第三章 全纯函数的积分表示

§3.1习题

1.计算积分\(\int_{\gamma}^{}\frac{2z - 3}{z}\text{dz}\),其中,\(\gamma\)为 (iii)沿圆周\(\{ z \in \mathbb{C}:|z| = 2\}\)的正向.

\(\int_{\gamma}^{}\frac{2z - 3}{z}\text{dz}\)=2\(\int_{\gamma}^{}{dz - \int_{\gamma}^{}\frac{3}{z}}\text{dz}\)=0-6\(\pi i = - 6\pi i\)

2.计算\(\int_{|z|\text{=1}}^{}\frac{\text{dz}}{z + 2},\)并证明,\(\int_{0}^{\pi}\frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}\)=0.

奇点在区域外积分为0

\(\int_{|z|\text{=1}}^{}\frac{\text{dz}}{z + 2}\)=\(\int_{0}^{2\pi}\frac{de^{\text{iθ}}}{e^{\text{iθ}} + 2}\)=\(\int_{0}^{2\pi}\frac{ie^{\text{iθ}}}{e^{\text{iθ}} + 2}\text{dθ}\)=\(i\int_{0}^{2\pi}\frac{e^{\text{iθ}}(e^{- \text{iθ}} + 2)}{(e^{\text{iθ}} + 2)(e^{- \text{iθ}} + 2)}d\theta = 0\)

\[{\Rightarrow \int_{0}^{2\pi}\frac{1 + 2e^{\text{iθ}}}{1 + 2e^{\text{iθ}} + 2e^{- \text{iθ}} + 4}d\theta = 0 }{\Rightarrow \int_{0}^{2\pi}\frac{1 + 2\cos\theta + 2i\sin\theta}{5 + 2\cos\theta + 2\sin\theta + 2i\cos( - \theta) + 2i\sin( - \theta)}d\theta = 0 }{\Rightarrow \int_{0}^{2\pi}\frac{1 + 2\cos\theta + 2i\sin\theta}{5 + 4\cos\theta}d\theta = 0 }{\Rightarrow \int_{0}^{2\pi}\frac{1 + 2\cos\theta}{5 + 4\cos\theta}d\theta + \int_{0}^{2\pi}\frac{2\sin\theta}{5 + 4\cos\theta}d\theta i = 0 }{(\int_{0}^{\pi}{+ \int_{\pi}^{2\pi}{)\frac{1 + 2\cos\theta}{5 + 4\cos\theta}}}d\theta = 0}\]

由对称性

\[\int_{0}^{\pi}\frac{1 + 2\cos\theta}{5 + 4\cos\theta}d\theta = \int_{\pi}^{2\pi}\frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}.^{}所以\int_{0}^{\pi}\frac{1 + 2\cos\theta}{5 + 4\cos\theta}d\theta = 0\]

3.计算积分\(\int_{|z| = 3}^{}\frac{2z - 1}{z(z - 1)}\text{dz}\).

\[\int_{|z| = 3}^{}\frac{2z - 1}{z(z - 1)}\text{dz=}\int_{|z| = 3}^{}\frac{z - 1 + z}{z(z - 1)}dz = \int_{|z|\text{=3}}^{}\frac{1}{z}dz + \int_{|z| = 3}^{}\frac{1}{z - 1}dz = 4\pi i\]

4.如果多项式Q(z)比多项式P(z)高两次,试证: \(\lim_{R \rightarrow \infty}\int_{|Z| = R}^{}\frac{P(z)}{Q(z)}dz = 0\)

证明:\(\lim_{R \rightarrow \infty}\frac{P(z)z^{2}}{Q(z)} = A\) (A为某个常数)

从而存在M>0,R\(0\)>0使得当R\(\geq\)R\(0\)时,有\(\left| \frac{P(z)z^{2}}{Q(z)} \right| \leq M\)

因此\(|\int_{|z| = R}^{}\frac{P(z)}{Q(z)}\text{dz} \leq \int_{|z| = R}^{}{|\frac{P(z)}{Q(z)}}||z| \leq \int_{|z| = R}^{}\frac{M}{|z|^{2}}|dz| = \frac{2\pi M}{R} \rightarrow 0(R \rightarrow \infty)\)

所以\(\lim_{R \rightarrow \infty}\int_{}^{}\frac{P(z)}{Q(z)}dz = 0\)

6.设f\(\in C^{1}(D),\gamma\)是域D中分别以a和b为起点和终点的可求长曲线.证明:

\[\int_{\gamma}^{}\frac{\partial f(z)}{\partial z}dz + \frac{\partial f(z)}{\partial\overline{z}}d\overline{z} = f(b) - f(a)\]

\(\frac{\partial f(z)}{\partial z} = \frac{1}{2}(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}), dz = dx + idy\), \(\frac{\partial f(z)}{\partial z} = \frac{1}{2}(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}),d\overline{z} = dx - \text{idy}\)

\[{\frac{\partial f(z)}{\partial z}dz + \frac{\partial f(z)}{\partial\overline{z}}d\overline{z} = \frac{1}{2}(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y})(dx + idy) + \frac{1}{2}(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y})(dx - dy) }{= \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = df}\]

故\(\int_{\gamma}^{}{\{\frac{\partial f(z)}{\partial z}}dz + \frac{\partial f(z)}{\partial\overline{z}}d\overline{z}\} = \int_{\gamma}^{}\text{df} = f(b) - f(a)\)

8.设\(\gamma\)是域D中以a为起点,以b为终点的可求长曲线,f,g\(\in H(D) \cap C^{1}(D).\)证明分部积分公式:\(\int_{\gamma}^{}{f(z)g^{'}}(z)dz = f(z)g(z)|_{a}^{b} - \int_{\gamma}^{}f^{’}(z)dz.\)

\(\int_{\gamma}^{}{f(z)g'(z)dz + \int_{\gamma}^{}{f'(z)g(z)dz = \int_{\gamma}^{}{\lbrack f'(z)g(z) + f(z)g'(z)\rbrack dz = \int_{\gamma}^{}{\lbrack f(z)g(z)\rbrack'dz = f(z)g(z)|_{a}^{b}}}}}\)故\(\int_{\gamma}^{}{f(z)g'(z)dz = f(z)g(z)|_{a}^{b}} - \int_{\gamma}^{}{f'(z)g(z)dz}\)

9.设\(\gamma\)是正向可求简单曲线,证明:\(\gamma\)内部的面积为\(\frac{1}{2i}\int_{r}^{}\overline{z}\text{dz}\).

证明:由公式得\(\int_{r}^{}\overline{z}\text{dz}\)=\(\int_{D}^{}{( - \frac{\partial\overline{z}}{\partial z}})\text{dz} \land d\overline{z}\)=-\(\int_{D}^{}{\text{dz} \land d\overline{z}}\)

= -\(\int_{D}^{}{(dx + idy) \land (dx - \text{idy}) = \int_{D}^{}{2idx \land dy = \int_{D}^{}{2idA}}}\) = 2iA

所以A=\(\frac{1}{2i}\int_{r}^{}\overline{z}\text{dz}\)

11.设\(f\)在z\(0\)处连续,证明:

(i)\(\lim_{r \rightarrow 0}\frac{1}{2\pi}\int_{0}^{2\pi}{f(z_{0}} + re^{\text{iθ}})d\theta = f(z_{0});\)(ii) \(\lim_{r \rightarrow 0}\frac{1}{2\pi i}\int_{|z - z_{0}| = r}^{}\frac{f(z)}{z - z_{0}}dz = f(z_{0})\).

证明:(i)\(\left| \frac{1}{2\pi}\int_{0}^{2\pi}{f(z_{0}} + re^{\text{iθ}})d\theta - f(z_{0}) \right| \leq \frac{1}{2\pi}\int_{0}^{2\pi}{|f(z_{0}} + re^{\text{iθ}}) - f(z_{0})|d\theta\)

\(\leq \sup_{|z - z_{0}| = r}|f(z) - f(z_{0})| \rightarrow 0(r \rightarrow 0^{+})\). 所以\(\lim_{r \rightarrow 0}\frac{1}{2\pi}\int_{0}^{2\pi}{f(z_{0}} + re^{\text{iθ}})d\theta = f(z_{0})\).

(ii) \(\frac{1}{2\pi i}\int_{|z - z_{0}| = r}^{}\frac{f(z)}{z - z_{0}}dz = \frac{1}{2\pi}\int_{0}^{2\pi}{f(z_{0} + re^{\text{iθ}})}\text{dθ}\). 故\(\lim_{r \rightarrow 0}\frac{1}{2\pi i}\int_{|z - z_{0}| = r}^{}\frac{f(z)}{z - z_{0}}dz = f(z_{0})\)

12.设D={\(z\mathbb{\in C:}\theta_{0} < \arg(z - a) < \theta_{0} + \alpha\}(0 < a \leq 2\pi),\) \(f\)在\(\overline{D}\backslash\{ a\}\)上连续,证明:

(i)如果\(\lim_{\begin{matrix} \& z \rightarrow a \\ \& z \in \overline{D}\backslash\{ a\} \\ \end{matrix}}(z - a)f(z) = A,\)那么\(\lim_{r \rightarrow 0}\int_{\begin{matrix} \&|z - a| = r \\ \& z \in \overline{D} \\ \end{matrix}}^{}{f(z)dz = i\alpha A;}\)

(ii)如果\(\lim_{z \rightarrow \infty}\int_{\begin{matrix} \&|z - a| = R \\ \& z \in \overline{D} \\ \end{matrix}}^{}{f(z)dz = i\alpha B.}\)

证明:(i)\(\left| \int_{\begin{matrix} \&|z - a| = r \\ \& z \in \overline{D} \\ \end{matrix}}^{}{f(z)dz - \text{iαA}} \right| = \left| \int_{\begin{matrix} \&|z - a| = r \\ \& z \in \overline{D} \\ \end{matrix}}^{}{f(z) - \frac{A}{z - a}}\text{dz} \right| \leq \frac{1}{r}\int_{\begin{matrix} \&|z - a| = r \\ \& z \in \overline{D} \\ \end{matrix}}^{}{|(z - a)f(z) - A}||dz|\)

\[\leq \sup_{\begin{matrix} \&|z - a| = r \\ \& z \in \overline{D} \\ \end{matrix}}|(z - a)f(z) - A| \rightarrow 0(r \rightarrow 0^{+})\]

(ii)略

§3.2习题

1.计算积分:

(i)\(\int_{|z| = r}^{}\frac{|dz|}{|z - a|^{2}},|a| \neq r\)

\[\int_{|z| = r}^{}\frac{|dz|}{|z - a|^{2}} = \int_{0}^{2\pi}\frac{\text{rdθ}}{(re^{\text{iθ}} - a)(re^{- \text{iθ}} - \overline{a})} = \frac{r}{i}\int_{|z| = r}^{}\frac{\text{dz}}{(z - a)(r^{2} - \overline{a}z)}\]

\[= \frac{- \text{ir}}{r^{2} - |a|^{2}}\int_{|z| = r}^{}\left( \frac{1}{z - a} - \frac{1}{z - r^{2}/\overline{a}} \right) dz = \frac{2\pi r}{|r^{2} - |a|^{2}|}\]

(iii)\(\int_{|z| = 5}^{}\frac{\text{zdz}}{z^{4} - 1};\)

\[\int_{|z| = 5}^{}\frac{\text{zdz}}{z^{4} - 1} = \frac{1}{2}\int_{|z| = 5}^{}\frac{dz^{2}}{(z^{2} - 1)(z^{2} + 1)} = \frac{1}{4}\int_{|z| = 5}^{}\left( \frac{1}{z^{2} - 1} - \frac{1}{z^{2} + 1} \right)dz^{2} = 0\]

(iv)\(\int_{|z| = 2a}^{}\frac{e^{z}}{z^{2} + a^{2}}dz,a > 0.\)

\[\int_{|z| = 2a}^{}\frac{e^{z}}{z^{2} + a^{2}}dz = \frac{1}{2\text{ai}}\int_{|z| = 2a}^{}\frac{e^{z}}{z - \text{ai}}\text{dz} - \frac{1}{2ai}\int_{|z| = 2a}^{}\frac{e^{z}}{z + ai}\text{dz}\]

=\(\frac{1}{2\text{ai}} \cdot 2\pi i \cdot e^{\text{ai}} - \frac{1}{2ai}2\pi i \cdot e^{- \text{ai}} = \frac{2\pi i}{a}\sin a\)

  1. 设\(f\)在\(\{ z\mathbb{\in C:}r < |z| < \infty\}\)中全纯,且\(\lim_{z \rightarrow \infty}zf(z) = A.\)

证明:\(\int_{|z| = R}^{}{f(z)dz = 2\pi iA,}\) 这里\(R > r\)

证:设\(r < R < R' < \infty,\)利用\(\text{Cauc}hy\)积分公式得

\[\left| \int_{|z| = R}^{}{f(z)}\text{dz} - 2\text{πiA} \right| = \left| \int_{|z| = R'}^{}{f(z)dz - \int_{|z| = R^{'}}^{}\frac{A}{z}}\text{dz} \right| \leq \frac{1}{R^{'}}\int_{|z| = R'}^{}{|zf(z) - A}||dz|\]

\[\leq 2\pi \cdot \sup_{|z| = R^{'}}|zf(z) - A| \rightarrow 0(R^{'} \rightarrow \infty)\]

4.设\(0 < r < R, f\)在\(B(0,R)\)中全纯.证明:

\[(i)f(0) = \frac{1}{2\pi}\int_{0}^{2\pi}{f(re^{\text{iθ}}})d\theta;(\text{ii})f(0) = \frac{1}{\pi r^{2}}\int_{|z| < r}^{}{f(z)dxdy.}\]

证明:(i)\(\forall R > \delta > 0,\) 由\(\text{Cauchy}\)积分公式得\(\frac{1}{2\pi i}\int_{|z| = r}^{}\frac{f(z)}{z}dz = \frac{1}{2\pi i}\int_{|z| = \delta}^{}\frac{f(z)}{z}\text{dz}\)

即\(\frac{1}{2\pi}\int_{0}^{2\pi}{f(re^{\text{iθ}}})d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}{f(\delta e^{\text{iθ}}})d\theta\),令\(\delta \rightarrow 0\),则有 \(\frac{1}{2\pi}\int_{0}^{2\pi}{f(re^{\text{iθ}}})d\theta = f(0)\)

(ii)\(\frac{1}{\pi r^{2}}\int_{|z| < r}^{}{f(z)dxdy}\)=\(\frac{1}{\pi r^{2}}\int_{0}^{r}{\text{ρdρ}\int_{0}^{2\pi}{f(\rho e^{\text{iθ}})}}\text{dθ}\) (1)

利用(i)得\(\int_{0}^{2\pi}{f(\rho e^{\text{iθ}}})d\theta = 2\pi f(0)\) (2)

由(1)和(2)式得\(f(0) = \frac{1}{\pi r^{2}}\int_{|z| < r}^{}{f(z)dxdy}\)

5.设u是B(0,R)中的调和函数,\(0 < r < R.\)证明:\(u(0) = \frac{1}{2\pi}\int_{0}^{2\pi}{u(re^{\text{iθ}}})d\theta.\)

证明:单连通域上的调和函数是某个全纯函数\(f\)的实部:\(u = \text{Re}f\)

\(f(0) = \frac{1}{2\pi}\int_{0}^{2\pi}{f(re^{\text{iθ}}})d\theta \Rightarrow u(0) = \text{Re}f(0) = \frac{1}{2\pi}\int_{0}^{2\pi}{u(re^{\text{iθ}}})d\theta\)

6.设\(\int_{0}^{\pi}{\text{log}(1 - \text{2rcos}\theta + r^{2}})d\theta = 0\)

证明:令U(z)=\(\ln|z - 1|^{2}\),则U在B(0,1)上调和,由题5,0=U(0)= \(\frac{1}{2\pi}\int_{0}^{2\pi}{u(re^{\text{iθ}}})d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}{\log(1 - 2r\cos\theta + r^{2})}d\theta = \frac{1}{\pi}\int_{0}^{\pi}{\log(1 - 2r\cos\theta + r^{2})}\text{dθ}\).

§3.3习题

1.设\(D\)是域,\(f,g \in H(D)\).如果\(fg^{'}\)在\(D\)上有原函数\(\phi\).证明:\(fg^{'}\)在上\(D\)有原函数\(\text{fg} - \phi\)

证明:\(f,g \in H(D)\).由\(fg^{'} = \phi\),得\((fg - \phi)^{'} = fg^{'}\).

3.设\(f \in C^{n}\mathbb{(C) \cap}H\mathbb{(C)}\),并且\(f^{(n)}(z) \equiv 0\).证明:\(f\)必为次数不大于\(n\)-1的多项式.

证明:归纳法 k=1时显然,设k=n-1时成立,取定\(\xi \in C\),由\(f^{(n)}(z) \equiv 0 \Rightarrow f^{(n - 1)}(z) = C\)

故\(\left\lbrack f(z) - \frac{Cz^{n - 1}}{(n - 1)!} \right\rbrack^{(n - 1)} \equiv 0\),从而\(f(z) - \frac{Cz^{n - 1}}{(n - 1)!}\)是次数不超过n-2的多项式.

5.设\(f\)是凸域\(D\)上的全纯函数,如果对每点\(z \in D\),有\(\text{Re}\left\{ f^{'}(z) \right\} > 0\),那么\(f\)是\(D\)上的单叶函数.

证明:\(\forall z_{1},z_{2} \in D\),\(z_{1} \neq z_{2}\) 则

\(f(z_{2}) - f(z_{1}) = \int_{z_{1}}^{z_{2}}{f^{'}(\xi)}d\xi = \int_{0}^{1}f^{'}(z_{1} + t(z_{2} - z_{1}))(z_{2} - z_{1})dt\)

则\(\frac{f(z_{2}) - f(z_{1})}{z_{2} - z_{1}} = \int_{0}^{1}{f^{'}(z_{1} + t(z_{2} - z_{1}))}\text{dt}\)

从而\(\left| \frac{f(z_{2}) - f(z_{1})}{z_{2} - z_{1}} \right| \geq \int_{0}^{1}{\text{Re}\left\{ f^{'}(z_{1} + t(z_{2} - z_{1})) \right\}}dt > 0\)

故\(f(z_{1}) \neq f(z_{2})\),这表明\(f\)是\(D\)上的单叶函数.

§3.4习题

  1. 计算下列积分:

(1)\(\int_{|z| = 1}^{}\frac{\sin z}{z^{2} - 1}\text{dz}\)=\(\int_{|z| = 1}^{}\frac{\sin z}{z^{2} - 1}dz = \int_{|z - 1| = 1}^{}\frac{1}{z - 1} \cdot \frac{\sin z}{z + 1}dz = 2\pi i\left. \ \left( \frac{\sin z}{z + 1} \right) \right|_{z = 1} = i\pi\sin 1\)

(2) \(\int_{|z| = 2}^{}\frac{\text{dz}}{1 + z^{2}} = \frac{1}{2i}\int_{|z| = 2}^{}\left( \frac{1}{z - i} - \frac{1}{z + i} \right)dz = 0\)

(4)\(\int_{|z| = \frac{3}{2}}^{}\frac{\text{dz}}{(z^{2} + 1)(z^{2} + 4)};\) 与第二题类似,答案为0.

(6)\(\int_{|z| = R}^{}\frac{\text{dz}}{(z - a)^{n}(z - b)}\), n为正整数,a\(\neq\)b不在圆周|z|=R上.解:原式\(= \left\{ \begin{matrix} \&{{{{{{{{{{{{{{{0^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}a,b均在圆外. \\ \&{{{{{\frac{2\pi i}{(b - a)^{n}}^{}}^{}}^{}}^{}}^{}}^{}a在圆外,b在圆内. \\ \& - {{{\frac{2\pi i}{(b - a)^{n}}^{}}^{}}^{}}^{}a在圆内,b在圆外. \\ \&{{{{{{{{{{{{{{{0^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}}^{}a,b均在圆内. \\ \end{matrix} \right.\ \)

3.设D是由有限条可求长简单闭曲线围成的域,\(z_{1},\ldots,z_{n}\)是D中n个彼此不同的点.如果\(f \in H(D) \cap C(\overline{D})\),证明:\(P(z) = \frac{1}{2\pi i}\int_{\partial D}^{}{\frac{f(\zeta)}{\omega_{n}(\zeta)}\frac{\omega_{n}(\zeta) - \omega_{n}(z)}{\zeta - z}}\text{dζ}\)是次数不超过n-1的多项式,并且\(P(z_{k}) = f(z_{k}),k = 1,2,\ldots,n.\text{\!\!}其中,\text{\!\!\!}\omega_{n}(z) = (z - z_{1})\ldots(z - z_{n}).\)

证明:由于\(\omega_{n}(z) = (z - z_{1})\ldots(z - z_{n})\),从而\(\frac{\omega_{n}(\zeta) - \omega_{n}(z)}{\zeta - z}\)是z的n-1次多项式,

记\(h(\xi,z) = f(\xi)\frac{\omega_{n}(\zeta) - \omega_{n}(z)}{\zeta - z}\), 取\(\varepsilon > 0\)充分小,由Cauchy积分公式

\[P(z) = \sum_{k = 1}^{n}\frac{1}{2\pi i}\int_{|\xi - z_{k}| = \varepsilon}^{}\frac{h(\xi,z)}{\overset{\underset{n}{j = 1}}{\Pi}(\xi - z_{j})}d\xi = \sum_{k = 1}^{n}{h(z_{k},z)}\overset{\underset{n}{\begin{matrix} \& j = 1 \\ \& j \neq k \\ \end{matrix}}}{\Pi}(z_{k} - z_{j})^{- 1}\]

因为\(h(\xi,z) = f(\xi)\frac{- \omega_{n}(z)}{\zeta - z}\)是次数不超过n-1的多项式,故P(z)是次数不超过n-1的多项式 又\(- \omega_{n}(z) = (z_{k} - z)\overset{\underset{n}{\begin{matrix} \& j = 1 \\ \& j \neq k \\ \end{matrix}}}{\Pi}(z_{j} - z),\)故\(P(z_{k}) = f(z_{k})\)是次数不超过n-1的多项式.

  1. 设\(f \in H(B(0,1)) \cap C(\overline{B(0,1)})\).证明:

\[{(1)\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})}\cos^{2}\left( \frac{\theta}{2} \right)d\theta = 2f(0) + f^{'}(0); }{(2)\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})}\sin^{2}\left( \frac{\theta}{2} \right)d\theta = 2f(0) - f^{'}(0)}\]

(提示:分别计算积分\(\frac{1}{2\pi i}\int_{|\zeta| = 1}^{}\left( 2 + \zeta + \frac{1}{\zeta} \right)f(\zeta)\frac{\text{dζ}}{\zeta}和\frac{1}{2\pi i}\int_{|\zeta| = 1}^{}\left( 2 - \zeta - \frac{1}{\zeta} \right)f(\zeta)\frac{\text{dζ}}{\zeta}\)即可.)

证明:由Cauchy公式,得

\(f(0) = \frac{1}{2\pi i}\int_{|\zeta| = 1}^{}\frac{f(\zeta)}{\zeta}d\zeta = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{f(e^{\text{iθ}})}{e^{\text{iθ}}}ie^{\text{iθ}}d\theta = \frac{1}{2\pi i}\int_{0}^{2\pi}{f(e^{\text{iθ}})}\text{dθ}\),\(f^{'}(0) = \frac{1}{2\pi i}\int_{|\zeta| = 1}^{}\frac{f(\zeta)}{\zeta^{2}}d\zeta = \frac{1}{2\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})e^{- \text{iθ}}}\text{dθ}\) ②

又由Cauchy定理,\(\int_{|\zeta| = 1}^{}{f(\zeta)d\zeta = 0}\)即\(\frac{1}{2\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})e^{\text{iθ}}}d\theta = 0\) ③

②+③得\(f^{'}(0) = \frac{1}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\cos\theta}d\theta = \frac{1}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})(2\cos^{2}\frac{\theta}{2} - 1)}\text{dθ}\)

即\(\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\cos^{2}\frac{\theta}{2}}d\theta = f^{'}(0) + \frac{1}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})}d\theta = f^{'}(0) + 2f(0)\)

6.利用上题结果证明:

设\(f \in H(B(0,1)) \cap C(\overline{B(0,1)})\),且\(f(0) = 1,\text{Re}f(z) \geq 0\),那么 \(- 2 \leq {\text{Re}f}^{'}(0) \leq 2\).

证明:\(\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\cos^{2}\frac{\theta}{2}}d\theta = f^{'}(0) + 2f(0)\)

两边取实部,即\(\text{Re}\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\cos^{2}\frac{\theta}{2}}d\theta = 2 + {\text{Re}f}^{'}(0) \geq 0\)

同理

\(\frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\sin^{2}\frac{\theta}{2}}d\theta = 2f(0) - f^{'}(0) \Rightarrow {\text{Re}f}^{'}(0) = 2 - \frac{2}{\pi}\int_{0}^{2\pi}{f(e^{\text{iθ}})\sin^{2}\frac{\theta}{2}}d\theta \leq 2 - 0 = 2\)所以\(- 2 \leq {\text{Re}f}^{'}(0) \leq 2\).

§3.5习题

  1. 设\(f\)是有界整函数,\(z_{1},z_{2}\)是\(B(0,r)\)中任意两点.

证明:\(\int_{|z| = r}^{}{\frac{f(z)}{(z - z_{1})(z - z_{2})}dz = 0}\)并由得出Liouyille定理.

证明:利用Cauchy积分公式得

\[\int_{|z| = r}^{}{\frac{f(z)}{(z - z_{1})(z - z_{2})}dz =}\int_{|z| = r}^{}{\frac{1}{z_{1} - z_{2}}\left( \frac{f(z)}{z - z_{1}} - \frac{f(z)}{z - z_{2}} \right)dz = 2\pi i\frac{f(z_{1}) - f(z_{2})}{z_{1} - z_{2}}}\]

另一方面, 由于\(f\)有界,\(\exists M > 0,s.t.\left| f(z) \right| \leq M,\forall z \in C\)

由Cauchy积分定理

\(\left| \int_{|z| = r}^{}{\frac{f(z)}{(z - z_{1})(z - z_{2})}\text{dz}} \right| = \left| \int_{|z| = R > r}^{}{\frac{f(z)}{(z - z_{1})(z - z_{2})}\text{dz}} \right| \leq \frac{M}{(R - \left| z_{1} \right|)(R - \left| z_{2} \right|)} \cdot 2\pi R \rightarrow 0(R \rightarrow \infty)\)从而 \(2\pi i\frac{f(z_{1}) - f(z_{2})}{z_{1} - z_{2}} = 0 \Rightarrow f(z_{1}) = f(z_{2}) \Rightarrow f(z) = C\)

2.设\(f\)是整函数, 如果当\(z \rightarrow \infty\)时,\(f(z) = O(|z|^{\alpha}),\alpha \geq 0,\)

证明\(f\)是次数不超过\(\lbrack\alpha\rbrack\)的多项式.

解:令\(n = \lbrack\alpha\rbrack + 1\)

\[\left| f^{(n)}(z) \right| = \left| \frac{n!}{2\pi i}\int_{|\zeta - z| = R}^{}{\frac{f(\zeta)}{(\zeta - z)^{n + 1}}\text{dζ}} \right| \leq \frac{n!}{2\pi}\int_{|\zeta - z| = R}^{}{\left| \frac{f(\zeta)}{(\zeta - z)^{n + 1}} \right|\left| \text{dζ} \right|} \leq \frac{n!}{2\pi}\int_{|\zeta - z| = R}^{}{\frac{M|\zeta|^{\alpha}}{R^{n + 1}}\left| \text{dζ} \right|} \]

故\(f^{(n)}(z) \equiv 0\)

4.设\(f\)是整函数,如果\(f\mathbb{(C) \subset}\left\{ z\mathbb{\in C;}\text{Im}z > 0 \right\}\),证明\(f\)是一个常值函数.

证明:令\(g(z) = \frac{f(z) - i}{f(z) + i}\),则|g(z)|<1,g是整函数.

从而\(g(z) = \frac{f(z) - i}{f(z) + i} = c \Rightarrow f(z) = \widetilde{c}\)

5.设\(f\)是整函数,如果\(f\mathbb{(C) \subset C\backslash\lbrack}0,1\rbrack\)证明\(f\)是一个常值函数.

证明:令\(g(z) = \frac{f(z)}{1 - f(z)}\),则\(g\left( \mathbb{C} \right)\mathbb{\subset C\backslash}\left. \ 0,\infty \right)\).再令\(h(z) = \sqrt{g(z)}\)

则\(\left| h\left( \mathbb{C} \right) \right| \subset \left\{ z\mathbb{\in C:}\text{Im}z > 0 \right\}\) 由上题知\(h\)常值,故\(f\)常值.

  1. 设\(f\)在域D上全纯,\(z_{0} \in D\)定义

\(F(z) = \left\{ \begin{matrix} \&\frac{f(z) - f(z_{0})}{z - z_{0}}\begin{matrix} , & & z \in D\backslash\{ z_{0}\}; \\ \end{matrix} \\ \& f^{'}(z_{0})\begin{matrix} , & & \\ \end{matrix}\begin{matrix} & \\ \end{matrix}z = z_{0}. \\ \end{matrix} \right.\ \)

证明:\(F \in H(D)\)

证明:\(F(z_{0}) = f^{'}(z_{0}) = \lim_{z \rightarrow z_{0}}\frac{f(z) - f(z_{0})}{z - z_{0}} = \lim_{z \rightarrow z_{0}}F(z_{0})\)

故F在\(z_{0}\)点也连续.将F限制在\(B(z_{0},\varepsilon)\)上,则\(|F| \leq M\),对D内任一简单闭曲线\(\gamma\),可取一含于\(B(z_{0},\varepsilon)\)的简单闭曲线使得\(\int_{\gamma}^{}{f(z)dz = \int_{\gamma_{\varepsilon}}^{}{f(z)dz}}\),对\(\forall\varepsilon > 0,\)由此易得\(\int_{\gamma}^{}{f(z)dz = 0}\),从而F在D上全纯.

7.设\(\gamma\)是可求长曲线,\(f\)在域D上连续,在\(D\backslash\gamma\)上全纯.证明:\(f\)在D上全纯.

证明:任取D中简单闭曲线\(\gamma_{0}\)

  1. 当\(\gamma\)含于\(\gamma_{0}\)内部时,延长\(\gamma\),交\(\gamma_{0}\)于A,B两点.

\[\int_{\gamma_{0}}^{}{f(z)dz = \int_{\gamma_{1} + \overline{\text{AB}}}^{}{f(z)dz +}}\int_{\gamma_{2} + \overline{\text{BA}}}^{}{f(z)dz} = 0\]

  1. 同理,当\(\gamma\),\(\gamma_{0}\)相交时,\(\int_{\gamma_{0}}^{}{f(z)dz} = 0\)

故由Morera定理知\(f\)在D上全纯.

§3.6习题

1、设D是由有限条可求长简单闭曲线围成的域,\(f \in C^{1}(\overline{D})\).证明:

(i)\(\iint_{D}^{}{\frac{\partial f(\xi)}{\partial\bar{\xi}}d\overline{\xi} \land}d\xi = \int_{\partial D}^{}{f(\xi)d\xi}\)

(ii)\(\iint_{D}^{}{\frac{\partial f(\xi)}{\partial\xi}\text{dξ} \land}d\overline{\xi} = \int_{\partial D}^{}{f(\xi)d\overline{\xi}}\)

证明:\(f(\xi)\)是域上的一个0次微分形式\(f \in C^{1}(\overline{D})\),根据定理3.6.1\(\int_{\partial D}^{}f = \int_{D}^{}\text{df}\)

可得\(\int_{\partial D}^{}{f(\xi)d\overline{\xi}} = \int_{D}^{}{df(\xi) \land d\overline{\xi}} = \int_{D}^{}{(\frac{\partial f(\xi)}{\partial\xi}}d\xi + \frac{\partial f(\xi)}{\partial\overline{\xi}}d\overline{\xi}) \land d\overline{\xi} = \int_{D}^{}\frac{\partial f(\xi)}{\partial\overline{\xi}}d\overline{\xi} \land \text{dξ}\)

同理\(\int_{\partial D}^{}{f(\xi)d\overline{\xi}} = \int_{D}^{}{(\frac{\partial f(\xi)}{\partial\xi}}d\xi + \frac{\partial f(\xi)}{\partial\overline{\xi}}d\overline{\xi}) \land d\overline{\xi} = \int_{D}^{}\frac{\partial f(\xi)}{\partial\xi}\text{dξ} \land d\overline{\xi}\)

4、设D是由有限条可求长简单闭曲线围成的域,\(f\)在\(\overline{D}\)上全纯,\(z \in D\).证明:

\[\iint_{D}^{}{\frac{f^{'}(\xi)}{\overline{\xi} - \overline{z}}\text{dξ} \land}d\overline{\xi} = \int_{\partial D}^{}{(\frac{f(\xi)}{\xi - z} + \frac{f(\xi)}{\overline{\xi} - \overline{z}}d\overline{\xi}})\]

证:因\(f \in H(\overline{D})\),从而\(\overline{f} \in C^{1}(\overline{D})\),则对\(\overline{f}\)用Pompeiu公式

\[\overline{f}(z) = \frac{1}{2\pi i}\int_{\partial D}^{}{\frac{\overline{f}(\xi)}{\xi - z}\text{dξ}} + \frac{1}{2\pi i}\iint_{D}^{}{\frac{\partial\overline{f}(\xi)}{\partial\bar{\xi}}\frac{1}{\xi - z}\text{dξ} \land}d\overline{\xi}\]

从而两边取共轭得

\[f(z) = - \frac{1}{2\pi i}\int_{\partial D}^{}{\frac{f(\xi)}{\overline{\xi} - \overline{z}}d\overline{\xi}} - \frac{1}{2\pi i}\iint_{D}^{}{\frac{f^{'}(\xi)}{\overline{\xi} - \overline{z}}d\overline{\xi} \land}\text{dξ}\]

\(f(z) = - \frac{1}{2\pi i}\int_{\partial D}^{}{\frac{f(\xi)}{\bar{\xi} - \bar{z}}d\overline{\xi}} + \frac{1}{2\pi i}\iint_{D}^{}\frac{f^{'}(\xi)}{\overline{\xi} - \overline{z}}\text{dξ} \land d\overline{\xi}\) (1)

\(f(z) = \frac{1}{2\pi i}\int_{\partial D}^{}{\frac{f(\xi)}{\xi - z}\text{dξ}}\) (2)

由(1)(2)可得\(\iint_{D}^{}{\frac{f^{'}(\xi)}{\overline{\xi} - \overline{z}}\text{dξ} \land}d\overline{\xi} = \int_{\partial D}^{}{(\frac{f(\xi)}{\xi - z} + \frac{f(\xi)}{\overline{\xi} - \overline{z}}d\overline{\xi}})\).

§3.7习题

2.设D是域,\(f \in C^{\infty}(D)\), 证明:若\(u_{0} \in C^{\infty}(D)\)是非齐次\(\bar{\partial}\)方程的解,即\(\frac{\partial u_{0}}{\partial\bar{z}} = f\), 则该方程的解的全体为\(u_{0} + H(D)\).

提示:显然\(u_{0} + H(D)\)中的元都是解.

另一方面,设\(u\)是方程的一个解,则\(\frac{\partial\left( u - u_{0} \right)}{\partial\bar{z}} = f - f = 0\),即\(u - u_{0}\)满足C-R方程,

又\(u \in C^{\infty}(D)\),故\(u - u_{0} \in H(D)\).