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\(\sum_{n=0}^{\infty }c_n(z-z_0)^n\) β a power series. Properties:
There exists a unique \(R \in [0, +\infty ]\) such that the series converges for \(\left|z - z_0\right| < R\) and diverges for \(\left|z - z_0\right| > R\) (\(R\) is the radius of convergence, the disk \(\left|z - z_0\right| < R\) is the disk of convergence);
The series converges uniformly on every compact subset of the disk of convergence;
A power series may be differentiated term by term, and the radius of convergence does not change.
Corollary: the sum of the series is holomorphic in the disk of convergence.
Formula:
\[
R = \left(\varlimsup _{n \to \infty }\left|c_n\right|^{1/n}\right)^{-1}
\]
Find the radius of convergence of \[
\sum _{m=0}^{\infty } \left(\frac{27 - 26i^m}{27 + 26i^m}\right)^{m^2} z^{m^2}
\]
We have a power series \(\sum_{n=0}^{\infty }c_n z^n\), where \[
c_n = \begin{cases} \left(\frac{27 - 26i^m}{27 + 26i^m}\right)^{m^2}, & n = m^2 \text{ for some }m \in \mathbb {Z}_+ \\ 0, & n \text{ is not a perfect square} \end{cases}
\] Note that for \(n = m^2\) \[
\left|c_n\right|^{1/n} = \left|\left(\frac{27 - 26i^m}{27 + 26i^m}\right)^{m^2}\right|^{1/n} = \left|\frac{27 - 26i^m}{27 + 26i^m}\right| = \begin{cases} \frac{1}{53}, & m \equiv _4 0 \\ 1, & m \equiv _4 1 \\ 53, & m \equiv _4 2 \\ 1 , & m \equiv _4 3 \end{cases}
\] Consequently, \[
\varlimsup _{n \to \infty }\left|c_n\right|^{1/n} = 53, \qquad \text{i.e. } R = \frac{1}{53}
\]
A function is called analytic in a domain \(D \subset \mathbb {C}\) if for every point \(z_0 \in D\) there exists a neighborhood \(U(z_0)\) in which the function can be represented as a power series (in powers of \((z - z_0)\)).
Theorem 1 A function \(f\) is holomorphic in \(D\) \(\iff\) \(f\) is analytic in \(D\).
Find the radii of convergence of the power series:
\(\sum_{n = 1}^{\infty }\left(\frac{1-i n}{1+i n}\right)^{n^{2}} z^{n}\);
\(\sum_{n = 1}^{\infty }\left(1+i^{n}\right)^{n} z^{n}\);
\(\sum_{n = 1}^{\infty } 2^{n} z^{n!}\);
\(\sum_{n = 1}^{\infty }\left((1+i)^{n}+(1-i)^{n}\right) z^{n}\).
The complex exponential, trigonometric functions
\(e^{z} = \exp (z) = \sum_{n=0}^{\infty }\frac{z^n}{n!}\).
\[
\begin{align} \sin (z) & := \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos (z) = \frac{e^{iz} + e^{-iz}}{2} \\ \operatorname {sh}(z) & := \frac{e^{z} - e^{-z}}{2}, \quad \operatorname {ch}(z) := \frac{e^{z} + e^{-z}}{2} \end{align}
\]
Find all solutions of the following equations:
- \(\operatorname {ctg} z = -\frac{3 i}{5}\);
Essentially we need to find \(\operatorname {Arcctg}\left(-\frac{3}{5}i\right)\).
\[
\begin{align} -\frac{3}{5}i & = \operatorname {ctg}z = \frac{\cos z}{\sin z} = \frac{2i}{2} \frac{\operatorname {exp}\left(iz\right) + \operatorname {exp}\left(-iz\right)}{\operatorname {exp}\left(iz\right) - \operatorname {exp}\left(-iz\right)} \\ 3\operatorname {exp}\left(-iz\right) - 3\operatorname {exp}\left(iz\right) & = 5\operatorname {exp}\left(iz\right) + 5\operatorname {exp}\left(-iz\right) \\ 8\operatorname {exp}\left(2iz\right) & = -2 \\ \operatorname {exp}\left(-2y + 2ix\right) & = -\frac{1}{4} = \frac{1}{4} \cdot \operatorname {exp}\left(i\pi \right) \end{align}
\]
I.e. \(y = -\frac{1}{2}\ln \frac{1}{4} = \ln (2)\), \(2x = \pi + 2\pi k\),
\[
z = \frac{\pi }{2} + \pi k + i\ln (2), \quad k \in \mathbb {Z}
\]
\(z = i \ln 2+\pi \left(k+\frac{1}{2}\right), k \in \mathbb {Z}\);
Formulas:
\[
\begin{align} \cos (z_1 + z_2) & = \cos (z_1)\cos (z_2) - \sin (z_1)\sin (z_2) \\ \sin (z_1 + z_2) & = \sin (z_1)\cos (z_2) + \cos (z_1)\sin (z_2) \\ \operatorname {ch}(z_1 + z_2) & = \operatorname {ch}z_1 \operatorname {ch}z_2 + \operatorname {sh}z_1\operatorname {sh} z_2 \\ \operatorname {sh}(z_1 + z_2) & = \operatorname {sh}z_1 \operatorname {ch}z_2 + \operatorname {ch}z_1\operatorname {sh} z_2 \end{align}
\]
Easy to prove if one knows the multiplicativity of the exponential.
Also:
\[
\begin{align} \operatorname {sh} z & = -i \sin (iz) & \operatorname {ch}z & = \cos (iz) \\ \operatorname {th} z & = -i \operatorname {tg}(iz), & \operatorname {cth}z & = i\operatorname {ctg}(iz) \end{align}
\]
Let \(x = \operatorname {Re} z, y = \operatorname {Im} z\). Find \(\operatorname {Re} \cos z\) and \(\operatorname {Im} \cos z\).
We have \[
\cos z = \cos (x + iy) = \cos (x)\cos (iy) -\sin (x) \sin (iy) = \cos x \operatorname {ch}y - i \sin x \operatorname {sh}y
\]
\(\operatorname {Re} \cos z = \cos x \cdot \operatorname {ch} y, \quad \operatorname {Im} \cos z = -\sin x \cdot \operatorname {sh} y\).
Let \(x = \operatorname {Re} z, y = \operatorname {Im} z\). Find \(\left|\cos z\right|\).
\[
\begin{align} \left|\cos z\right|^2 & = \cos ^2(x) \operatorname {ch}^2(y) + \sin ^2(x) \operatorname {sh}^2(y) = \operatorname {ch}^2(y) - \sin ^2(x)\operatorname {ch}^2(y) + \sin ^2(x) \operatorname {sh}^2(y) = \\ & =\operatorname {ch}^2(y) - \sin ^2(x) \left(\operatorname {ch}^2(y) - \operatorname {sh}^2(y)\right) = \operatorname {ch}^2(y) - \sin ^2(x) \end{align}
\]
\(\left|\cos z\right| = \sqrt{\operatorname {ch}^2 y - \sin^2 x}\).
Prove the formulas:
\(\operatorname {ch}^{2} z-\operatorname {sh}^{2} z = 1\);
\(\operatorname {ch}\left(z_{1}+z_{2}\right) = \operatorname {ch} z_{1} \operatorname {ch} z_{2}+\operatorname {sh} z_{1} \operatorname {sh} z_{2}\);
Prove the formulas:
\(\operatorname {sh} z = -i \sin (i z)\);
\(\cos (i z) = \operatorname {ch} z\);
Let \(x = \operatorname {Re} z, y = \operatorname {Im} z\). Prove that:
\(\operatorname {Re} \cos z = \cos x \cdot \operatorname {ch} y, \quad \operatorname {Im} \cos z = -\sin x \cdot \operatorname {sh} y\);
\(\operatorname {Re} \operatorname {sh} z = \operatorname {sh} x \cdot \cos y, \quad \operatorname {Im} \operatorname {sh} z = \operatorname {ch} x \cdot \sin y\);
We have
\[
\cos z = \cos (x + iy) = \cos (x)\cos (iy) -\sin (x) \sin (iy) = \cos x \operatorname {ch}y - i \sin x \operatorname {sh}y
\]
We have
\[
2\operatorname {sh}z = e^{z} - e^{-z} = \underbrace{\left(e^{x} - e^{-x}\right)}_{=2\operatorname {sh}x}\cos (y) + i\underbrace{\left(e^{x} + e^{-x}\right)}_{=2\operatorname {ch}x}\sin (y)
\]
Let \(x = \operatorname {Re} z, y = \operatorname {Im} z\). Prove that:
- \(\operatorname {Re} \sin z = \sin x \cdot \operatorname {ch} y, \quad \operatorname {Im} \sin z = \cos x \cdot \operatorname {sh} y ;\)
Find all solutions of the following equations:
- \(\cos z = \frac{3 i}{4}\);
\(\frac{3}{2}i = e^{iz} + e^{-iz} = e^{-y + ix} + e^{y - ix} = e^{-y}\cdot \left(\cos (x) + i\sin (x)\right) + e^{y} \cdot \left(\cos (x) - i\sin (x)\right)\), i.e.
\[
\begin{aligned}
\begin{cases}
(e^{-y} + e^{y}) \cos(x) = 0 \\
(e^{-y} - e^{y}) \sin(x) = \frac{3}{2}
\end{cases} &\iff
\begin{cases}
\cos(x) = 0 \\
(e^{-y} - e^{y}) \sin(x) = \frac{3}{2}
\end{cases}
\left[ \begin{array}{c}
\begin{cases}
x = \frac{\pi}{2} + 2\pi k \\
e^{-y} - e^{y} = \frac{3}{2}
\end{cases}
\\
\begin{cases}
x = -\frac{\pi}{2} + 2\pi k \\
e^{-y} - e^{y} = -\frac{3}{2}
\end{cases}
\end{array}\right. \iff \\
&\iff \left[ \begin{array}{c}
\begin{cases}
x = \frac{\pi}{2} + 2\pi k \\
y = -\ln(2)
\end{cases}
\\
\begin{cases}
x = -\frac{\pi}{2} + 2\pi k \\
y = \ln(2)
\end{cases}
\end{array} \right.
\end{aligned}
\] (the first system used \(e^{-y} - e^{y} = -2\operatorname {sh}(y) = \frac{3}{2}\) \(\iff\) \(\operatorname {sh}(y) = -\frac{3}{4}\) \(\iff\) \(y = \ln \left(-\frac{3}{4} + \sqrt{\frac{9}{16}+1}\right) = -\ln 2\); the second is solved symmetrically). In total
\[
z = \pm \left(\frac{\pi }{2} - i \ln 2\right) + 2\pi k, \quad k \in \mathbb {Z}
\]
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the solution set of cos z = 3i/4 β two series of points on the lines Im z = βln2 with step 2pi. Interactive: input/slider for the right-hand side w (complex), regenerating the series of solutions of cos z = w; highlight the symmetry z -> -z and the periodicity z -> z + 2pi.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/120/010_cos_equation_solutions_fig.tex
\(z = \pm \left(-i \ln 2+\frac{\pi }{2}\right)+2 k \pi , k \in \mathbb {Z}\);
Find all solutions of the following equations:
\(\sin z = \frac{4 i}{3}\);
\(\cos z = \frac{3+i}{4}\);
\(\operatorname {ctg} z = -\frac{3 i}{5}\);
Let the function \(f(z) = u(x, y)+i v(x, y)\) be differentiable in a domain \(G \subset \mathbb {C}\), and suppose one of the functions
- \(r(x, y) = |f(z)|\); keeps a constant value in the domain \(G\). Prove that \(f(z) \equiv \operatorname {const}\).
We have \(u^2 + v^2 \equiv \operatorname {const}\) in the domain \(G\). If this constant equals \(0\), then \(f \equiv 0\), and we are done. Otherwise we differentiate the identity with respect to \(x\) and \(y\), and also apply the Cauchy-Riemann conditions:
\[
\begin{cases} 0 = 2uu_x + 2vv_x = 2uv_y + 2vv_x \\ 0 = 2uu_y + 2vv_y = -2uv_x + 2vv_y \end{cases}
\]
This is a linear system for the pair \((v_x, v_y)\) with determinant
\[
\begin{vmatrix} v & u \\ -u & v \end{vmatrix} = u^2 + v^2 \neq 0,
\]
i.e. \(v_x = v_y = 0\) throughout the domain \(G\). Taking into account the differentiability of \(v\), we obtain that \(v \equiv \operatorname {const}\) on \(G\). By the Cauchy-Riemann conditions also \(u_x = v_y = 0\) and \(u_y = -v_x = 0\), i.e. \(u \equiv \operatorname {const}\) as well.
Let the function \(f(z) = u(x, y)+i v(x, y)\) be differentiable in a domain \(G \subset \mathbb {C}\), and suppose one of the functions
\(u(x, y)\);
\(\varphi (x, y) = \arg f(z)\) keeps a constant value in the domain \(G\). Prove that \(f(z) \equiv \operatorname {const}\).
Recover the regular function \(f(z)\) from the condition
- \(\operatorname {Im} f(z) = x \sin y \operatorname {ch} x+y \operatorname {sh} x \cos y, f(0) = 0\);
\(f(z) = u(x,y) + iv(x,y)\), and moreover
\[
\begin{align} v(x,y) & = x \sin y \operatorname {ch} x+y \operatorname {sh} x \cos y = \\ & =x \cdot \operatorname {Im}\operatorname {sh}z + y \cdot \operatorname {Re}\operatorname {sh}z \end{align}
\]
The first Cauchy-Riemann condition:
\[
\begin{align} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} & = x \operatorname {ch}x \cos y + \operatorname {sh}x \cos y - y \operatorname {sh}x \sin y = \\ & = \cos y \cdot x\operatorname {ch}x + \left(\cos y - y \sin y\right)\operatorname {sh}x \end{align}
\]
Consequently,
\[
\begin{align} u(x,y) & = \cos y \cdot \left(x \operatorname {sh}x - \operatorname {ch}x\right) + \left(\cos y - y \sin y\right)\operatorname {ch}x + \varphi (y) = \\ & = x \cos y \operatorname {sh}x - y \operatorname {sin} y \operatorname {ch}x + \varphi (y) = \\ & = x \operatorname {Re}\operatorname {sh}z - y \operatorname {Im}\operatorname {sh}z + \varphi (y) \end{align}
\]
The second Cauchy-Riemann condition:
\[
\begin{align} -\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} & = \sin y \operatorname {ch}x + x \sin y \operatorname {sh}x + y \cos y \operatorname {ch}x \end{align}
\]
At the same time
\[
\begin{align} -\frac{\partial u}{\partial y} & =\sin y \cdot x \operatorname {sh}x + \sin y \operatorname {ch}x + y \cos y \operatorname {ch}x - \varphi '(y) \end{align}
\]
I.e. \(\varphi '(y) = 0\), \(\varphi (y) = C\),
\[
u(x,y) = \cos y \cdot \left(x \operatorname {sh}x - \operatorname {ch}x\right) + \left(\cos y - y \sin y\right)\operatorname {ch}x + C
\]
Further, \(f(0) = 0\), i.e. \(0 = u(0,0) = -1 + 1 + C\), \(C = 0\). In total,
\[
\begin{align} f(z) & = u + iv = x \operatorname {Re}\operatorname {sh}z - y \operatorname {Im}\operatorname {sh}z + i \cdot \left(x \cdot \operatorname {Im}\operatorname {sh}z + y \cdot \operatorname {Re}\operatorname {sh}z\right) =\\ & =x \cdot \left(\operatorname {Re}\operatorname {sh}z + i \operatorname {Im}\operatorname {sh}z\right) + iy \cdot \left(\operatorname {Re}\operatorname {sh}z + i \operatorname {Im}\operatorname {sh}z\right) = x \cdot \operatorname {sh}z + iy \cdot \operatorname {sh}z = \\ & = z \cdot \operatorname {sh}z \end{align}
\]
\(z \operatorname {sh} z\);
Recover the regular function \(f(z)\) from the condition
\(\operatorname {Im} f(z) = y \operatorname {ch} x \cos y+x \sin y \operatorname {sh} x, f(0) = 1\);
\(\operatorname {Re} f(z) = x e^{x} \cos y-(y+1) e^{x} \sin y, \quad f(0) = i\);
Find the values \(\sin i, \operatorname {tg}(1+i), \operatorname {ctg}(\pi / 4-i \ln 2)\).
The complex logarithm, complex powers
\(\operatorname {Ln}(z)\) denotes all possible solutions of the equation \(e^{w} = z\). We have
\[
e^{w} = z \iff e^{\operatorname {Re}(w)} \cdot e^{i\operatorname {Im}(w)} = \left|z\right| \cdot e^{i\operatorname {arg}(z)}
\]
I.e.
\[
\begin{cases} \operatorname {Re}(w) = \ln (\left|z\right|) \\ \operatorname {Im}(w) = \operatorname {arg}(z) + 2 \pi k, \; k \in \mathbb {Z} \end{cases}
\]
Thus,
\[
\operatorname {Ln}(z) = \ln (\left|z\right|) + i \cdot \operatorname {Arg}z
\]
where
\[
\operatorname {Arg}z := \underbrace{\operatorname {arg}(z)}_{\in (-\pi , \pi ]} + 2 \pi k
\]
is the generalized argument. Both \(\operatorname {Arg}z\) and \(\operatorname {Ln}z\) are multivalued functions.
Power: \(a^b := \operatorname {exp}\left(b \cdot \operatorname {Ln}a\right)\), \(a,b \in \mathbb {C}\), \(a \neq 0\). A multivalued function.
Compute:
\(1^i\);
\(i^1\);
\(i^i\).
\(1^i = \operatorname {exp}\left(i\operatorname {Ln}1\right) = \operatorname {exp}\left(i \cdot 2\pi k i\right) = \operatorname {exp}\left(-2\pi k\right)\), \(k \in \mathbb {Z}\);
\(i^1 = \operatorname {exp}\left(\operatorname {Ln} i\right) = \operatorname {exp}\left(i\left(\frac{\pi }{2} + 2\pi k\right)\right) = i\);
\(i^i = \operatorname {exp}\left(i \operatorname {Ln} i\right) = \operatorname {exp}\left(-\frac{\pi }{2} - 2\pi k\right)\), \(k \in \mathbb {Z}\).
\(e^z\) can be understood in two ways: either as the single-valued function \(\operatorname {exp}z = \sum_{n=0}^{\infty } \frac{z^n}{n!}\), or as exponentiation, that is, as the multivalued function
\[
e^z = \operatorname {exp}\left(\operatorname {Ln}e \cdot z\right) = \operatorname {exp}\left(\left(1 + 2\pi k i\right) \cdot z\right) = \operatorname {exp}(z) \cdot \operatorname {exp}(2\pi k i z)
\]
For example, \(e^{i\pi } = - \operatorname {exp}\left(-2\pi^2 k\right)\), \(k \in \mathbb {Z}\), but \(\operatorname {exp}(i \pi ) = -1\).
As a rule, \(e^z\) is understood as \(\operatorname {exp}(z)\), i.e. the single-valued function, while \(a^{z}\), \(a \neq e\) is understood as the multivalued function, exponentiation.
Inverse trigonometric functions
\(\operatorname {Arctg}(z)\) denotes all possible solutions of the equation \(z = \operatorname {tg} w\). Let us solve it. We have
\[
z = \operatorname {tg}w = \frac{\sin w}{\cos w} = \frac{e^{i w} - e^{-iw}}{i(e^{iw} + e^{-iw})} = \frac{e^{2iw} - 1}{i(e^{2iw} + 1)}
\]
Then \(e^{2iw} (zi - 1) = -(1 + zi)\), i.e.
\[
w = \frac{1}{2i} \operatorname {Ln} \left(\frac{1 + zi}{1 - zi}\right) =: \operatorname {Arctg}(z)
\]
Find all values of \(2^{i}, i^{i},(-1)^{2 i}, \operatorname {Arctg}(5 i / 3)\).
Find the sum of the trigonometric series \[
\sum _{n = 0}^{\infty } \frac{\cos n \varphi }{n!}, \quad \varphi \in \mathbb {R}
\]