University of Toronto – MAT334H1-F – LEC0101
Complex Variables
11 - Isolated singularities
Jean-Baptiste Campesato October 23rd, 2020
Definitions 1. Let𝑈 ⊂ ℂbe open and𝑧0∈ 𝑈. Assume that𝑓 ∶ 𝑈 ⧵ {𝑧0} → ℂis holomorphic/analytic.
• Either𝑓 is bounded in a neighborhood of𝑧0,
i.e.∃𝑀 ∈ ℝ, ∃𝑟 > 0, ∀𝑧 ∈ 𝐷𝑟(𝑧0) ∩ 𝑈 , |𝑓 (𝑧)| ≤ 𝑀, then we say that𝑧0is aremovable singularityof𝑓,
• or lim
𝑧→𝑧0|𝑓 (𝑧)| = +∞, then we say that𝑧0is apoleof𝑓,
• otherwise, if none of the above occurs, we say that𝑧0is anessential singularityof𝑓. Examples 2(Removable singularities).
• 𝑓 (𝑧) = 𝑧+𝑖
𝑧2+1onℂ ⧵ {0}with𝑧0= 0.
• 𝑓 (𝑧) = sin𝑧
𝑧 onℂ ⧵ {0}with𝑧0= 0.
Examples 3(Poles).
• 𝑓 (𝑧) = 1
𝑧 onℂ ⧵ {0}with𝑧0= 0.
• 𝑓 (𝑧) = 1
𝑧2−1on𝐷1(1)with𝑧0 = 1.
Examples 4(Essential singularities).
• 𝑓 (𝑧) =cos1𝑧 onℂ ⧵ {0}with𝑧0 = 0.
• 𝑓 (𝑧) = 𝑒1𝑧 onℂ ⧵ {0}with𝑧0 = 0.
Theorem 5(Theorem: Riemann’s removable singularity theorem).
Let𝑈 ⊂ ℂbe open and𝑧0∈ 𝑈. Assume that𝑓 ∶ 𝑈 ⧵ {𝑧0} → ℂis holomorphic/analytic. Then TFAE:
1. 𝑧0is a removable singularity of𝑓 (i.e. 𝑓 is bounded in a neighborhood of𝑧0) 2. lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑓 (𝑧) = 0
3. 𝑓 can be holomorphically/analytically extended on𝑈
(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ holomorphic/analytic such that𝑓|𝑈 ⧵{𝑧̃
0}= 𝑓) 4. 𝑓 can be continuously extended on𝑈
(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ continuous such that𝑓|𝑈 ⧵{𝑧̃
0}= 𝑓)
2 Isolated singularities
Proof. (1)⟹(2)⟹(3)⟹(4)⟹(1). The only non-trivial part is (2)⟹(3).
Define𝑔 ∶ 𝑈 → ℂby𝑔(𝑧0) = 0and𝑔(𝑧) = (𝑧 − 𝑧0)2𝑓 (𝑧)otherwise. Then
𝑧→𝑧lim0
𝑔(𝑧) − 𝑔(𝑧0) 𝑧 − 𝑧0 = lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑓 (𝑧) = 0 Hence𝑔is holomorphic on𝑈.
Since𝑔(𝑧0) = 𝑔′(𝑧0) = 0we may write 𝑔(𝑧) =
+∞
∑𝑛=2
𝑎𝑛(𝑧 − 𝑧0)𝑛 = (𝑧 − 𝑧0)2
+∞
∑𝑛=2
𝑎𝑛(𝑧 − 𝑧0)𝑛−2
where𝑧 ∈ 𝐷𝑟(𝑧0) ∩ 𝑈 for some𝑟 > 0.
Hence𝑓 (𝑧) =̃
+∞
∑𝑛=2
𝑎𝑛(𝑧 − 𝑧0)𝑛−2is a suitable holomorphic extension of𝑓 on𝐷𝑟(𝑧0) ∩ 𝑈. ■ Remark 6. Note that if𝑓 admits a continuous extension at𝑧0then it is holomorphic.
Examples 7.
• 𝑓 (𝑧) = 𝑧2+1
𝑧+𝑖 onℂ ⧵ {−𝑖}may be holomorphically extended toℂby𝑓 (𝑧) = 𝑧 − 𝑖.̃
• 𝑓 (𝑧) = sin𝑧
𝑧 onℂ ⧵ {0}may be continuous extended at0by setting𝑓 (0) = 1.̃
Theorem 8. Let𝑈 ⊂ ℂbe open and𝑧0 ∈ 𝑈. Assume that𝑓 ∶ 𝑈 ⧵ {𝑧0} → ℂis holomorphic/analytic. Then the followings are equivalent:
1. 𝑧0is a pole of𝑓, i.e. lim
𝑧→𝑧0|𝑓 (𝑧)| = +∞.
2. There exist𝑛 ∈ ℕ>0and𝑔 ∶ 𝑈 → ℂanalytic such that𝑔(𝑧0) ≠ 0and𝑓 (𝑧) = 𝑔(𝑧)
(𝑧 − 𝑧0)𝑛 on𝑈 ⧵ {𝑧0}.
3. 𝑧0is not a removable singularity of𝑓 and there exists𝑛 ∈ ℕ>0such that lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑛+1𝑓 (𝑧) = 0.
Proof. • (1)⟹(3):
Then lim
𝑧→𝑧0
1
𝑓 (𝑧) = 0and 𝑧0 is a removable singularity of 1/𝑓, so that we can extend it to a holomorphic functionℎ ∶ 𝑈 → ℂdefined byℎ(𝑧) =
{
1
𝑓 (𝑧) if𝑧 ≠ 𝑧0
0 otherwise (See Remark6).
Denote by 𝑛 ≔ 𝑚ℎ(𝑧0) ∈ ℕ>0the order of vanishing of ℎat𝑧0 (sinceℎ(𝑧0) = 0), thenℎ(𝑧) = (𝑧 − 𝑧0)𝑛 ̃ℎ(𝑧) where ̃ℎ ∶ 𝑈 → ℂis holomorphic and ̃ℎ(𝑧0) ≠ 0.
Then lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑛+1𝑓 (𝑧) = lim
𝑧→𝑧0
𝑧 − 𝑧0
̃ℎ(𝑧) = 0.
• (3)⟹(2):
Pick the smallest𝑛such that lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑛+1𝑓 (𝑧) = 0.
Define𝑔 ∶ 𝑈 ⧵ {𝑧0} → ℂby𝑔(𝑧) = (𝑧 − 𝑧0)𝑛𝑓 (𝑧). Then lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑔(𝑧) = 0and𝑧0is a removable singularity of𝑔, so that𝑔may be extended to a holomorphic function𝑔 ∶ 𝑈 → ℂby Theorem5.
Besides𝑔(𝑧0) = lim
𝑧→𝑧0𝑔(𝑧) = lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑛𝑓 (𝑧) ≠ 0by definition of𝑛.
• (2)⟹(1) Obvious.
■
MAT334H1-F – LEC0101 – J.-B. Campesato 3
Definition 9. The integer𝑛 ∈ ℕ>0in (2) is uniquely defined and we say that𝑓 admits apole of order𝑛at 𝑧0.
We saw in the previous proof that the order of the pole𝑧0is also:
• The order of vanishing of1/𝑓 at𝑧0.
• The smallest𝑛such that lim
𝑧→𝑧0(𝑧 − 𝑧0)𝑛+1𝑓 (𝑧) = 0.
Examples 10.
• 1is a pole of order1of𝑓 (𝑧) = 1
𝑧2−1 =
1 𝑧+1
𝑧−1
• 0is a pole of order3of𝑓 (𝑧) = 1
𝑧3
Theorem 11(Great Picard’s Theorem –Not part of MAT334).
Let𝑈 ⊂ ℂbe open,𝑧0 ∈ 𝑈 and𝑓 ∶ 𝑈 ⧵ {𝑧0} → ℂbe holomorphic/analytic.
If𝑧0 is an essential singularity of𝑓 then, on any punctured neighborhood of𝑧0,𝑓 takes all possible complex values, with at most a single exception, infinitely many times.
Example 12. 𝑓 ∶ ℂ ⧵ {0} → ℂdefined by𝑓 (𝑧) = 𝑒1𝑧 has an essential singularity at0.
It takes the value𝑤 ∈ ℂ ⧵ {0}at𝑧 = 1
Log(𝑤)+2𝑖𝜋𝑛, 𝑛 ∈ ℤ.
Remark 13. In the above proofs, we used in an essential manner that the function𝑓 was holomorphic in a punctured neighborhood of𝑧0, i.e. that there exists𝑟 > 0such that𝑓 is holomorphic on
𝐷𝑟(𝑧0) ⧵ {𝑧0} = {𝑧 ∈ ℂ ∶ 0 < |𝑧 − 𝑧0| < 𝑟}
Therefore, when we will work with functions having several singularities, we will need to assume that they areisolated.
Formally, let𝑈 ⊂ ℂbe open,𝑆 ⊂ 𝑈be thesingular locusand𝑓 ∶ 𝑈 ⧵ 𝑆 → ℂbe holomorphic.
We need that if𝑧0 ∈ 𝑆then𝑓 is holomorphic on𝐷𝑟(𝑧0) ⧵ {𝑧0}for a small𝑟 > 0.
Otherwise stated, that there exists a small disk centered at𝑧0which doesn’t contain another singular point.
To summarize,𝑆needs to satisfy∀𝑧0∈ 𝑆, ∃𝑟 > 0, 𝐷𝑟(𝑧0) ∩ 𝑆 = {𝑧0}.
If you take MAT327, it simply means that𝑆is discrete in𝑈.
We will only study isolated singularities, we won’t study wilder singular loci.
Examples 14.
• The function𝑓 ∶ ℂ ⧵ {±1} → ℂdefined by𝑓 (𝑧) = 1
𝑧2−1is holomorphic and has 2 isolated singularities at
−1and+1.
• Let𝑓 ∶ ℂ ⧵ ({𝜋𝑛1, 𝑛 ∈ ℤ} ∪ {0}) → ℂbe defined by𝑓 (𝑧) =cot1
𝑧. Then0is not an isolated singularity of𝑓:
ℜ ℑ
4 Isolated singularities
Definitions 15.Let𝑈 ⊂ ℂbe an open neighborhood of infinity, i.e. there exists𝑟 > 0such that{𝑧 ∈ ℂ ∶ |𝑧| > 𝑟} ⊂ 𝑈. Let𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.
Then∞is an isolated singularity of𝑓 (i.e.𝑓 is defined in a neighborhood of∞but not at∞).
1. Either𝑓 is bounded in a neighborhood of∞,
i.e.∃𝑀 ∈ ℝ, ∃𝑟 > 0, ∀𝑧 ∈ 𝑈 , |𝑧| > 𝑟 ⟹ |𝑓 (𝑧)| ≤ 𝑀, then we say that∞is aremovable singularityof𝑓,
2. or lim
𝑧→∞|𝑓 (𝑧)| = +∞, then we say that∞is apoleof𝑓,
3. otherwise, if none of the above occurs, we say that∞is anessential singularityof𝑓. Remark 16. Recall that the inversionℂ → ̂̂ ℂ,𝑧 ↦ 1
𝑧, swaps0and∞.
Hence, if we set𝑔(𝑧) = 𝑓 (1𝑧)then the type of singularity of𝑓at∞coincides with the type of singularity of 𝑔at0.
Example 17. Let𝑓 ∶ ℂ ⧵ {𝜋𝑛 ∶ 𝑛 ∈ ℤ} → ℂbe defined by𝑓 (𝑧) =cot𝑧.
Then∞is not an isolated singularity of𝑓:
ℜ ℑ
Spoiler: a first introduction to Laurent series Assume that𝑧0is a pole of order𝑛 ∈ ℕ>0of𝑓.
Then there exists𝑔 ∶ 𝑈 → ℂholomorphic/analytic such that𝑔(𝑧0) ≠ 0and𝑓 (𝑧) = 𝑔(𝑧)
(𝑧 − 𝑧0)𝑛 in a neighbor- hood of𝑧0.
Since𝑔is analytic at𝑧0, it may be expressed as a power series in a small neighborhood of𝑧0: 𝑔(𝑧) =
+∞
∑𝑘=0
𝑎𝑘(𝑧 − 𝑧0)𝑘 and since𝑔(𝑧0) ≠ 0we know that𝑎0≠ 0.
Therefore, in a punctured neighborhood of𝑧0, we may express𝑓 as 𝑓 (𝑧) =
+∞
𝑘≥−𝑛∑
𝑎𝑘+𝑛(𝑧 − 𝑧0)𝑘
= 𝑎0(𝑧 − 𝑧0)−𝑛+ 𝑎1(𝑧 − 𝑧0)−𝑛+1+ ⋯ + 𝑎𝑛+ 𝑎𝑛+1(𝑧 − 𝑧0) + 𝑎𝑛+2(𝑧 − 𝑧0)2+ ⋯
Note that the above expression has some negative exponents: it is a first example ofLaurent series, notion that we will study next week.
