$z^n=z_0=r_0{\mathrm{e}}^{\mathrm{i}{\theta }_0}$$z^n=z_0=r_0\newcommand{\e}{{\rm e}}\newcommand{\i}{{\rm i}}\e^{\i \theta_0}$ 的全部根为 $z=\sqrt[n]{r_0}{\mathrm{e}}^{\mathrm{i}(\frac{{\theta }_0}{n}+\frac{2\pi k}{n})}$$z=\root{n}\of{r_0}\e^{\i\left( {\theta_0\over n}+{2\pi k\over n} \right)}$ 柯西-黎曼方程：$f(z)=u(x,y)+\mathrm{i}v(x,y)$$f(z)=u(x,y)+\i v(x,y)$ 可导$\Rightarrow u_x=v_y,u_y=-v_x$$\Rarr~u_x=v_y,u_y=-v_x$。逆命题：$u_x=v_y,u_y=-v_x$$u_x=v_y,u_y=-v_x$ 且 $u_x,u_y,v_x,v_y$$u_x,u_y,v_x,v_y$ 都连续$\Rightarrow f(z)$$\Rarr~f(z)$可导。极坐标为 $r{u}_r=v_{\theta },r{v}_r=-u_{\theta }$$ru_r=v_\theta,rv_r=-u_\theta$。 反射原理：解析函数在实轴上取实值$\iff f(\overline{z})=\overline{f(z)}$$\Lrarr~f(\bar z)=\overline{f(z)}$

$\log z=\ln r+\mathrm{i}(\theta +2n\pi ),\mathrm{L}\mathrm{o}\mathrm{g}z=\ln r+\mathrm{i}\mathrm{\Theta }$$\log z=\ln r+\i(\theta+2n\pi),~{\rm Log}z=\ln r+\i\Theta$，其中 $\mathrm{\Theta }=\mathrm{A}\mathrm{r}\mathrm{g}z$$\Theta={\rm Arg} z$ 为主值。 $A\log B=\log (B^A)$$A\log B=\log(B^A)$ 不总是成立，但 $\log A+\log B=\log AB,\log A-\log B=\log (A/B)$$\log A+\log B=\log AB,~\log A-\log B=\log(A/B)$ 总是成立。 $\cos z=({\mathrm{e}}^{\mathrm{i}z}+{\mathrm{e}}^{-\mathrm{i}z})/2,\sin z=({\mathrm{e}}^{\mathrm{i}z}-{\mathrm{e}}^{-\mathrm{i}z})/2$$\cos z=(\e^{\i z}+\e^{-\i z})/2,~\sin z=(\e^{\i z}-\e^{-\i z})/2$ $\cosh z=({\mathrm{e}}^z+{\mathrm{e}}^{-z})/2,\sinh z=(e^z-e^{-z})/2$$\cosh z=(\e^z+\e^{-z})/2,~\sinh z=(e^z-e^{-z})/2$ $\cos \mathrm{i}y=\cosh y,\sin \mathrm{i}y=\mathrm{i}\sinh y$$\cos \i y=\cosh y,~\sin \i y=\i \sinh y$ $\cos z=\cos x\cosh y-\mathrm{i}\sin x\sinh y,\sin z=\sin x\cosh y+\mathrm{i}\cos x\sin y$$\cos z=\cos x\cosh y-\i\sin x\sinh y,~~\sin z=\sin x\cosh y+\i\cos x\sin y$ $\cosh z=\cosh x\cos y+\mathrm{i}\sinh x\sin y,\sinh z=\sinh x\cos y+\mathrm{i}\cosh x\sin y$$\cosh z=\cosh x\cos y+\i\sinh x\sin y,~~\sinh z=\sinh x\cos y+\i\cosh x\sin y$ $|\sin z{|}^2={\sin }^{2}x+{\sinh }^{2}y,|\cos z{|}^2={\cos }^{2}x+{\sinh }^{2}y$$|\sin z|^2=\sin^2x+\sinh^2y,~~|\cos z|^2=\cos^2x+\sinh^2y$ $|\sinh z{|}^2={\sinh }^{2}x+{\sin }^{2}y,|\cosh z{|}^2={\sinh }^{2}x+{\cos }^{2}y$$|\sinh z|^2=\sinh^2x+\sin^2y,~~|\cosh z|^2=\sinh^2x+\cos^2y$

柯西积分：当 $f$$f$ 在 $C$$C$ 内解析时，$f(z_0)=\frac{1}{2\pi \mathrm{i}}{\int }_C\frac{f(z)}{z-z_0}\mathrm{d}z,f^{(n)}(z_0)=\frac{n!}{2\pi \mathrm{i}}{\int }_C\frac{f(z)}{(z-z_0{)}^n}\mathrm{d}z$$f(z_0)={1\over2\pi\i}\int_C{f(z)\over z-z_0}\newcommand{\dd}{{\rm d}}\dd z,~~f^{(n)}(z_0)={n!\over2\pi\i}\int_C{f(z)\over (z-z_0)^n}\dd z$ 柯西不等式：$|f^{(n)}(z_0)|\le \frac{n!}{2\pi }\frac{M_R}{R^{n+1}}2\pi R$$|f^{(n)}(z_0)|\le{n!\over 2\pi}{M_R\over R^{n+1}}2\pi R$ 刘维尔定理：有界整函数必是常函数。

$\tan z=z+z^3/3+2z^5/15+17z^7/315+62z^9/2835+...$$\tan z=z+z^3/3+2z^5/15+17z^7/315+62z^9/2835+...$ $\arcsin z=z+z^3/6+3z^5/40+5z^7/112+35z^9/1152+...$$\arcsin z=z+z^3/6+3z^5/40+5z^7/112+35z^9/1152+...$ arcsinh一正一负 $\arccos z=\pi /2-z-z^3/6-3z^5/40-5x^7/112-35x^9/1152-...$$\arccos z=\pi/2-z-z^3/6-3z^5/40-5x^7/112-35x^9/1152-...$

洛朗级数：$f(z)=\sum c_n(z-z_0{)}^n,c_n=\frac{1}{2\pi \mathrm{i}}{\int }_c\frac{f(z)\mathrm{d}z}{(z-z_0{)}^{n+1}}$$f(z)=\sum c_n(z-z_0)^n,~~c_n={1\over 2\pi \i}\int_c{f(z)\dd z\over (z-z_0)^{n+1}}$

$\underset{z=\mathrm{\infty }}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)=-\underset{z=0}{\mathrm{R}\mathrm{e}\mathrm{s}}(f(1/z)/z^2),\underset{z=0}{\mathrm{R}\mathrm{e}\mathrm{s}}(f(1/z)/z^2)=\sum _{\mathrm{a}\mathrm{l}\mathrm{l}}\underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)$$\newcommand\Res[1]{\mathop{\rm Res~}\limits_{#1}}\Res{z=\infty}f(z)=-\Res{z=0}(f(1/z)/z^2),~~\Res{z=0}(f(1/z)/z^2)=\sum_{\rm all}\Res{}f(z)$ m阶极点：$f(z)=\frac{\phi (z)}{(z-z_0{)}^m}$$f(z)={\varphi(z)\over (z-z_0)^m}$ 其中φ(z)解析。 $\underset{z=z_0}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)=\frac{{\phi }^{(m-1)}(z_0)}{(m-1)!}$$\Res{z=z_0}f(z)={\varphi^{(m-1)}(z_0)\over(m-1)!}$ $p(z_0)\ne 0,q(z_0)=0$$p(z_0)\ne 0,~q(z_0)=0$ 且为一阶零点 $\Rightarrow \underset{z=z_0}{\mathrm{R}\mathrm{e}\mathrm{s}}\frac{p}{q}=\frac{p}{q^{\prime }}$$\Rarr~\Res{z=z_0}{p\over q}={p\over q'}$

若当引理：$f(z)$$f(z)$ 在 $C_R$$C_R$ 上最大值趋向于零，则 ${\int }_{C_R}f(z){\mathrm{e}}^{\mathrm{i}az}\mathrm{d}z\to 0$$\int_{C_R}f(z)\e^{\i a z}\dd z\to 0$ 绕一阶极点半圈积分 ${\int }_{c}f(z)\mathrm{d}z=\frac{1}{2}\underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)$$\int_c f(z)\dd z={1\over 2}\Res{}f(z)$ ${\int }_0^{2\pi }F(\cos x,\sin x)\mathrm{d}x={\int }_{c}F(\frac{z+z^{-1}}{2},\frac{z-z^{-1}}{2\mathrm{i}})\frac{\mathrm{d}z}{\mathrm{i}z}$$\int_0^{2\pi}F(\cos x,\sin x)\dd x=\int_cF\left( {z+z^{-1}\over 2},{z-z^{-1}\over 2\i} \right){\dd z\over \i z}$

$\stackrel{\mathrm{\infty }}{\sum _{-\mathrm{\infty }}}f(n)\to \sum \underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)\pi \cot \pi z,\stackrel{\mathrm{\infty }}{\sum _{-\mathrm{\infty }}}(-1{)}^{n}f(n)\to \sum \underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}\frac{f(z)\pi }{\sin \pi z}$$\overset{\infty}{\mathop\sum\limits_{-\infty}} f(n)\to \sum\Res{}f(z)\pi\cot \pi z,~~ \overset{\infty}{\mathop\sum\limits_{-\infty}}(-1)^n f(n)\to \sum\Res{}{f(z)\pi\over \sin \pi z}$ $\stackrel{\mathrm{\infty }}{\sum _{-\mathrm{\infty }}}f(n+\frac{1}{2})\to -\sum \underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}f(z)\pi \tan \pi z,\stackrel{\mathrm{\infty }}{\sum _{-\mathrm{\infty }}}(-1{)}^{n}f(n+\frac{1}{2})\to -\sum \underset{}{\mathrm{R}\mathrm{e}\mathrm{s}}\frac{f(z)\pi }{\cos \pi z}$$\overset{\infty}{\mathop\sum\limits_{-\infty}} f(n+{1\over 2})\to -\sum\Res{}f(z)\pi\tan \pi z,~~ \overset{\infty}{\mathop\sum\limits_{-\infty}}(-1)^n f(n+{1\over 2})\to -\sum\Res{}{f(z)\pi\over \cos \pi z}$

幅角原理：C内亚纯，C上非零：$\frac{1}{2\pi }{\mathrm{\Delta }}_c\arg f(z)=Z-P$${1\over 2\pi}\Delta_c\arg f(z)=Z-P$ 儒歇定理：f,g在C内解析，C上|f(z)|>|g(z)|.则f与f+g在C内有同数零点

$\mathrm{\Gamma }(z)=\underset{n\to \mathrm{\infty }}{lim}\frac{1.2.3...n}{z(z+1)(z+2)...(z+n)}{n}^z,z\ne 0,-1,-2,...\frac{1}{\mathrm{\Gamma }(z)}=z{\mathrm{e}}^{\gamma z}\stackrel{\mathrm{\infty }}{\prod _{n=1}}(1+\frac{z}{n}){\mathrm{e}}^{-z/n}$$\Gamma(z)=\underset{n\to \infty}{\lim}{1.2.3...n\over z(z+1)(z+2)...(z+n)}n^z,~~z\ne 0,-1,-2,...~~{1\over\Gamma(z)}=z\e^{\gamma z}\overset{\infty}{\underset{n=1}\prod}(1+{z\over n})\e^{-z/n}$ $\mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-t}{t}^{z-1}\mathrm{d}t=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-t^2}{t}^{2z-1}\mathrm{d}t={\int }_0^1{(\ln \frac{1}{t})}^{z-1}\mathrm{d}t,\mathrm{\Re }z>0$$\Gamma(z)=\int_0^\infty\e^{-t}t^{z-1}\dd t=2\int_0^\infty\e^{-t^2}t^{2z-1}\dd t=\int_0^1\left(\ln{1\over t}\right)^{z-1}\dd t,~~\Re z>0$ $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z),\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)=\frac{\pi }{\sin z\pi },\mathrm{\Gamma }(n)=(n-1)!,\mathrm{\Gamma }(1/2)=\sqrt{\pi }$$\Gamma(z+1)=z\Gamma(z),~~\Gamma(z)\Gamma(1-z)={\pi\over \sin z\pi},~~\Gamma(n)=(n-1)!,~~\Gamma(1/2)=\sqrt\pi$ $\mathrm{\Gamma }(1+z)\mathrm{\Gamma }(\frac{1}{2}+z)=2^{-2z}\sqrt{\pi }\mathrm{\Gamma }(2z+1)$$\Gamma(1+z)\Gamma({1\over 2}+z)=2^{-2z}\sqrt\pi\Gamma(2z+1)$ $\mathrm{\Gamma }(z)$$\Gamma(z)$在负整数处简单极点留数为 $\frac{(-1{)}^n}{n!}$${(-1)^n\over n!}$ $\psi (z)=\frac{{\mathrm{\Gamma }}^{\prime }(z)}{\mathrm{\Gamma }(z)},\psi (z+1)=\underset{n\to \mathrm{\infty }}{lim}(\ln n-\frac{1}{z+1}-\frac{1}{z+2}-...-\frac{1}{z+n})$$\psi(z)={\Gamma'(z)\over\Gamma(z)},~\psi(z+1)=\underset{n\to\infty}{\lim}\left( \ln n-{1\over z+1}-{1\over z+2}-...-{1\over z+n} \right)$ $\psi (z+1)=-\gamma +\sum _{m=1}^{\mathrm{\infty }}\frac{z}{m(m+z)},\psi (z)=-\gamma$$\psi(z+1)=-\gamma+\sum_{m=1}^\infty{z\over m(m+z)},~~\psi(z)=-\gamma$ ; ${\psi }^{(m)}(z+1)=(-1{)}^{m+1}m!\stackrel{\mathrm{\infty }}{\sum _{n=1}}\frac{1}{(z+n{)}^{m+1}}$$\psi^{(m)}(z+1)=(-1)^{m+1}m!\overset{\infty}{\mathop\sum\limits_{n=1}}{1\over(z+n)^{m+1}}$ $\ln \mathrm{\Gamma }(z+1)=\stackrel{\mathrm{\infty }}{\sum _{n=1}}\frac{z^n}{n!}{\psi }^{(n-1)}(1)=-\gamma z+\stackrel{\mathrm{\infty }}{\sum _{n=2}}(-1{)}^n\frac{z^n}{n}\zeta (n)$$\ln\Gamma(z+1)=\overset{\infty}{\mathop\sum\limits_{n=1}}{z^n\over n!}\psi^{(n-1)}(1)=-\gamma z+\overset{\infty}{\mathop\sum\limits_{n=2}}(-1)^n{z^n\over n}\zeta(n)$

$\mathrm{B}(p,q)=\frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{\mathrm{\Gamma }(p+q)}=2{\int }_0^{\pi /2}{\cos }^{2p-1}\theta {\sin }^{2q-1}\theta \mathrm{d}\theta$$\Beta(p,q)={\Gamma(p)\Gamma(q)\over \Gamma(p+q)}=2\int_0^{\pi/2}\cos^{2p-1}\theta\sin^{2q-1}\theta\dd \theta$ $\mathrm{B}(p+1,q+1)={\int }_0^1{t}^p(1-t{)}^q\mathrm{d}t=2{\int }_0^1{x}^{2p+1}(1-x^2{)}^q\mathrm{d}x={\int }_0^{\mathrm{\infty }}\frac{u^p}{(1+u{)}^{p+1+2}}\mathrm{d}u$$\Beta(p+1,q+1)=\int_0^1t^p(1-t)^q\dd t=2\int_0^1x^{2p+1}(1-x^2)^q\dd x=\int_0^\infty{u^p\over(1+u)^{p+1+2}}\dd u$

$\sum _{n=0}^N{C}_{N+1}^n{B}_n={\delta }_N^0,B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=-\frac{1}{30},...$$\sum_{n=0}^NC_{N+1}^nB_n=\delta_N^0,~~B_0=1,B_1=-{1\over 2},B_2={1\over 6},B_3=0,B_4=-{1\over 30},...$ $\frac{t}{{\mathrm{e}}^t-1}=\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{B_n}{n!}{t}^n,\frac{t}{2}\coth \frac{t}{2}-1=\stackrel{\mathrm{\infty }}{\sum _{n=2}}\frac{B_n}{n!}{t}^n,\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{B_n}{n!}(P^{(n)}(1)-P^{(n)}(0))=P^{\prime }(0)$${t\over\e^t-1}=\overset{\infty}{\mathop\sum\limits_{n=0}} {B_n\over n!}t^n ,~~{t\over 2}\coth{t\over 2}-1=\overset{\infty}{\mathop\sum\limits_{n=2}} {B_n\over n!}t^n,~~\overset{\infty}{\mathop\sum\limits_{n=0}}{B_n\over n!}(P^{(n)}(1)-P^{(n)}(0))=P'(0)$ ${\int }_0^{\mathrm{\infty }}f(x)\mathrm{d}x=\frac{f(0)}{2}+\stackrel{\mathrm{\infty }}{\sum _{n=1}}f(n)+\stackrel{\mathrm{\infty }}{\sum _{n=2}}\frac{B_n}{n!}{f}^{(n-1)}(0)$$\int_0^\infty f(x)\dd x={f(0)\over 2}+\overset{\infty}{\mathop\sum\limits_{n=1}}f(n)+\overset{\infty}{\mathop\sum\limits_{n=2}}{B_n\over n!}f^{(n-1)}(0)$ $\log \mathrm{\Gamma }(z+1)=\frac{\ln (2\pi )}{2}+(z+\frac{1}{2})\log z-z+\frac{1}{12z}..,\mathrm{\Gamma }(z+1)=\sqrt{2\pi }\frac{z^{z+\frac{1}{2}}}{{\mathrm{e}}^{-z}}(1+\frac{1}{12}+..)$$\log\Gamma(z+1)={\ln(2\pi)\over 2}+(z+{1\over2})\log z-z+{1\over12z}..,~\Gamma(z+1)=\sqrt{2\pi}{z^{z+{1\over 2}}\over\e^{-z}}(1+{1\over 12}+..)$

$\zeta (z)=\frac{1}{\mathrm{\Gamma }(z)}{\int }_0^{\mathrm{\infty }}\frac{t^{z-1}\mathrm{d}t}{{\mathrm{e}}^t-1},(\mathrm{\Re }z>1),\zeta (z)$$\zeta(z)={1\over\Gamma(z)}\int_0^\infty{t^{z-1}\dd t\over \e^t-1},~~(\Re z>1),~~~\zeta(z)$在$z=1$$z=1$ 处简单极点留数为1 $\mathrm{\Gamma }(\frac{z}{2}){\pi }^{-z/2}\zeta (z)=\mathrm{\Gamma }(\frac{z-1}{2}){\pi }^{-\frac{1-z}{2}}\zeta (1-z)$$\Gamma({z\over 2})\pi^{-z/2}\zeta(z)=\Gamma({z-1\over 2})\pi^{-{1-z\over 2}}\zeta(1-z)$ $n\ge 1:\zeta (2n)=(-1{)}^{n+1}\frac{(2\pi {)}^{2n}{B}_{2n}}{2(2n)!},\zeta (-n)=\frac{-B_n}{n+1}$$n\ge 1:~\zeta(2n)=(-1)^{n+1}{(2\pi)^{2n}B_{2n}\over2(2n)!},~\zeta(-n)={-B_n\over n+1}$

傅里叶变换：$g(w)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t){\mathrm{e}}^{\mathrm{i}wt}\mathrm{d}t,g(\overrightarrow{k})=\frac{1}{{\sqrt{2\pi }}^3}\int f(\overrightarrow{r}){\mathrm{e}}^{\mathrm{i}\overrightarrow{k}\cdot \overrightarrow{r}}{\mathrm{d}}^{3}r$$g(w)={1\over \sqrt{2\pi}}\int_{-\infty}^\infty f(t)\e^{\i wt}\dd t,~~g(\vec k)={1\over \sqrt{2\pi}^3}\int f(\vec r)\e^{\i\vec k\cdot\vec r }\dd^3r$ ${\delta }_n(x)=\frac{\sin nx}{\pi x}=\frac{1}{2\pi }{\int }_{-n}^n{\mathrm{e}}^{\mathrm{i}wt}\mathrm{d}t,\delta (x)=\underset{n\to \mathrm{\infty }}{lim}{\delta }_n(x)$$\delta_n(x)={\sin nx\over \pi x}={1\over 2\pi}\int_{-n}^n\e^{\i wt}\dd t,~\delta(x)=\lim_{n\to \infty}\delta_n(x)$ ${\left[f(\overrightarrow{r}-\overrightarrow{R})\right]}^T={\mathrm{e}}^{\mathrm{i}\overrightarrow{k}\cdot \overrightarrow{R}}g(\overrightarrow{k}),{\left[f(a\overrightarrow{r})\right]}^T=\frac{g(\overrightarrow{k}/a)}{a^3}(a>0),{\left[f(-\overrightarrow{r})\right]}^T=g(-\overrightarrow{k}),{\left[f^{\ast }(-\overrightarrow{r})\right]}^T=g^{\ast }(\overrightarrow{k})$$\newcommand{\t}[1]{\left[{#1}\right]^T}\t{f(\vec r-\vec R)}=\e^{\i\vec k\cdot\vec R}g(\vec k),~~\t{f(a\vec r)}={g(\vec k/a)\over a^3}~~(a>0),~~\t{f(-\vec r)}=g(-\vec k),~~\t{f^*(-\vec r)}=g^*(\vec k)$ ${\left[\mathrm{\nabla }f(\overrightarrow{r})\right]}^T=-\mathrm{i}\overrightarrow{k}g(\overrightarrow{k}),{\left[{\mathrm{\nabla }}^{2}f(\overrightarrow{r})\right]}^T=-k^{2}g(\overrightarrow{k}),{\left[t^{n}f(t)\right]}^T={\mathrm{i}}^{-n}\frac{{\mathrm{d}}^{n}g(w)}{\mathrm{d}{w}^n}$$\t{\nabla f(\vec r)}=-\i\vec k g(\vec k),~~\t{\nabla^2 f(\vec r)}=-k^2g(\vec k),~~\t{t^nf(t)}=\i^{-n}{\dd^ng(w)\over\dd w^n}$ 对于卷积$(f\ast g)(x)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(y)g(x-y)\mathrm{d}y$$(f*g)(x)={1\over \sqrt{2\pi}}\int_{-\infty}^\infty f(y)g(x-y)\dd y$ 有 ${\left[f\ast g\right]}^T={\left[f\right]}^T{\left[g\right]}^T\Rightarrow f\ast g={\left[{\left[f\right]}^T{\left[g\right]}^T\right]}^{-T}$$\t{f*g}=\t{f}\t{g}~~\Rarr~~f*g=\left[\t{f}\t{g}\right]^{-T}$

拉普拉斯变换：$\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-st}f(t)\mathrm{d}t$$\newcommand{\l}[1]{\mathcal{L}\left\{#1\right\}}\l{f(t)}=\int_0^\infty\e^{-st}f(t)\dd t$ 默认当 $t<0$$t<0$ 有 $f(t)=0$$f(t)=0$ $\mathcal{L}\left\{1\right\}=\frac{1}{s}(\mathrm{\Re }s>0),\mathcal{L}\left\{{\mathrm{e}}^{kt}\right\}=\frac{1}{s-k}(\mathrm{\Re }s>\mathrm{\Re }k),\mathcal{L}\{\cosh kt\}=\frac{s}{s^2-k^2},\mathcal{L}\{\sinh kt\}=\frac{k}{s^2-k^2}$$\l{1}={1\over s}~~(\Re s>0),~~\l{\e^{kt}}={1\over s-k}~~(\Re s>\Re k),~~\l{\cosh kt}={s\over s^2-k^2},~~\l{\sinh kt}={k\over s^2-k^2}$ $\mathcal{L}\{\cos kt\}=\frac{s}{s^2+k^2},\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2},\mathcal{L}\left\{t^n\right\}=\frac{\mathrm{\Gamma }(n+1)}{s^{n+1}},\mathcal{L}\{\delta (t)\}=1$$\l{\cos kt}={s\over s^2+k^2},~~\l{\sin kt}={k\over s^2+k^2},~~\l{t^n}={\Gamma(n+1)\over s^{n+1}},~~\l{\delta(t)}=1$ $\mathcal{L}\{f(t)\}=F(s)\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0^{+})\mathcal{L}\{f^{″}(t)\}=s^{2}F(s)-sf(0)+f^{\prime }(0)$$\l{f(t)}=F(s)~~~\l{f'(t)}=sF(s)-f(0^+)~~~\l{f''(t)}=s^2F(s)-sf(0)+f'(0)$ $\mathcal{L}\{{\int }^{t}f(x)\mathrm{d}x\}=\frac{F(s)}{s},\mathcal{L}\{f(at)\}=\frac{1}{a}F(\frac{s}{a}),\mathcal{L}\{{\mathrm{e}}^{at}f(t)\}=F(s-a),\mathcal{L}\{f(t-b)\}={\mathrm{e}}^{-bs}F(s)$$\l{\int^tf(x)\dd x}={F(s)\over s},~~\l{f(at)}={1\over a}F({s\over a}),~~\l{\e^{at}f(t)}=F(s-a),~~\l{f(t-b)}=\e^{-bs}F(s)$ $\mathcal{L}\{(-t{)}^{n}f(t)\}=F^{(n)}(s),\mathcal{L}\left\{\frac{f(t)}{t}\right\}={\int }_{s}F(x)\mathrm{d}x$$\l{(-t)^nf(t)}=F^{(n)}(s),~~\l{{f(t)\over t}}=\int_sF(x)\dd x$ 对于卷积 $(f\ast g)(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(y)g(x-y)\mathrm{d}y$$(f*g)(x)=\int_{-\infty}^\infty f(y)g(x-y)\dd y$ 有 $\mathcal{L}\{f\ast g\}=\mathcal{L}\left\{f\right\}\mathcal{L}\left\{g\right\}$$\l{f*g}=\l{f}\l{g}$ ${\mathcal{L}}^{-1}\{F(s)\}=f(t)=\frac{1}{2\pi \mathrm{i}}\mathrm{P}\mathrm{V}{\int }_{\beta -\mathrm{i}\mathrm{\infty }}^{\beta +\mathrm{i}\mathrm{\infty }}{\mathrm{e}}^{st}F(s)\mathrm{d}s$$\mathcal{L}^{-1}\{F(s)\}=f(t)={1\over 2\pi \i}{\rm PV}\int_{\beta-\i\infty}^{\beta+\i\infty}\e^{st}F(s)\dd s$，其中$\beta$$\beta$满足极点均在其左。

$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$${(w-w_1)(w_2-w_3)\over(w-w_3)(w_2-w_1)}={(z-z_1)(z_2-z_3)\over(z-z_3)(z_2-z_1)}$ 将 $z_1,z_2,z_3$$z_1,z_2,z_3$ 分别映到 $w_1,w_2,w_3$$w_1,w_2,w_3$。 $w={\mathrm{e}}^{\mathrm{i}\alpha }\frac{z-z_0}{z-\overline{z_0}},\mathrm{\Im }{z}_0>0$$w=\e^{\i\alpha}{z-z_0\over z-\overline{z_0}},~~\Im z_0>0$ 将上半平面映到单位圆内。

$w=\log \frac{z-1}{z+1}\Rightarrow v=\arctan \left(\frac{2y}{x^2+y^2-1}\right)$$w=\log{z-1\over z+1}~~\Rarr~~v=\arctan\left({2y\over x^2+y^2-1}\right)$

$w=\exp z$$w=\exp z$	$w=\exp z$$w=\exp z$	$w=\exp z$$w=\exp z$
$w=\sin z$$w=\sin z$	$w=\sin z$$w=\sin z$	$w=z+\frac{1}{z}$$w=z+{1\over z}$
$w=z+\frac{1}{z}$$w=z+{1\over z}$	$w=\mathrm{L}\mathrm{o}\mathrm{g}\frac{z-1}{z+1}$$w={\rm Log}~{z-1\over z+1}$