Taylor series and the uniqueness theorem
18 October 2026
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1 Taylor series
As we recall, holomorphy in a domain implies analyticity at every point \(z_0\) of the domain: the function expands into a power series in a neighborhood of \(z_0\). It turns out that this series is necessarily the Taylor series.
Theorem 1 If \(f\) is holomorphic in \(U(z_0)\) and \(f = \sum_{n=0}^{\infty }c_n(z-z_0)^n\) in \(U(z_0)\), then the coefficients of this series are determined uniquely: \[ c_n = \frac{f^{(n)}(z_0)}{n!} \] I.e. this is necessarily the Taylor series.
Expand \(\frac{1}{(2 - z)^3}\) into a power series about zero.
By the formula for the sum of a geometric series we have \[ \frac{1}{2- z} = \frac{1}{2} \frac{1}{1 - z/2} = \sum _{n=0}^{\infty } \frac{1}{2^{n+1}}z^n, \qquad \left|z\right| < 2 \] Power series may be differentiated term by term, and the radius of convergence does not change. We have \[ \begin{align} \frac{1}{(2-z)^2} & = \left(\frac{1}{2- z}\right)' = \sum _{n=0}^{\infty } \frac{1}{2^{n+1}}\left(z^n\right)' = \sum _{n=1}^{\infty } \frac{n}{2^{n+1}}z^{n-1} \\ 2\frac{1}{(2-z)^3} & = \left(\frac{1}{2- z}\right)^{''} = \sum _{n=0}^{\infty } \frac{1}{2^{n+1}}\left(z^n\right)'' = \sum _{n=2}^{\infty } \frac{n(n-1)}{2^{n+1}}\left(z^{n-2}\right) = \sum _{n=0}^{\infty } \frac{(n+2)(n+1)}{2^{n+3}}\left(z^{n}\right) \end{align} \]
\(\frac{1}{(2 - z)^3} = \sum_{n=0}^{\infty }\frac{(n+1)(n+2)}{2^{n+4}}z^n\), \(\left|z\right| < 2\).
Expand \(e^{2z} \sin (4z)\) into a power series about zero.
We have \[ e^{2z} \sin (4z) = e^{2z} \cdot \frac{e^{4iz} - e^{-4iz}}{2i} = \frac{e^{z(2 + 4i)} - e^{z(2 - 4i)}}{2i} \] It remains to expand each of the exponentials into a power series. Their radii of convergence are \(+\infty\), so the radius of convergence of the resulting series will also be \(+\infty\).
\(\sum_{n=0}^{\infty } \left(\frac{(4i + 2)^n - (2-4i)^n}{2in!}\right) z^{n}\), \(z \in \mathbb {C}\).
Expand the function \[ f(z) = \frac{z^{2}+2 z}{(z+1)^{3}} \] into a Taylor series about the point \(z = 1\).
Expand into a Taylor series about the point \(z = 0\) the functions:
- \(\cos^{3} z\);
Find the Taylor expansion about the point \(z = 0\) of the function satisfying the conditions given below:
- \(\left(1-z^{2}\right) f^{\prime \prime }(z)-5 z f^{\prime }(z)-4 f(z) = 0, \quad f(0) = 0, \quad f^{\prime }(0) = 1\).
2 Orders of zeros
Let \(f \in H(D)\), i.e. \(f\) is holomorphic in some domain \(D\). A point \(a\in D\) is called a zero of the function \(f\) if \(f(a) = 0\).
Let \(a\) be a zero of \(f\), and \(f \neq 0\). In some neighborhood \(U(a)\) the function \(f\) expands into a power series, and this is necessarily the Taylor series. Consider it:
\[ f(z) = c_{N}(z-a)^N + c_{N+1}(z-a)^{N+1} + \ldots , \]
where \(c_N \neq 0\), \(N \in \mathbb {N}\). \(N\) is called the order of the zero of the point \(z = a\) for the function \(f\). We also say that \(a\) is a zero of order \(N\), or an order-\(N\) zero.
Let \(z = a\) be a zero of order \(m\) for \(f\) and a zero of order \(k\) for \(g\). Prove that then the point \(a\) is a zero of order \((m+k)\) for \(fg\).
\(f(z) = (z-a)^m h(z)\), where \(h\) is holomorphic in a neighborhood of \(a\) and does not vanish at \(z = a\): \(h(z) = c_m + c_{m+1}(z - a) + \ldots\), \(h(a) = c_m \neq 0\). (This is in fact an equivalent definition of the order of a zero.) Similarly \(g(z) = (z-a)^{k}\tilde{h}(z)\), \(\tilde{h}(a) \neq 0\). Then \[ f(z)g(z) = (z-a)^{m+k} \cdot h(z)\tilde{h}(z), \qquad h(a)\tilde{h}(a) \neq 0, \] i.e. \(a\) is a zero of order \((m+k)\) for \(fg\).
Let \(f(z) = (e^{i\pi z} + 1)^3\). Find the order of the zero at \(a = -1\).
Substitute \(w = z + 1\). We need to find the order of the zero at \(w = 0\) for the function \((e^{i\pi (w-1)} + 1)^3\). We have \[ g(w) := e^{i\pi (w-1)} + 1 = e^{-i\pi }e^{i\pi w} + 1 = -e^{i\pi w} + 1 = -\sum _{n=1}^{\infty }\frac{(i\pi )^n}{n!}w^n \] I.e. \(w = 0\) is a zero of order \(1\) for \(g(w)\). Hence, by the previous example, \(w=0\) is a zero of order \(3\) for \(g^3(w)\).
\(3\).
Let \(f(z) = \left(\cos 3z - \cos 5z\right)^2\left(1 - \cos 2z\right)^3\). Find the order of the zero at \(a = 0\).
Let us expand the factors into Taylor series: \[ \begin{align} \cos 3z - \cos 5z & = \left(1 - \frac{(3z)^2}{2} + \ldots \right) - \left(1 - \frac{(5z)^2}{2} + \ldots \right) = \frac{5^2 - 3^2}{2}z^2 + \ldots = 8z^2 + \ldots \\ 1 - \cos 2z & = \frac{(2z)^2}{2} + \ldots = 2z^2 + \ldots \end{align} \] I.e. \(0\) is a zero of order 2 for both factors. The order of the zero of a product equals the sum of the orders (previous example), so the answer is \(2 \cdot 2 + 2 \cdot 3 = 10\).
\(10\).
All zeros of \(\sin , \cos , \operatorname {ch}, \operatorname {sh}\) have order 1.
Determine the order \(m\) of the zero \(z = a\) of the function \(f(z)\), if:
\(f(z) = (\cos 3 z-\cos 5 z)^{2}(1-\cos 2 z)^{3}, \quad a = 0\);
\(f(z) = \left(z^{4}+2 z^{3}-2 z-1\right)^{2}\left(e^{i \pi z}+1\right)^{3}, a = -1\).
3 The uniqueness theorem and its corollaries
Theorem 2 (The uniqueness theorem) Let \(D\) be a domain, \(E \subset D\), and let \(E\) have a limit point \(a\) in \(D\). Let \(f, g \in H(D)\). Then if they coincide on \(E\), they coincide on \(D\) as well: \[ f|_E \equiv g|_E \; \Rightarrow \; f \equiv g \]
That is, for two functions holomorphic in the same domain to coincide, it suffices that they coincide on some sequence in the domain having a limit point.
Does the uniqueness theorem remain valid if \(a \not\in D\)?
The key idea in the proof of the uniqueness theorem is expanding the function \(f - g\) into a power series in powers of \((z - a)\) in some neighborhood of \(a\). If \(a\) is not in the domain, this cannot be done, and the proof breaks down.
As a concrete example: let \(f(z) = \sin \left(\frac{1}{z}\right)\), \(g \equiv 0\). The functions \(f\), \(g\) are holomorphic in the domain \(D = \mathbb {C} \setminus \left\{ 0\right\}\), and coincide at least on the set
\[ E = \left\{ z = \frac{1}{\pi k}, \; k \in \mathbb {Z}\right\} \]
Yet these functions are different.
The uniqueness theorem is not applicable in this case: the limit point of \(E\) is the point \(0\), but it does not belong to \(D\).
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the graph of sin(1/x) on the real axis and the zeros 1/(pi k), accumulating at the punctured point 0. Interactive: zoom toward zero (log scale in x), a slider for the number of zeros shown, highlighting that the limit point 0 does not belong to the domain of holomorphy C\{0}.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/130/010_uniqueness_counterexample_fig.tex
Does there exist \(f \in H(D)\) for some neighborhood of zero \(D = U(0)\) such that \[ f\left(\frac{1}{n}\right) = \frac{\sin ^2 \frac{\pi n}{2}}{n}, \quad \forall n \in \mathbb {N} \]
We have \[ f\left(\frac{1}{2k}\right) = 0, \qquad f\left(\frac{1}{2k + 1}\right) = \frac{1}{2k + 1} \] Consider \(E_1 = \left\{ \frac{1}{2k}, \; k \in \mathbb {N}\right\}\), \(E_2 = \left\{ \frac{1}{2k+1}, \; k \in \mathbb {N}\right\}\). By the uniqueness theorem for \(g_1 \equiv 0\) and \(E_{1}\) we obtain that \(f \equiv 0\) on \(D\). By the uniqueness theorem for \(g_2(z) = z\) and \(E_{2}\) we obtain that \(f(z) = z\) on \(D\). Contradiction.
No.
Does there exist a function \(f(z)\), regular in some neighborhood of the point \(z = 0\), satisfying one of the following conditions (for all \(n = 1,2, \ldots )\):
\(f\left(\frac{1}{n}\right) = \frac{\cos^{2} \pi n}{2 n+1}\);
\(2^{-n} < \left|f\left(\frac{1}{n}\right)\right| < 2^{1-n}\);
\(\left|f\left(\frac{1}{n}\right)-\frac{\cos \pi n}{2 n+1}\right| < \frac{1}{n^{2}}\).
\(\cos^2 \pi n = 1\) for all \(n \in \mathbb {N}\), i.e. \(f\left(\frac{1}{n}\right) = \frac{1}{2n + 1} = \frac{1/n}{2 + 1/n}\). Consider the function
\[ f(z) = \frac{z}{2 + z} \]
Note that
\[ f(z) = z \cdot \frac{1}{2} \cdot \frac{1}{1 + z/2} = z \cdot \frac{1}{2} \cdot \sum _{n=0}^{\infty } \frac{(-1)^n}{2^n}z^n, \quad \left|z\right| < 2 \]
\(f(z)\) is represented as the sum of a power series in the open ball of radius \(2\) about zero. Hence it is regular there.
If \(f\) is regular in a neighborhood of \(z= 0\), then it is analytic there. I.e. there is some neighborhood \(U(0)\) in which \(f\) expands into the sum of a power series:
\[ f(z) = \sum _{n=N}^{\infty } a_n z^n = z^{N} \cdot \underbrace{\left(\sum _{k=0}^{\infty } a_{k+N} z^k\right)}_{:= h(z)}, \]
where \(a_N \neq 0\) (\(N \in \mathbb {Z}_+\)) is the first nonzero coefficient in the series (\(f \equiv 0\) in any neighborhood of zero is impossible in view of \(2^{-n} < \left|f(1/n)\right|\)). Note that \(h(z)\) is a function holomorphic in the same domain \(U(0)\), with \(h(0) = a_{N} \neq 0\). Moreover, we have
\[ 2^{-n} < \left|\frac{1}{n^{N}} \cdot h\left(\frac{1}{n}\right)\right| < 2 \cdot 2^{-n} \]
Further,
\[ n^{N} \cdot 2^{-n} < \left| h\left(\frac{1}{n}\right)\right| < 2 \cdot n^{N} \cdot 2^{-n} \xrightarrow [n \to \infty ]{} 0, \]
i.e. \(\left|h(1/n)\right| \to 0\). Holomorphy implies continuity. We arrive at a contradiction with \(h(0)\neq 0\).
We have \(\cos \pi n = (-1)^n\). The condition:
\[ \left|f\left(\frac{1}{n}\right)-\frac{\cos \pi n}{2 n+1}\right| < \frac{1}{n^{2}} \iff \left|f\left(\frac{1}{n}\right)-\frac{(-1)^n \cdot \frac{1}{n}}{2+\frac{1}{n}}\right| < \left(\frac{1}{n}\right)^2 \]
Consider the functions
\[ g_1(z) = f(z) + \frac{z}{2 + z}, \qquad g_2(z) = f(z) - \frac{z}{2+z} \]
If \(f\) is regular in a neighborhood of \(z= 0\), then it is analytic there. I.e. there is some neighborhood \(U_{\varepsilon }(0)\) in which \(f\) expands into the sum of a power series in powers of \(z\). Without loss of generality, let \(\varepsilon < 2\). We know that in \(U_{2}(0)\) the function \(\frac{z}{2 + z}\) expands into a power series. By the properties of sums of power series we conclude that \(g_1\), \(g_2\) must also expand into power series in \(U_{\varepsilon }(0)\).
Let \(g_1(z) = \sum_{n=0}^{\infty }a_nz^n\) for \(z \in U_{\varepsilon }(0)\). For odd \(n\) we have \(\cos \pi n = -1\), i.e. the condition becomes \(\left|g_1\left(\frac{1}{n}\right)\right| < \frac{1}{n^2}\). We have
\[ \left|g_1(z)\right| < z^2 \qquad \text{for } z = \frac{1}{2n+1} \qquad \; \Rightarrow \; \qquad \left|\sum _{k=0}^{\infty }a_kz^{k-2}\right| < 1 \qquad \text{for } z = \frac{1}{2n+1} \]
It follows that \(a_0 = a_1 = 0\) (otherwise the sum of the series would not be bounded).
Similarly for \(g_2(z) = \sum_{n=0}^{\infty }b_nz^n\) (even \(n\), \(\cos \pi n = 1\)): we have
\[ \left|g_2(z)\right| < z^2 \qquad \text{for } z = \frac{1}{2n} \]
Consequently, \(b_0=b_1 = 0\).
Thus, in the expansion of \(g_1(z) - g_2(z)\) about zero, the coefficient of the first power is also absent. However
\[ g_1(z) - g_2(z) = \frac{2z}{2+z} = z \cdot \sum _{n=0}^{\infty } \frac{(-1)^n}{2^n}z^n = z - \frac{1}{2}z^2 + \frac{1}{4}z^3 - \ldots , \quad z \in U_{\varepsilon }(0) \]
We arrive at a contradiction, since the power-series expansion of a holomorphic function in a neighborhood of a point of the domain is determined uniquely.
yes.
no;
no.
Does there exist a function \(f(z)\), regular in some neighborhood of the point \(z = 0\), satisfying one of the following conditions (for all \(n = 1,2, \ldots )\):
\(f\left(\frac{1}{n}\right) = \sin \frac{\pi n}{2}\);
\(f\left(\frac{1}{n}\right) = \frac{1}{2 n+\cos \pi n}\);
\(\left|f\left(\frac{1}{n}\right)\right| < e^{-n}\);
\(n^{-5 / 2} < \left|f\left(\frac{1}{n}\right)\right| < 2 n^{-5 / 2}\);
A function of a complex argument \(f\) is called entire if it is holomorphic in the entire complex plane:
\[ f \in H(\mathbb {C}) \]
Prove that \(e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}\) for \(z_1, z_2 \in \mathbb {C}\).
What do we know? \(e^z = \sum_{n=0}^{\infty }\frac{1}{n!}z^n\). This power series converges everywhere, i.e. \(e^{z}\) is an entire function. We also know that \(e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}\) for \(z_1, z_2 \in \mathbb {R}\).
Fix \(z_1 \in \mathbb {R}\) and consider \(f(z) = e^{z_1 + z} - e^{z_1}e^{z}\). We know that \(f\) is an entire function, and that \(f \equiv 0\) on \(\mathbb {R}\). \(\mathbb {R} \subset D = \mathbb {C}\), and \(\mathbb {R}\) has a limit point in \(\mathbb {C}\). By the uniqueness theorem we obtain that \(f \equiv 0\) on \(\mathbb {C}\), i.e. \(e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}\) for \(z_1 \in \mathbb {R}, z_2 \in \mathbb {C}\).
Fix \(z_2 \in \mathbb {C}\) and consider \(f(z) = e^{z + z_2} - e^{z}e^{z_2}\). We know that \(f\) is an entire function, and that \(f \equiv 0\) on \(\mathbb {R}\) (by the previous point). By the uniqueness theorem we obtain that \(f \equiv 0\) on \(\mathbb {C}\), i.e. \(e^{z_1 + z_2} = e^{z_1} \cdot e^{z_2}\) for \(z_1 \in \mathbb {C}, z_2 \in \mathbb {C}\).
Relying on the validity of the formulas given below for real values of the variables, and on the uniqueness theorem, prove their validity for arbitrary complex values of these variables:
\(\operatorname {ch} 2 z = \operatorname {ch}^{2} z+\operatorname {sh}^{2} z\);
\(\sin \left(z_{1}+z_{2}\right) = \sin z_{1} \cos z_{2}+\cos z_{1} \sin z_{2}\);
Theorem 3 (Liouvilleβs theorem) \(f\) entire and bounded (i.e. \(\exists M > 0\): \(\left|f(z)\right| \leq M\) \(\forall z \in \mathbb {C}\)) \(\; \Rightarrow \;\) \(f \equiv \operatorname {const}\) on all of \(\mathbb {C}\).
Describe all entire doubly periodic \(f\), i.e. entire functions \(f\) such that \[ f(z+i) = f(z+1) =f(z), \qquad z \in \mathbb {C} \]
Note that \(f\) is uniquely determined by its values on the set
\[ z = x + iy \; : \; x,y \in [0,1) \]
Indeed, for all \(z \in \mathbb {C}\) we have
\[ f(z) = f(x + iy) =f((x - \lfloor x \rfloor ) + i(y - \lfloor y \rfloor )), \]
where \((x - \lfloor x \rfloor ), (y - \lfloor y \rfloor ) \in [0,1)\).
Consider \(f\) on the square \(z = x + iy\), \(x,y \in [0,1]\). This is a compact set, \(f\) is continuous, hence \(f\) is bounded on this set. By periodicity it is bounded on the entire complex plane as well. By Liouvilleβs theorem, it is identically constant.
\(f \equiv \operatorname {const}\)
Suppose all values of the entire function \(f\) lie in the upper half-plane. Find the form of this function.