九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2015年8月実施 複素関数論 Author Zero, 祭音Myyura
Description 複素関数 f ( z ) = π cot π z z 2 + a 2 f(z) = \frac{\pi\cot\pi z}{z^2 + a^2} f ( z ) = z 2 + a 2 π c o t π z a > 0 a > 0 a > 0
(1) f ( z ) f(z) f ( z )
(2) 図に示す閉路 C N C_N C N ∮ C N f ( z ) d z \oint_{C_N} f(z)dz ∮ C N f ( z ) d z N N N lim N → ∞ ∮ C N f ( z ) d z \lim_{N \rightarrow \infty}\oint_{C_N}f(z)dz lim N → ∞ ∮ C N f ( z ) d z
(3) lim N → ∞ ∮ C N f ( z ) d z \lim_{N \rightarrow \infty}\oint_{C_N}f(z)dz lim N → ∞ ∮ C N f ( z ) d z ∑ n = 1 ∞ 1 n 2 + a 2 = π 2 a coth π a − 1 2 a 2 \sum_{n=1}^{\infty}\frac{1}{n^2 + a^2} = \frac{\pi}{2a}\coth \pi a - \frac{1}{2a^2} ∑ n = 1 ∞ n 2 + a 2 1 = 2 a π coth πa − 2 a 2 1
Consider the complex function f ( z ) = π cot π z z 2 + a 2 f(z) = \frac{\pi\cot\pi z}{z^2 + a^2} f ( z ) = z 2 + a 2 π c o t π z a > 0 a > 0 a > 0
(1) Find the residues of f ( z ) f(z) f ( z )
(2) Consider the complex integral ∮ C N f ( z ) d z \oint_{C_N} f(z)dz ∮ C N f ( z ) d z C N C_N C N N N N lim N → ∞ ∮ C N f ( z ) d z \lim_{N \rightarrow \infty}\oint_{C_N}f(z)dz lim N → ∞ ∮ C N f ( z ) d z
(3) Using the value of lim N → ∞ ∮ C N f ( z ) d z \lim_{N \rightarrow \infty}\oint_{C_N}f(z)dz lim N → ∞ ∮ C N f ( z ) d z ∑ n = 1 ∞ 1 n 2 + a 2 = π 2 a coth π a − 1 2 a 2 \sum_{n=1}^{\infty}\frac{1}{n^2 + a^2} = \frac{\pi}{2a}\coth \pi a - \frac{1}{2a^2} ∑ n = 1 ∞ n 2 + a 2 1 = 2 a π coth πa − 2 a 2 1
题目描述
设 a > 0 a>0 a > 0
f ( z ) = π cot π z z 2 + a 2 . f(z)=\frac{\pi\cot\pi z}{z^2+a^2}. f ( z ) = z 2 + a 2 π cot π z . 回答下列问题：
找出 f ( z ) f(z) f ( z )
令 N N N C N C_N C N
lim N → ∞ ∮ C N f ( z ) d z . \lim_{N\to\infty}\oint_{C_N}f(z)\,dz. N → ∞ lim ∮ C N f ( z ) d z . 原图给出的 C N C_N C N ( N + 1 2 ) ( ± 1 ± i ) (N+\frac12)(\pm1\pm i) ( N + 2 1 ) ( ± 1 ± i ) cot ( π z ) \cot(\pi z) cot ( π z )
利用第 2 问的极限证明
∑ n = 1 ∞ 1 n 2 + a 2 = π 2 a coth ( π a ) − 1 2 a 2 . \sum_{n=1}^{\infty}\frac1{n^2+a^2}
=\frac{\pi}{2a}\coth(\pi a)-\frac1{2a^2}. n = 1 ∑ ∞ n 2 + a 2 1 = 2 a π coth ( πa ) − 2 a 2 1 .
Kai (1)
f ( z ) = π cot π z z 2 + a 2 = π cos π z ( z 2 + a 2 ) sin π z = π cos π z ( z + a i ) ( z − a i ) sin π z \begin{aligned}
f(z) &= \frac{\pi \cot \pi z}{z^2 + a^2} \\
&= \frac{\pi \cos \pi z}{(z^2 + a^2)\sin\pi z} \\
&= \frac{\pi \cos \pi z}{(z + ai)(z - ai)\sin \pi z}
\end{aligned} f ( z ) = z 2 + a 2 π cot π z = ( z 2 + a 2 ) sin π z π cos π z = ( z + ai ) ( z − ai ) sin π z π cos π z z = a i , − a i , n z = ai,-ai,n z = ai , − ai , n n n n
Res z = a i f ( z ) \text{Res}_{z = ai}f(z) Res z = ai f ( z ) g ( z ) = π cos π z ( z + a i ) sin π z g(z) = \frac{\pi \cos \pi z}{(z + ai)\sin \pi z} g ( z ) = ( z + ai ) s i n π z π c o s π z
Res z = a i f ( z ) = g ( a i ) = π cos π ( a i ) 2 a i ⋅ sin π ( a i ) = π ⋅ cosh ( π a ) 2 a i ⋅ i sinh ( π a ) = − π 2 a coth ( π a ) \begin{aligned}
\text{Res}_{z = ai}f(z) &= g(ai)\\
&= \frac{\pi \cos \pi(ai)}{2ai \cdot \sin\pi(ai)} \\
&= \frac{\pi \cdot \cosh(\pi a)}{2ai \cdot i\sinh(\pi a)} \\
&= -\frac{\pi}{2a} \coth(\pi a)
\end{aligned} Res z = ai f ( z ) = g ( ai ) = 2 ai ⋅ sin π ( ai ) π cos π ( ai ) = 2 ai ⋅ i sinh ( πa ) π ⋅ cosh ( πa ) = − 2 a π coth ( πa ) 次に、 Res z = − a i f ( z ) \text{Res}_{z = -ai}f(z) Res z = − ai f ( z ) g ( z ) = π cos π z ( z − a i ) sin π z g(z) = \frac{\pi \cos\pi z}{(z - ai)\sin\pi z} g ( z ) = ( z − ai ) s i n π z π c o s π z
Res z = − a i f ( z ) = g ( − a i ) = π cos π ( − a i ) ( − 2 a i ) sin π ( − a i ) = π cosh ( − π a ) − 2 a i ⋅ i sinh ( − π a ) = π 2 a coth ( − π a ) = − π 2 a coth π a \begin{aligned}
\text{Res}_{z = -ai}f(z) &= g(-ai)\\
&= \frac{\pi\cos\pi(-ai)}{(-2ai)\sin\pi(-ai)} \\
&= \frac{\pi\cosh(-\pi a)}{-2ai \cdot i \sinh(-\pi a)} \\
&= \frac{\pi}{2a}\coth(-\pi a) \\
&= -\frac{\pi}{2a}\coth \pi a
\end{aligned} Res z = − ai f ( z ) = g ( − ai ) = ( − 2 ai ) sin π ( − ai ) π cos π ( − ai ) = − 2 ai ⋅ i sinh ( − πa ) π cosh ( − πa ) = 2 a π coth ( − πa ) = − 2 a π coth πa 最後に、 Res z = n f ( z ) \text{Res}_{z = n}f(z) Res z = n f ( z )
Res z = n f ( z ) = lim z → n π ⋅ 1 { ( z 2 + a 2 ) tan π z } ′ = lim z → n π ⋅ 1 2 z tan π z + ( z 2 + a 2 ) ⋅ 1 cos 2 π z ⋅ π = lim z → n π ⋅ cos 2 π z 2 z sin π z cos π z + π ( z 2 + a 2 ) = lim z → n π ⋅ cos 2 π z z sin 2 π z + π ( z 2 + a 2 ) = lim z → n π ⋅ 1 π ( n 2 + a 2 ) = 1 n 2 + a 2 \begin{aligned}
\text{Res}_{z = n}f(z) &= \lim_{z \rightarrow n} \pi \cdot \frac{1}{\{(z^2 + a^2) \tan \pi z\}'} \\
&= \lim_{z \rightarrow n} \pi \cdot \frac{1}{2z\tan \pi z + (z^2 + a^2) \cdot \frac{1}{\cos^2\pi z} \cdot \pi} \\
&= \lim_{z \rightarrow n} \pi \cdot \frac{\cos^2\pi z}{2z\sin\pi z \cos\pi z + \pi(z^2 + a^2)} \\
&= \lim_{z \rightarrow n} \pi \cdot \frac{\cos^2 \pi z}{z \sin 2\pi z + \pi(z^2 + a^2)} \\
&= \lim_{z \rightarrow n} \pi \cdot \frac{1}{\pi(n^2 + a^2)} \\
&= \frac{1}{n^2 + a^2}
\end{aligned} Res z = n f ( z ) = z → n lim π ⋅ {( z 2 + a 2 ) tan π z } ′ 1 = z → n lim π ⋅ 2 z tan π z + ( z 2 + a 2 ) ⋅ c o s 2 π z 1 ⋅ π 1 = z → n lim π ⋅ 2 z sin π z cos π z + π ( z 2 + a 2 ) cos 2 π z = z → n lim π ⋅ z sin 2 π z + π ( z 2 + a 2 ) cos 2 π z = z → n lim π ⋅ π ( n 2 + a 2 ) 1 = n 2 + a 2 1 (2)
図より R N = N + 1 2 R_N=N+\frac12 R N = N + 2 1 C N C_N C N ∣ Re z ∣ ≤ R N , ∣ Im z ∣ ≤ R N |\operatorname{Re}z|\leq R_N,\ |\operatorname{Im}z|\leq R_N ∣ Re z ∣ ≤ R N , ∣ Im z ∣ ≤ R N
∣ cot π ( ± R N + i y ) ∣ = ∣ tanh π y ∣ ≤ 1 , |\cot\pi(\pm R_N+iy)|=|\tanh\pi y|\leq1, ∣ cot π ( ± R N + i y ) ∣ = ∣ tanh π y ∣ ≤ 1 , 水平辺では ∣ cot π ( x ± i R N ) ∣ ≤ coth ( π R N ) |\cot\pi(x\pm iR_N)|\leq\coth(\pi R_N) ∣ cot π ( x ± i R N ) ∣ ≤ coth ( π R N ) z ∈ C N z\in C_N z ∈ C N ∣ z ∣ ≥ R N |z|\geq R_N ∣ z ∣ ≥ R N R N > a R_N>a R N > a
∣ z 2 + a 2 ∣ ≥ ∣ z ∣ 2 − a 2 ≥ R N 2 − a 2 . |z^2+a^2|\geq |z|^2-a^2\geq R_N^2-a^2. ∣ z 2 + a 2 ∣ ≥ ∣ z ∣ 2 − a 2 ≥ R N 2 − a 2 . したがって ML 評価より
∣ ∮ C N f ( z ) d z ∣ ≤ 8 π R N coth ( π R N ) R N 2 − a 2 ⟶ 0. \left|\oint_{C_N}f(z)\,dz\right|
\leq\frac{8\pi R_N\coth(\pi R_N)}{R_N^2-a^2}
\longrightarrow0. ∮ C N f ( z ) d z ≤ R N 2 − a 2 8 π R N coth ( π R N ) ⟶ 0. (3)
留数定理から、
lim N → ∞ ∮ C N f ( z ) d z = 2 π i ( ∑ n = − ∞ ∞ 1 n 2 + a 2 − π a coth π a ) 0 = 2 π i ( ∑ n = − ∞ ∞ 1 n 2 + a 2 − π a coth π a ) ∑ n = − ∞ ∞ 1 n 2 + a 2 = π a coth π a ∑ n = − ∞ − 1 1 n 2 + a 2 + 1 a 2 + ∑ n = 1 ∞ 1 n 2 + a 2 = π a coth π a 2 ∑ n = 1 ∞ 1 n 2 + a 2 = π a coth π a − 1 a 2 ∑ n = 1 ∞ 1 n 2 + a 2 = π 2 a coth π a − 1 2 a 2 \begin{aligned}
\lim_{N \rightarrow \infty}\oint_{C_N}f(z)dz &= 2\pi i \bigg(\sum_{n = -\infty}^{\infty} \frac{1}{n^2 + a^2} - \frac{\pi}{a}\coth \pi a\bigg) \\
0 &= 2\pi i \bigg(\sum_{n = -\infty}^{\infty}\frac{1}{n^2 + a^2} - \frac{\pi}{a}\coth\pi a\bigg) \\
\sum_{n = -\infty}^{\infty} \frac{1}{n^2 + a^2} &= \frac{\pi}{a}\coth \pi a \\
\sum_{n = -\infty}^{-1}\frac{1}{n^2 + a^2} + \frac{1}{a^2} + \sum_{n = 1}^{\infty}\frac{1}{n^2 + a^2} &= \frac{\pi}{a} \coth\pi a \\
2\sum_{n = 1}^{\infty}\frac{1}{n^2 + a^2} &= \frac{\pi}{a}\coth\pi a - \frac{1}{a^2} \\
\sum_{n = 1}^{\infty}\frac{1}{n^2 + a^2} &= \frac{\pi}{2a}\coth\pi a - \frac{1}{2a^2}
\end{aligned} N → ∞ lim ∮ C N f ( z ) d z 0 n = − ∞ ∑ ∞ n 2 + a 2 1 n = − ∞ ∑ − 1 n 2 + a 2 1 + a 2 1 + n = 1 ∑ ∞ n 2 + a 2 1 2 n = 1 ∑ ∞ n 2 + a 2 1 n = 1 ∑ ∞ n 2 + a 2 1 = 2 πi ( n = − ∞ ∑ ∞ n 2 + a 2 1 − a π coth πa ) = 2 πi ( n = − ∞ ∑ ∞ n 2 + a 2 1 − a π coth πa ) = a π coth πa = a π coth πa = a π coth πa − a 2 1 = 2 a π coth πa − 2 a 2 1 