Isolated singular points
18 October 2026
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Let \(f\) be holomorphic in a punctured neighborhood \(\dot{U}_{\varepsilon }(z_{0})\) of a point \(z_0 \in \mathbb {C} \cup \left\{ \infty \right\}\), but not holomorphic in any unpunctured neighborhood of \(z_0\). Then \(z_0\) is an isolated singular point (of single-valued character).
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Classification of isolated singular points
1 Remark: the Laurent series in a neighborhood of infinity
Consider the function \(f(z)\) in a punctured neighborhood of infinity
\[ \dot{U}_{\varepsilon }(\infty ) = \left\{ z \in \mathbb {C} \; : \; \left|z\right| > 1/\varepsilon \right\} . \]
The Laurent series in this neighborhood: \(\zeta := \frac{1}{z}\),
\[ f(z) = \sum _{-\infty }^{+\infty } c_n \zeta ^n = \underbrace{\sum _{n=0}^{+\infty } c_n \zeta ^n}_{\text{regular part}} + \underbrace{\sum _{n=1}^{+\infty } c_{-n}\frac{1}{\zeta ^n}}_{\text{principal part}} = \underbrace{\sum _{n=0}^{+\infty } c_n \frac{1}{z^n}}_{\text{regular part}} + \underbrace{\sum _{n=1}^{+\infty } c_{-n}z^n}_{\text{principal part}} \]
That is, expanding a function into a Laurent series in a neighborhood of \(\infty\) is the same as expanding it in powers of \(z\), except the terminology changes: the part with positive powers is now the principal part, and the part with nonpositive powers is the regular part.
Expand \(e^z\) into a Laurent series
in a neighborhood of zero;
in a neighborhood of \(\infty\).
\[ e^z = \underbrace{1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \ldots }_{\text{regular part}} \]
the principal part is zero here
\[ e^z = \underbrace{1}_{\substack {\text{regular}\\ \text{part}}} + \underbrace{z + \frac{z^2}{2!} + \frac{z^3}{3!} + \ldots }_{\text{principal part}} \]
Determine the character of the singular point \(z_0\) for the function
\(\dfrac {\operatorname {tg}z}{z}\), \(z_0 = 0\);
\(\dfrac {1}{z^3 \operatorname {tg}z}\), \(z_0 = 0\);
\(\dfrac {1}{z^3 \operatorname {tg}z}\), \(z_0 = \infty\).
The Laurent series (in this case simply the Taylor series) in some neighborhood of zero \(U_{\varepsilon }(0)\) (it does not even matter which \(\varepsilon\)) is known for \(\operatorname {tg}z\): \[ \operatorname {tg}(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 + \frac{17}{315}z^7 + \ldots \]
In \(\dot{U}_{\varepsilon }(0)\) we have \(\frac{\operatorname {tg}z}{z} = 1 + \frac{1}{3}z^2 + \frac{2}{15}z^4 + \frac{17}{315}z^6 + \ldots\). There are no negative powers, the principal part is zero, i.e. for this function \(0\) is a removable singularity.
We have
\[ \operatorname {tg}(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 + \frac{17}{315}z^7 + \ldots = z \cdot \left(1 + \underbrace{\frac{1}{3}z^2 + \frac{2}{15}z^4 + \frac{17}{315}z^6 + \ldots }_{=: h(z)}\right) \]
Further,
\[ \frac{1}{\operatorname {tg}(z)} = \frac{1}{z} \cdot \frac{1}{1 + h(z)} \]
Note that \(h(z)\) is holomorphic in some neighborhood of zero, with \(h(0) = 0\). Holomorphy implies continuity, and continuity implies boundedness. Hence there is some neighborhood of zero in which \(\left|h(z)\right| < 1\). In this neighborhood, by the geometric series formula, we have
\[ \begin{align} \frac{1}{1 + h(z)} & = \frac{1}{1 - (-h(z))} = 1 - h(z) + h^2(z) - h^3(z) + \ldots = \\ & = 1 - \left(\frac{1}{3}z^2 + \frac{2}{15}z^4 + \frac{17}{315}z^6 + \ldots \right) + \left(\frac{1}{3}z^2 + \frac{2}{15}z^4 + \ldots \right)^2 - \left(\frac{1}{3}z^2 + \ldots \right)^3 + \ldots = \\ & = 1 - \frac{1}{3}z^2 + \left(-\frac{2}{15} + \left(\frac{1}{3}\right)^2\right)z^4 + \left(-\frac{17}{315} + 2 \cdot \frac{1}{3} \cdot \frac{2}{15} - \left(\frac{1}{3}\right)^3\right)z^6 + \ldots = \\ & = 1 - \frac{1}{3}z^2 - \frac{1}{45}z^4 - \frac{2}{945}z^6 + \ldots \end{align} \]
Then the Laurent series for \(\frac{1}{z^3 \operatorname {tg}z}\) in the same neighborhood, but now punctured:
\[ \frac{1}{z^3 \operatorname {tg}(z)} = \frac{1}{z^4} \cdot \frac{1}{1 + h(z)} = \underbrace{\frac{1}{z^4}- \frac{1}{3}\frac{1}{z^2}}_{\text{principal part}} \underbrace{ - \frac{1}{45}- \frac{2}{945}z^2 + \ldots }_{\text{regular part}} \]
\(0\) is a pole of order \(4\) for \(\frac{1}{z^{3}\operatorname {tg}z}\).
The function \(\frac{1}{z^3\operatorname {tg}z} = \frac{\operatorname {ctg}z}{z^3}\) has poles at all zeros of sine: \(z = \pi k\), \(k \in \mathbb {Z}\). These accumulate toward \(\infty\), so no punctured neighborhood \(\dot{U}(\infty )\) is free of singular points: \(\infty\) is not an isolated singular point, but a limit point of poles. (The expansion from part 2 is valid only for \(0 < \left|z\right| < \frac{\pi }{2}\) β it cannot be βcontinuedβ into a neighborhood of infinity; this function simply has no Laurent series in \(\dot{U}(\infty )\).)
removable singularity;
pole of order \(4\);
\(\infty\) is not an isolated singular point (a limit point of the poles \(\pi k\)).
Prove that the point \(z = a\) is a removable singularity for the following functions:
\(\frac{\sin z}{z} \quad (a = 0)\);
\(\frac{1}{e^{z}-1}-\frac{1}{\sin z} \quad (a = 0)\);
Prove that the point \(z = a\) is a pole of the following functions:
- \(\frac{z}{\left(e^{z}-1\right)^{2}} \quad (a = 0)\);
Find all isolated singular points of single-valued character for the following functions and determine their type:
\(\frac{1}{z^{2}-1} \cos \frac{\pi z}{z+1}\);
\(e^{\operatorname {ctg} \frac{\pi }{z}}\);
Let \(z = z_0\) be a zero of order \(n_1\) for \(f_1\), and a zero of order \(n_2\) for \(f_2\), i.e.
\[ f_1(z) = (z-z_0)^{n_1} \cdot h_1(z), \quad f_2(z) = (z-z_0)^{n_2} \cdot h_2(z) \]
where \(h_1, h_2\) are holomorphic in some neighborhood of \(z_0\), and nonzero at the point \(z_0\) itself. Then
\[ \frac{f_1}{f_2} = (z-z_0)^{n_1-n_2} \cdot \frac{h_1(z)}{h_2(z)}, \]
where \(\frac{h_1}{h_2}\) is holomorphic in some neighborhood of \(z_0\), and nonzero at \(z_0\). Hence \(z = z_0\) for the function \(f_1/f_2\) is
a zero of order \((n_1-n_2)\), if \(n_1 > n_2\); also a removable singularity;
a removable singularity, if \(n_1 = n_2\);
a pole of order \((n_2-n_1)\), if \(n_1 < n_2\);
Let \(z = a\) be a removable singularity or a pole for \(f\) and for \(g\). Then for \(fg\) this is also a removable singularity or a pole. And for \(f/g\) as well (if \(g \not\equiv 0\)).
Let \(z = a\) be a removable singularity or a pole for \(f\), and an essential singularity for \(g\), with \(f \not\equiv 0\). Then for \(fg\) it is an essential singularity.
Theorem 1 (Sokhotskiβs theorem) Let \(a\) be an essential singularity for \(f\). Then for every \(A \in \bar{\mathbb {C}}\) there is a sequence from the punctured neighborhood \(\left\{ z_n\right\} \subset \dot{U}(a)\) such that \[ z_n \to a, \qquad f(z_n) \to A \]
Theorem 2 (Picardβs great theorem) Let \(a\) be an essential singularity for \(f\). Then for every \(A \in \mathbb {C}\) (with possibly one exception), and for every \(\dot{U}(a)\), the equation \(f(z) = A\) has infinitely many solutions in \(\dot{U}(a)\).
Example: \(e^z\), \(a = \infty\). \(e^z \neq 0\).
Find the singular points of the function \(f(z) = e^{\sin z}\), and classify them.
\(a = \infty\). This is an essential singularity: along the 2 sequences \(z^{(1)}_n = 2\pi n\), \(z^{(2)}_n = \frac{\pi }{2} + 2\pi n\), the limit is different: \(e^{\sin z^{(1)}_n} = e^0 = 1\), \(e^{\sin z^{(2)}_n} = e^1 = e\).
\(z = \infty\) is an essential singularity; there are no other singular points (\(f\) is entire).
Find the singular points of the function \(f(z) = \frac{1 - \cos (z)}{z^3}\), and classify them.
\(a = \infty\). This is an essential singularity: an essential singularity for \(1 - \cos (z)\), a pole of order \(3\) for \(z^3\).
\(a = 0\). \(1 - \cos (z) = 1 - 1 + \frac{z^2}{2} - \ldots\). For the numerator, \(a\) is a zero of order \(2\). For the denominator, it is a zero of order \(3\). Hence this is a pole of order \(1\).
\(0\) is a pole of order \(1\), \(\infty\) is an essential singularity.
Find the singular points of the function \(f(z) = \frac{1}{z^2 - 1} \cos \frac{\pi z}{z + 1}\) and classify them.
\(a = \infty\). \(\frac{\pi z}{z+1} = \frac{\pi }{1 + \frac{1}{z}} \to \pi\), i.e. the cosine tends to \(\cos \pi = -1\), while \(\frac{1}{z^2-1} \to 0\).
\(f(z) \to 0\). \(\infty\) is a removable singularity.
\(a = 1\). For \(\frac{1}{z^2 - 1}\) this is a pole of order \(1\).
For \(g(z) = \cos \frac{\pi z}{z+1}\): \(g(1) = \cos \frac{\pi }{2} = 0\), and moreover
\[ g'(z) = -\sin \left(\frac{\pi z}{z+1}\right) \cdot \frac{\pi }{(z+1)^2}, \qquad g'(1) = -\sin \left(\frac{\pi }{2}\right) \cdot \frac{\pi }{4} = -\frac{\pi }{4} \neq 0, \]
i.e. \(a = 1\) is a zero of order \(1\) for \(g\). A first-order zero cancels a first-order pole:
removable singularity.
\(a = -1\). For \(\frac{1}{z^2 - 1}\) this is a pole of order \(1\).
For the cosine: consider \(z = -1 + \varepsilon\).
\(\cos \frac{\pi (\varepsilon - 1)}{\varepsilon } = \cos (\pi - \frac{\pi }{\varepsilon }) = -\cos (\pi /\varepsilon )\) β the principal part in powers of \(\frac{1}{\varepsilon } = \frac{1}{z+1}\) is infinite. I.e. this is an essential singularity.
Multiplying by \(\frac{1}{z^2 - 1} \not\equiv 0\) (a pole) preserves the essential singularity (rule 3 above). I.e. it is an essential singularity for \(f(z)\).
\(z = 1\) and \(z = \infty\) are removable singularities, \(z = -1\) is an essential singularity.
Find the singular points of the function \[ f(z) = \frac{1}{z\sin \frac{1}{z}} \] and classify them.
\(a = \infty\). In some punctured neighborhood of infinity we have
\[ \begin{align} f(z) & = \frac{1}{z} \cdot \frac{1}{\frac{1}{z} - \frac{1}{3!}\frac{1}{z^3} + \frac{1}{5!}\frac{1}{z^5} - \ldots } = \frac{1}{1 - \left(\frac{1}{3!}\frac{1}{z^2} - \frac{1}{5!}\frac{1}{z^4} + \ldots \right)} = \\ & = \sum _{k=0}^{\infty }\left(\frac{1}{3!}\frac{1}{z^2} - \frac{1}{5!}\frac{1}{z^4} + \ldots \right)^k = 1 + \frac{1}{3!}\frac{1}{z^2} + \ldots \end{align} \]
β only nonpositive powers, no principal part. Removable singularity.
Alternatively, one could simply note that as \(z \to \infty\), \(\sin \frac{1}{z} \sim \frac{1}{z}\), i.e.
\[ f(z) = \frac{1}{z\sin \frac{1}{z}} \xrightarrow [z \to \infty ]{} 1 \]
The points where sine vanishes: \(\frac{1}{z} = \pi k\). I.e. \(a = a_k = \frac{1}{\pi k}\), \(k \in \mathbb {Z} \setminus \left\{ 0\right\}\). \(a_k\) is a zero of some order for \(\sin \frac{1}{z}\). Differentiating, we get \(-\frac{1}{z^2}\cos \frac{1}{z}\). At the point \(a_k\) this is not zero. I.e. \(a_k\) is a zero of order \(1\) for \(\sin \frac{1}{z}\). Hence it is a pole of order \(1\) for \(f\).
\(a = 0\). This is a non-isolated singular point: a limit point of the poles \(a_k\).
\(\infty\) is a removable singularity; \(\frac{1}{\pi k}\), \(k \in \mathbb {Z} \setminus \left\{ 0\right\}\), are poles of order \(1\); \(0\) is a non-isolated singular point (a limit point of poles).
Find all singular points of the function \(f(z)\) and determine their type, if:
- \(f(z) = \frac{8 z^{2}-6 z+1}{1-\cos \frac{\pi }{z}}\);
Let \(z = 0\) be a pole of the function \(f(z)\). Prove that for the function \(e^{f(z)}\) the point \(z = 0\) is essentially singular.
Find all singular points of the function \(f(z)\) and determine their type, if:
- \(f(z) = \frac{\operatorname {sh} \pi z-\cos \frac{i \pi }{z}}{\left(i-e^{\pi / z}\right)^{2}}\);
A typical exam-style problem:
Find the singular points of the function \[ f(z) = \frac{4z^2 + \pi ^2}{\operatorname {ch}3z}\left(\exp \left(\frac{z}{z-2}\right) - 1\right)\operatorname {ctg}^2z \] and classify them.
Denote
\[ \begin{align} f_1(z) & = 4z^2 + \pi ^2 & f_3(z) & = \exp \left(\frac{z}{z-2}\right) - 1 = \exp \left(1 + 2 \cdot \frac{1}{z-2}\right) - 1 \\ f_2(z) & = \operatorname {ch}3z& f_4(z) & = \operatorname {ctg}z \end{align} \]
We have
\[ f = \frac{f_1}{f_2} \cdot f_3 \cdot f_4^2 \]
Singular points of \(f\):
\(a = 2\). We have
\[ f_3(z) = e \cdot \exp \left(2\frac{1}{z-2}\right) - 1= e \cdot \sum _{k=0}^{\infty }\frac{2^k}{k!}\left(\frac{1}{z-2}\right)^k - 1 \]
\(a = 2\) is an essential singularity for \(f_3\).
Alternatively, one can justify that \(a = 2\) is an essential singularity for \(f_3\) as follows:
\[ f_3(z) \xrightarrow [z = x \to 2-]{} -1, \qquad f_3(z) \xrightarrow [z = x \to 2+]{} \infty \]
i.e. \(f_3(z)\) has no limit as \(z \to 2\).
The remaining functions \(f_1,f_2,f_4\) have neither zeros nor singularities at \(a=2\). In total, \(a = 2\) is an essential singularity for \(f\)
The points where the denominator \(f_2(z) = \operatorname {ch}3z\) vanishes.
\[ 0 = \operatorname {ch}3z = \frac{e^{3z} + e^{-3z}}{2} \]
We get
\[ e^{6z} + 1 = 0 \iff e^{6z} = -1 \iff 6z = \operatorname {Log}(-1) = i\left(\pi + 2\pi k\right) \]
\(a = a_k = i\left(\frac{1}{6}\pi + \frac{1}{3}\pi k\right)\). What order do these zeros have for \(f_2\)? Let us take the derivative: \(f_2'(z) = 3\operatorname {sh}3z\), which is nonzero at the points \(a\). Accordingly, \(a\) is a zero of order \(1\) for \(f_2\).
What will these points be for the remaining functions? For \(f_3\), \(f_4\) they are neither zeros nor singularities (see above). For \(f_1(z) = 4z^2 + \pi^2\) the zeros are \(\pm i \frac{\pi }{2}\), of order \(1\). Consider the cases:
\(a = a_k = i\left(\frac{1}{6}\pi + \frac{1}{3}\pi k\right)\), \(k \in \left\{ 1,-2\right\}\), i.e. \(a = \pm i \frac{\pi }{2}\). This is a removable singularity for \(f\).
\(a = a_k = i\left(\frac{1}{6}\pi + \frac{1}{3}\pi k\right)\), \(k \not\in \left\{ 1,-2\right\}\). This is a pole of order \(1\) for \(f\).
The points where \(f_4(z) = \operatorname {ctg}z\) is undefined. These are \(a_4 = \pi k\). For \(\operatorname {ctg}z = \frac{\cos z}{\sin z}\) this is a pole of order \(1\), since the numerator \(\cos\) is nonzero there, while for the denominator, \(\sin\), it is a zero of order \(1\).
For \(f_1\), \(f_2\) these are not zeros, and there are no singularities there. For \(f_3\) there are some subtleties:
for \(k \neq 0\), i.e. for \(a = \pi k \neq 0\), \(f_3\) has neither zeros nor singularities at \(a\). In total, in this case, for \(f\) the points of the form \(\pi k\), \(k \neq 0\), are poles of order \(2\).
for \(k = 0\), i.e. for \(a = 0\), \(f_3\) has a zero. Of what order? Let us differentiate:
\[ f_3'(z) = -2\frac{1}{(z-2)^2} \exp \left(1 + 2 \cdot \frac{1}{z-2}\right) \]
As we see, \(f'_3(z)|_{z = 0} \neq 0\), i.e. \(a = 0\) is a zero of order \(1\) for \(f_3\).
In total, for \(f\) the point \(a = 0\) is a pole of order \(1\).
\(a = \infty\). A non-isolated singular point, a limit of poles, in particular, of poles of the form \(i\left(\frac{1}{6}\pi + \frac{1}{3}\pi k\right)\)
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the singular points of the function, marked with colors.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/210/010_singular_points_colored_fig.tex
\(\infty\): non-isolated singular point, a limit point of the poles below
\(2\) is an essential singularity
\(\pm i \frac{\pi }{2}\) are removable singularities
\(0\) is a pole of order \(1\);
\(\frac{i}{3}\left(\frac{\pi }{2} + \pi k\right)\), \(k \not\in \left\{ 1,-2\right\}\), are poles of order \(1\).
\(\pi k\), \(k \neq 0\), are poles of order \(2\).
Find all singular points of the function \(f\) and determine their type, if \[ f(z) = \frac{\cos ^{2}\left(z+\frac{1}{z}\right)-\sin ^{2}\left(z-\frac{1}{z}\right)}{\sin (2 z) \cos \frac{2}{z}} \operatorname {th} \frac{1}{z+i} \]
Find the principal part of the Laurent series of the function \[ f(z) = \frac{1}{e^{z} - 1} \] in the annulus centered at the origin that contains the point \(z_0 = 10\).
Singular points: \(\operatorname {Ln}1 = 2\pi k i\), \(k \in \mathbb {Z}\). We must expand in the annulus
\[ D = \left\{ z \in \mathbb {C} \; : \; 2\pi < \left|z\right| < 4 \pi \right\} \]
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the singular points of the function (in red).
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/210/020_singular_points_fig.tex
How do we do this?
Consider \(f(z)\) in some \(\dot{U}(0)\):
\[ f(z) = \frac{1}{z} \cdot \overbrace{\frac{1}{1 + \frac{z}{2!} + \frac{z^2}{3!} + \ldots }}^{=: h(z)} = \frac{1}{z} \cdot h(z) \]
\(h\) is a function holomorphic at zero (unlike \(f\), it no longer has a singularity at zero), i.e. it expands in some neighborhood of zero simply into a Taylor series:
\[ h(z) = h(0) + h'(0)z + \ldots = 1 + h'(0)z + \ldots \]
Hence in \(\dot{U}(0)\) we have
\[ f(z) = \frac{1}{z} + \frac{1}{z}\left(h(z) - 1 \right) = \frac{1}{z} + \underbrace{h'(0) + \frac{h''(0)}{2}z + \ldots }_{=: f_1(z)} \]
where \(f_1(z)\) is holomorphic at zero.
In \(\dot{U}(2\pi i)\):
\[ f(z) = \frac{1}{e^{z} - 1} = \frac{1}{e^{z - 2\pi i} - 1} = \frac{1}{e^{w} - 1} = \frac{1}{w} + f_1(w) = \frac{1}{z - 2\pi i} + f_2(z) \]
where \(w = z - 2\pi i\), \(f_1\) is from the previous point, \(f_2(z) = f_1(z - 2\pi i)\), and \(f_2(z)\) is holomorphic in a neighborhood of \(2 \pi i\).
In \(\dot{U}(-2\pi i)\):
\[ f(z) = \frac{1}{z + 2\pi i} + f_3(z), \]
\(f_3\) is holomorphic in a neighborhood of \(-2\pi i\).
In total, the function
\[ g(z) = \frac{1}{e^z - 1} - \frac{1}{z} - \frac{1}{z + 2\pi i} - \frac{1}{z - 2\pi i} \]
is a function that, in the open disk \(\left\{ \left|z\right| < 4 \pi \right\}\), has only removable singular points. I.e. it is regular in this neighborhood (once redefined by continuity). It can be expanded in a Taylor series! Then in \(\left\{ \left|z\right| < 4 \pi \right\}\) (in particular, in the annulus \(\left\{ 2 \pi <\left|z\right| < 4 \pi \right\}\)), \(g(z)\) has no principal part (it is zero). Consequently, the principal part of the Laurent series for \(f(z)\) is the principal part of the Laurent series for
\[ \frac{1}{z} + \frac{1}{z + 2\pi i} + \frac{1}{z - 2\pi i} \]
Meaning: only the singular points contained within the annulus contribute to the principal part of the Laurent series.
Let us carry the problem through to the end:
\[ \begin{align} \frac{1}{z} + \frac{1}{z + 2\pi i} + \frac{1}{z - 2\pi i} & = \frac{1}{z} + \frac{2z}{z^2 + 4\pi ^2} = \frac{1}{z} + \frac{2z}{z^2} \cdot \frac{1}{1 - (- 4\pi ^2/z^2)} = \\ & = \frac{1}{z} + \frac{2}{z} \cdot \sum _{n=0}^{\infty } (-1)^n \cdot \left(4\pi ^2\right)^{n} \cdot \frac{1}{z^{2n}} = \\ & = \frac{3}{z} + \sum _{n=1}^{\infty }(-1)^n \cdot 2^{2n+1} \cdot \pi ^{2n} \frac{1}{z^{2n + 1}} \end{align} \]
\(\frac{3}{z} + \sum_{n=1}^{\infty }(-1)^n \cdot 2^{2n+1} \cdot \pi^{2n} \cdot \frac{1}{z^{2n + 1}}\).
Find the principal part of the Laurent series of the function \[ f(z) = \frac{1}{\sin z} \] in the annulus centered at the origin that contains the point \(z = 4\).
Let the function \(f\) be holomorphic in the punctured unit disk \(0 < \left|z\right| < 1\), and suppose that for some \(A > 0\) and \(\alpha \in [0,1]\) the inequality \(\left|f(z)\right| < \frac{A}{\left|z\right|^{\alpha }}\) holds. Determine the type of the isolated singular point \(z = 0\) for various values of \(\alpha\).