is simply connected.
11. Schwarz's Lemma:Hol (D) 3 f : D → D with f (0) = 0. Then |f (z)| ≤ |z| and only reaches
equality when f (z) = λz, |λ| = 1
12. Hol (D) 3 f : D → D, 1-1 and onto with f (0) = 0. Then there exists θ ∈ [0, 2π] such that
f (z) = e
iθ
z
13. f : D → D is conformal on D. Then there exist |a| < 1, θ ∈ [0, 2π] such that f (z) =
z−a
1−az
14. Riemann's thm: Let G ⊂ C, G 6= C and is simply connected. Then there exists a unique
holomorphic transform f : G → D, 1-1 and onto, with f (a) = 0, R 3 f
0
(a) > 0
.
15. Pick's Lemma : f : D → D is holomorphic, so |f
0
(z)| ≤
1−|f (z)|
2
1−|z|
2
, and
f (z)−f (w)
1−f (z)f (w)
≤
z−w
1−zw
Dierentiation
1. If f is dierentiable at z
0
, g is dierentiable at w
0
= f (z
0
)
, then
d
dt
g (f (z))
z=z
0
=
g
0
(w
0
) f (z
0
)
.
2. f is dierentiable, at x
0
⇐⇒
it veri es the Cauchy-Riemann equations.
Curves
1. Chain rule for curves: suppose γ is dierentiable at t
0
, f is holomorphic at γ (t
0
)
. So (f ◦ g)
0
=
f
0
(γ (t
0
)) ˙γ (t
0
)
Holomorphic Functions
1. Suppose f : G → C is holomorphic. Then f is constant ⇐⇒ ∀z ∈ G : f
0
(z) = 0
2. If f : G → C is holomorphic and real, then f is constant.
3. Suppose f : G → C is holomorphic such that |f| is constant in G, then f is globally constant.
4. Suppose that f : G → C is holomorphic and f
0
(t) 6= 0, ∀z ∈ G
. Then f is a conformal
transformation, such that ] (f (γ
1
) , f (γ
2
)) = ] (γ
1
, γ
2
)
.
5. Suppose f : C → C is an R-linear transformation. then ∀z ∈ C f (z) = az + b¯z, such that
a =
∂f
∂z
(0) ; b =
∂f
∂ ¯
z
(0)
.
6. Suppose f : G → C is R-dierentiable and conformal. Then f is holomorphic.
Harmonic Functions
1. f ∈ Hol (G). If f has continuous second partial derivatives, then f is harmonic (meaning
Im (f) , Re (f) are both harmonic),
2. u : G → C is harmonic, then
∂u
∂z
∈ Hol (G)
.
3. Let u : G → R be harmonic. Then:
(a) If G = C, there always exists a harmonic conjugate v.
(b) For each G, it is single up to ±c.
Möbius Transformations (ϕ (z) =
az+b
cz+d
, C is a clircle)
1. For each A ∈ GL
2
, h
A
(z)
transformations clircles to clircles.
2. If A, B ∈ GL
2
,
then h
A
◦ h
B
= h
AB
. Therefore:
(a) ∀λ 6= 0, h
λA
= h
A
.
(b) h
λd ¯
d
= h
I
.
3. ϕ has at most 2 xed points. If h
A
has 3 xed points, then it is h
I
.
4. z
1
, z
2
, z
3
, z
4
∈ C are distinct. Then [z
1
, z
2
, z
3,
z
4
] = [ϕ (z
1
) , ϕ (z
2
) , ϕ (z
3
) , ϕ (z
4
)]
.
Therefore, if w
1
, w
2
, w
3,
w
4
∈ C are distinct, there exists ϕ such that ∀
1≤j≤4
T z
j
= w
j
5. Suppose ϕ operates on C
1
7−→ C
2
. Then every z, [z]
∗
C
1
7→ z, [z]
∗
C
2
.
6. z
2
, z
3
, z
4
are distinct and on C. Then [z
1
]
∗
C
, z
2
, z
3
, z
4
= [z
1
, z
2
, z
3,
z
4
]
.
7. z
j
, 1 ≤ j ≤ 4
are all on the same clircle if (z
1
, z
2
, z
3
, z
4
) ∈ R.
Functions
1. Let G ⊆ C, u
1
, u
2
be branches of log. Then u
1
− u
2
≡
const.
2. Let G ⊆ C, l : G → C be a branch of log. Then l ∈ Hol (G) and l
0
(z) =
1
z
.
3. Let f (: G → C) ∈ Hol (G) and suppose that u is a branch of log (f). Then u is holomorphic
and ∀z ∈ G.u
0
(z) =
f
0
(z)
f (z)
.
4. Let f (: G → C) ∈ Hol (G) , z
0
∈ G, f (z
0
) 6= 0
. Then ∃δ > 0 such that D (z
0
, δ)
contains a
branch of log (f).
Series
1. If a series converges absolutely to a, it is invariant to any permutation of its order of summation.
2. Weierstrass' M-Test: let {u
n
(z)}
∞
n=0
be a series of functions in G, and {M
n
}
∞
n=0
a series of
positive numbers such that:
(a) sup
z∈G
|u
n
| < M
n
(b) P
∞
n=0
M
n
< +∞
Then P
∞
n=0
u
n
(z)
uniformly and absolutely in G.
3. D'alembert: For P
∞
n=0
c
n
z
n
, R = lim
n→∞
c
n
c
n+1
; Cauchy-Hadamard: R
−1
=
lim
n→∞
sup
n
p|a
n
|
4. Suppose f (z) = P
∞
n=0
c
n
(z − z
0
)
n
is a power series with convergence radius R ∈ [0, ∞].
(a) The series converges normally at D (z
0
, R)
.
(b) If |z − z
0
| > R
the series diverges.
(c) f (z) is holomorphic in D (z
0
, R)
and its derivative is f
0
(z) =
P
∞
n=1
c
n
n (z − z
0
)
n−1
. Its
convergence radius is the same as in f (z).
Therefore:
(a) f (z) = P
∞
n=0
c
n
(z − z
0
)
n
is C
∞
on D (z
0
, R)
, and f
(k)
(z) =
P
∞
n=k
n!
(n−k)!
c
n
(z − z
0
)
n−k
.
Speci cally for z = z
0
:
f
(k)
(z
0
)
k!
= c
k
.
(b) Suppose WLOG that z
0
= 0
. f has a primitive function with the same radius of conver-
gence: G (z) = P
∞
n=0
c
n
z
n+1
n+1
.
5. Tauber: Suppose n |c
n
|
n→∞
→
0
, and f (z) = P c
n
z
n
has a limit lim
R3n→1
f (x) = L
, so
P c
n
= L
.
6. Abel: Suppose P c
n
converges. Then if z
n
n→∞
→ 1
and sup
|z
n
−1|
1−|z
n
|
< ∞
, then lim
n→∞
f (z
n
) =
P c
n
.
Integrals
1. Newton-Leibnitz Formula: Suppose f : G → C is Holomorphic, γ : [a, b] → G is piecewise
dierentiable. Then
´
γ
f
0
(z) dz = f (γ (b)) − f (γ (a))
(a)
´
γ
f
0
depends only on the edges of γ
(b) If the curve closed, the integral is 0.
2. Suppose f : [a, b] → C, then
´
b
a
f (t) dt =
´
b
a
Ref (t) dt + i
´
b
a
Imf (t) dt.
3. Suppose γ
2
is a reparametrization of γ
1
and f is de ned on the image of γ
1
, then
´
γ
1
f =
´
γ
2
f
4. Suppose f
n
: G → C is a sequence of continuous functions such that f
n
⇒ f on G. Let
γ : [a, b] → C be piecewise C
1
. Then
´
γ
f
n
→
´
γ
f
.
5.
´
γ
f
≤
Length (γ) · sup
t∈[a,b]
|f (γ (t))|
6. Let G ⊂ C, f : G → C be continuous and γ : [a, b] be piecewise C
1
. Let ε > 0. So there exists
δ > 0
such that for every partition π = {a = t
0
< . . . < t
N
= b}
, such that λ (π) < δ:
(a)
´
γ
f −
P
N −1
i=0
f (z
i
) (z
i+1
− z
i
)
< ε
, z
i
= γ (t
i
)
.
(b) Let γ
π
be the polygonal curve connecting the vertices in f {π}, then
´
γ
π
−
´
γ
< ε
7. Goursat: Let G ⊂ C, f : G → C be holomorphic. Suppose T is a triangle in G, with
counterclockwise orientation. Then
´
γ
f = 0
.
8. If G ⊂ C is convex and f : G → C is holomorphic, then f has a primitive in G.
9. If G ⊂ C is convex and f : G → C is holomorphic, and suppose γ is closed and piecewise
continuous. Then
´
γ
f = 0
.
10. Cauchy's formula for a disk: See Section.
1
11. Intermediate Value Theorem: suppose f is holomorphic inside a circle containing D ⊂ C about
z
0
, then ∀z ∈ D: f (z) =
1
2πi
´
2π
0
f z + e
it
dt
. Therefore, any holomorphic function is locally
a Power Series.
12. The Cauchy integral is holomorphic outside the curve on which it is de ned, and is C
∞
there.
13. Power Series Coe cients:
Suppose f is holomorphic on D (z
0
, R)
.
So ∀z ∈
D (z
0
, R) , f (z) =
P
∞
n=0
a
n
(z − z
0
)
n
, and:
(a) a
n
=
n!
2πi
´
∂D(z
0
,R)
f (z)dz
(z−z
0
)
n+1
(b) f
(n)
(z
1
) =
n!
2πi
´
∂D(z
0
,R)
f (z)dz
(z−z
1
)
n+1
14. Morera (Convex): Suppose G is convex, then (see 20)
15. Liouville: If f is entire (harmonic/holomorphic) and bounded, it is xed.
16. Fundamental thm of Algebra: P ∈ C [z] is a nonconstant polynomial, then P has a root.
17. General Cauchy thm+formula: See Section
18. There Exists no nonconstant holomorphic function for the following cases:
(a) f : C → D (but any domain that isn't C is applicable)
(b) f : CP
1
→ C
(c) f : C → C with two linearly independent cycles.
19. Maximum Modulus Principle: G is bounded with a regular contour, f ∈ Hol (G) ∩ C G. So
the strict maximum is achieved on the boundary. Otherwise, the function is constant.
20. Morera (General): Suppose f is continuous in G and for every triangle T ⊂ G:
´
∂T
f (z) dz = 0
. Then f is holomorphic in G.
21. Suppose I ⊂ G is a closed contour. f ∈ Hol (G\I) ∩ C (G), Then f ∈ Hol (G).
Laurent Series
1. A nonconstant holomorphic function has a nite nullset.
2. Suppose f is holomorphic around z
0
. If f has a zero of order m there, then f (z) =
(z − z
0
)
m
g (z)
, where Hol (G) 3 g (z) =
(z − z
0
)
−m
f (z)
z = z
0
a
m
z 6= z
0
3. The nullset for f does not contain an accumulation point in G.
4. Suppose f, g ∈ Hol (G) and f = g on A ⊂ G. If A has an accumulation point, f ≡ g. Therefore
functions in R have a single analytic continuation.
5. Weierstraÿ convergence thm: Suppose f
n
: G → C is a sequence of holomorphic functions
converging uniformly to a limit function f.
(a) f ∈ Hol (G).
(b) For every k, f
(k)
n
→ f
(k)
locally uniformly.
6. Laurent Coe cients: Suppose f (z) = P
∞
n=−∞
a
n
z
n
in the annulus A = {R
1
< |z| < R
2
}
.
Then for every r ∈ (R
1
, R
2
)
, a
n
=
1
2πi
´
|z|=r
f (z)
z
n+1
dz
. Therefore:
(a) Laurent series can be reconstructed from one circle.
(b) Two laurent series which agree on one circle are equivalent.
7. Laurent's thm: f ∈ Hol (A) for A = {R
1
< |z| < R
2
}
. Then there exist {a
n
}
∞
n=−∞
such
that f = P
∞
n=−∞
a
n
z
n
Isolated Singularities
1. Riemann's Criterion : f is bounded in D\ {z
0
}
, then z
0
is removable. If z
0
= ∞
. it also
needs a bounded neigborhood.
2. Pole Criterion: f ∈ Hol (D (z
0
, r) \ {z
0
})
. Then z
0
is a pole ⇐⇒ lim
z→z
0
|f (z)| = ∞
.
3. Casorati-Weierstraÿ: f has an essential singularity at z
0
⇐⇒ ∀ε > 0
, f {D (z
0
, ε) \ {z
0
}}
is dense in C.
4. Picard's Thm: If f has an essential singualrity, then # {C\f {D (z
0
, ε) \ {z
0
}}} ≤ 1
.
5. f ∈ Hol (D (z
0
, r) \ {z
0
})
. Then ∀0 < ε < r,
´
|z−z
0
|=ε
f (z) dz = 2πi ·
res
z
0
f
.
6. Argument principle:f ∈ Mer (G) ∩ C G ⇒
´
∂G
f
0
f
= 2πi
P (Z
G
− P
G
)
. ∂G Must not vanish.
7. Rouché's thm: f, g ∈ Hol (G) ∩ C G, ∀z ∈ ∂G |f (z) − g (z)| < |f (z)|. Then Z
f
= Z
g
in
G
.
8. Open Mapping thm:f ∈ Hol (G) and nonconstant. Then f maps open sets to open sets.
9. Inverse function thm: f : G → C is holomorphic and 1-1. Then f
−1
is holomorphic and
f
−1
0
(z) =
1
f
0
(f
−1
(z))
.
10. Change of variables:
´
g(γ)
f (z) dz =
´
γ
f (g (w)) g
0
(w) dw
11. Residue at ∞: If f ∈ Hol ({z | |z| > R}) then ∀r > R, −
´
|z|=r
f (z) dz = 2πi ·
res
∞
(f )
12. The sum of residues on C is 0.
13. f ∈ Hol (C) with a pole at ∞ ⇒ f is a polynomial.
14. f ∈ Mer (C) with a pole/removable sg. at ∞ ⇒ f is rational.
15. Local Mapping thm: f ∈ Hol (G) and nonconstant. If z
0
∈ G, w
0
= f (z
0
)
with multiplicity
m
. Then for every su ciently small δ > 0, there exists ε > 0 such that every |w − w
0
| < ε
has
m
preimages in |z − z
0
| < δ
.
Cauchy Theorems:
1. General thm: Suppose G is bounded with a regular contour, f : G → C is continuous and
holomorphic in G. Then
´
∂G
f (z) dz = 0
2. Isolated Singularities: G\ {a
1
, . . . , a
n
}
,
´
∂G
f = 2πi
P
res
a
k
f
3. Residue thm: Γ is a contour which doesn't run through any singularities. Then
´
Γ
f =
2πi
P
ind
Γ
(a
k
)
res
a
k
f
4. Using Winding Numbers: γ ⊂ G is closed, z
0
∈ G\γ
. Then ind
γ
(z
0
) f (z
0
) =
1
2πi
´
γ
f (z)
z−z
0
dz
5. Disk formula: if D ⊆ C is open and f is holomorphic inside a disk containing ¯
D
, then ∀z
0
∈ D
,
f (z
0
) =
1
2πi
´
∂D
f (z)
z−z
0
6. General formula: For said G and Hol (G) ∩ C G ⊃ f : G → C:
1
2πi
´
∂G
f (ζ)
ζ−z
dζ =
0
z /
∈ G
f (z)
z ∈ G
Hyperbolic Geometry
1. Aut (D) 3 f : D → D, γ : [a, b] → D piecewise C
1
. Then L
H
(γ) = L
H
(f ◦ γ)
.
2. If said f is not an autmorphism, then L
H
[f ◦ γ] ≤ L
H
[γ]
3. There exists a uniqe Geodesic between two points, and it is contained in a clircle orthogonal
to ∂D.
Equations
Arithmetic
1
z
=
z
|z|
2
; zw = zw |z| |w| = |zw|
Residues
•
For P a
n
z
n
about p, res
p
f = a
−1
• g, h
are holomorphic in an open set with z
0
, and suppose h has a simple pole. Then res
z
0
g
h
=
g(z
0
)
h
0
(z
0
)
.
•
If f is holomorphic in an open set with z
0
and it has a zero of order m there, then res
z
0
f
0
f
= m
•
if c is a pole of order n:
res
c
f =
1
(n − 1)!
lim
z→c
d
n−1
dz
n−1
((z − c)
n
f (z))
Residues at ∞
•
If lim
z→∞
f (z) = 0
,then res
∞
f = lim
z→∞
z · f (z)
•
res
∞
f =
res
0
−
1
w
2
f
1
w
•
res
z
0
f ◦ g =
res
z
0
(f (g (z
0
)) g
0
(z
0
))
Laurent Series
• r = lim
n→∞
|a
−n
|
1/n
•
1
R
= lim
n→∞
|a
n
|
1/n
Written by Omer Shapira, using course notes from lectures by Prof. Bo'az Klartag, Tel Aviv University
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