Engineering Mathematic I -- Notes 1

Engineering Mathematic I – Notes 1

个人学习内容的总结（第一部分），对应课程为复变函数及其积分变换

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Contents:

1. Complex Calculation
2. Complex Function
3. Complex Line Integrals

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1. Complex Calculation

• Argument of a complex number
  For a complex number(z = x + yi):
  $${\theta }_z=\{\begin{array}{ll}arctan\frac{y}{x}& x>0\\ arctan\frac{y}{x}+\pi & x<0,y>0\\ arctan\frac{y}{x}-\pi & x<0,y<0\end{array}$$

• Polar representation of a complex number:
  $$z=r(cos\theta +\mathbf{i}sin\theta )$$
  , where $r$ is the modulus of z.
  
  

• Exponential form of a complex number:
  $$e^{\mathrm{i}\theta }=cos\theta +\mathrm{i}sin\theta$$ $$z=r{e}^{\mathrm{i}\theta }=r(cos\theta +\mathrm{i}sin\theta )$$

• Properties of polar/exponential forms:
  $$z_1\cdot z_2=r_1{r}_2[cos({\theta }_1+{\theta }_2)+\mathrm{i}sin({\theta }_1+{\theta }_2)]=r_1{r}_2{e}^{\mathrm{i}({\theta }_1+{\theta }_2)}$$

  $$\frac{z_1}{z_2}=\frac{r_1}{r_2}[cos({\theta }_1-{\theta }_2)+\mathrm{i}sin({\theta }_1-{\theta }_2)]=\frac{r_1}{r_2}{e}^{\mathrm{i}({\theta }_1-{\theta }_2)}$$

• The n^th power / roots of a complex number:
  $$(z{)}^n=r^n[cos(n\theta )+\mathrm{i}sin(n\theta )]=r^n{e}^{\mathrm{i}\mathrm{n}\theta }$$

  $$\sqrt[n]{z}=\sqrt[n]{r}(cos\frac{\theta +2k\pi }{n}+\mathrm{i}sin\frac{\theta +2k\pi }{n})=\sqrt[n]{r}{e}^{\mathrm{i}\frac{\theta +2k\pi }{n}}$$
  , where k = 1, 2, 3 …
  
  

• complex logarithm:
  $$\ln z=\ln |z|+\mathrm{i}arg(z)$$
  $$\ln z=\ln |z|+\mathrm{i}Arg(z)+\mathrm{i}2k\pi$$
  , where k = 1, 2, 3…
  Principal value of complex logarithm:
  $$\mathrm{Ln}z=\ln |z|+\mathrm{i}Arg(z)$$

2. Complex Function

For a complex function with the form of $f_{(z)}=u_{(x,y)}+\mathbf{i}{v}_{(x,y)}$ is defined on a domain $D$:

• Cauchy - Riemann Equations
  A complex equation is analytic on $D$ if it satisfies the C-R equations:
  $$u_x=v_y$$
  $$u_y=-v_x$$

• Cauchy - Riemann Equations in polar form
  $$u_r=\frac{1}{r}{v}_{\theta }$$
  $$v_r=-\frac{1}{r}{u}_{\theta }$$

• Harmonic functions and Laplace’s equation
  If a complex fucntion $f_{(z)}$ is analytic on $D$, the it satisfies the Laplace’s equation:
  $${\nabla }^{2}f=f_{xx}+f_{yy}=0$$

  Consequently, $u_{(x,y)}$, $v_{(x,y)}$ satisfy:
  $${\nabla }^{2}u=u_{xx}+u_{yy}=0$$
  $${\nabla }^{2}v=v_{xx}+v_{yy}=0$$
  Moreover, these derivatives are continuous.
  
  Additionally, a real function $u_{(x,y)}$ is called harmonic if it has continuous second order partial derivatives and $u_{xx}+u_{yy}=0$.
  
  

• Harmonic conjugate function
  Gievn the discribed situation, $v_{(x,y)}$ is said to the harmonic conjugate function of $u_{(x,y)}$ on $D$.

3. Complex Line Integrals

For a complex Integral with the form of ${\int }_{\gamma }{f}_{(z)}dz$ or with a closed path ${\oint }_{\gamma }{f}_{(z)}dz$

• the ML-inequality

  For a complex function $f_{(z)}$ satisfying the $|f_{(z)}|\le M$ for all z in domain, Then
  $$|{\oint }_{\gamma }{f}_{(z)}dz|\le ML$$
  , where $L$ is the length of $\gamma$ .
  
  

• Integration by parametrization
  Let $\gamma$ represented by $z_{(t)},t\in (a,b)$:
  $${\int }_{\gamma }{f}_{(z)}dz={\int }_a^b{f}_{(z(t))}{\dot{z}}_{(t)}$$
  , where ${\dot{z}}_{(t)}={\dot{\mathrm{x}}}_{(t)}+{\mathrm{i}\dot{\mathrm{y}}}_{(t)}$.

• Integration of analytic functions
  Let $f_{(z)}$ be an analytic function on a simply connected domain and $F^{\prime }(z)=f(z)$:
  $${\int }_{\gamma }f(z)dz=F(z_1)-F(z_2)$$
  , where $z_1$ is endpoint of $\gamma$ and $z_2$ is the starting point of $\gamma$.
  
  

• Cauthy’s Integral Theorem
  If a function $f(z)$ is analytic on a simply connected domain, then for every closed path $\gamma$:
  $${\oint }_{\gamma }f(z)dz=0$$

• Cauthy’s integral formula
  If $f(z)$ is analytic on a simply connected domain, then for any counterclockwise oriented simple closed path $C$ enclosing the point $z_0$:
  $${\oint }_C\frac{f(z)}{z-z_0}dz=2\pi \mathrm{i}f(z_0)$$
  For mutiply connected domain:
  $${\oint }_{C_1}\frac{f(z)}{z-z_0}dz+{\oint }_{C_2}\frac{f(z)}{z-z_0}dz=2\pi \mathrm{i}f(z_0)$$

• Cauchy’s integral formula for derivatives
  If f (z) is analytic on a domain, then for any point $z_0\in D$, any counterclockwise oriented simple closed path $\gamma$ enclosing $z_0$:
  $${\oint }_C\frac{f(z)}{(z-z_0{)}^{n+1}}dz=\frac{2\pi \mathrm{i}}{n!}{f}^{(n)}(z_0)$$