update and fix typo

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### Complex Integral
#### Definition 4.1
#### Definition 6.1
If $\phi(t)$ is a complex function defined on $[a,b]$, then the integral of $\phi(t)$ over $[a,b]$ is defined as
@@ -48,7 +48,7 @@ $$
\int_a^b \phi(t) dt = \int_a^b \text{Re}\{\phi(t)\} dt + i\int_a^b \text{Im}\{\phi(t)\} dt
$$
#### Theorem 4.3 (Triangle Inequality)
#### Theorem 6.3 (Triangle Inequality)
If $\phi(t)$ is a complex function defined on $[a,b]$, then
@@ -74,7 +74,7 @@ Assume $\phi$ is continuous on $[a,b]$, the equality means $\lambda(t)\phi(t)$ i
EOP
#### Definition 4.4 Arc Length
#### Definition 6.4 Arc Length
Let $\gamma$ be a curve in the complex plane defined by $\gamma(t)=x(t)+iy(t)$, $t\in[a,b]$. The arc length of $\gamma$ is given by
@@ -117,7 +117,7 @@ $$
\int_{\Gamma} f(\zeta) d\zeta=\int_0^{2\pi} f(\gamma(t))\gamma'(t) dt=\int_0^{2\pi} \frac{1}{Re^{-it}}iRe^{it} dt=2\pi i
$$
#### Theorem 4.11 (Uniform Convergence)
#### Theorem 6.11 (Uniform Convergence)
If $f_n(z)$ converges uniformly to $f(z)$ on $\Gamma$, assume length of $\Gamma$ is finite, then
@@ -144,7 +144,7 @@ $$
EOP
#### Theorem 4.6 (Integral of derivative)
#### Theorem 6.6 (Integral of derivative)
Suppose $\Gamma$ is a closed curve, $\gamma:[a,b]\to\mathbb{C}$ and $\gamma(a)=\gamma(b)$.

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# Math416 Lecture 11
## Continue on integration over complex plane
### Continue on last example
Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-\zeta^2}d\zeta=0$, however, the integral consists of four parts:
Path 1: $-a\to a$
$\int_{I_1}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-x^2}dx$
Path 2: $a+ib\to -a+ib$
$\int_{I_2}e^{-\zeta^2}d\zeta=\int_{a+ib}^{-a+ib}e^{-\zeta^2}d\zeta=\int_{0}^{b}e^{-(a+iy)^2}dy$
Path 3: $-a+ib\to -a-ib$
$-\int_{I_3}e^{-\zeta^2}d\zeta=-\int_{-a+ib}^{-a-ib}e^{-\zeta^2}d\zeta=-\int_{a}^{-a}e^{-(x-ib)^2}dx$
Path 4: $-a-ib\to a-ib$
$-\int_{I_4}e^{-\zeta^2}d\zeta=-\int_{-a-ib}^{a-ib}e^{-\zeta^2}d\zeta=-\int_{b}^{0}e^{-(-a+iy)^2}dy$
> #### The reverse of a curve 6.9
>
> If $\gamma:[a,b]\to\mathbb{C}$ is a curve, then the rever of $\gamma$ is the curve $-\gamma:[-b,-a]\to\mathbb{C}$ defined by $(-\gamma)(t)=\gamma(a+b-t)$. It is the curve one obtains from $\gamma$ by traversing it in the opposite direction.
>
> - If $\gamma$ is piecewise in $C^1$, then $-\gamma$ is piecewise in $C^1$.
> - $\int_{-\gamma}f(z)dz=-\int_{\gamma}f(z)dz$ for any function $f$ that is continuous on $\gamma([a,b])$.
If we keep $b$ fixed, and let $a\to\infty$, then
> #### Definition 6.10 (Estimate of the integral)
>
> Let $\gamma:[a,b]\to\mathbb{C}$ be a piecewise $C^1$ curve, and let $f:[a,b]\to\mathbb{C}$ be a continuous complex-valued function. Let $M$ be the maximum of $|f|$ on $\gamma([a,b])$. ($M=\max\{|f(t)|:t\in[a,b]\}$)
>
> Then
>
> $$\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)M$$
_Continue on previous example, we have:_
$$
\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)\max_{\zeta\in\gamma}|f(\zeta)|\to 0
$$
Since,
$$
\int_{-\infty}^{\infty}e^{-x^2}dx-\int_{-\infty}^{\infty}e^{-x^2+b^2}(\cos 2bx+i\sin 2bx)dx=0
$$
Since $\sin 2bx$ is odd, and $\cos 2bx$ is even, we have
$$
\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-x^2+b^2}\cos 2bxdx=\sqrt{\pi}e^{-b^2}
$$
##### Proof for the last step:
$$
\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}
$$
Proof:
Let $J=\int_{-\infty}^{\infty}e^{-x^2}dx$
Then
$$J^2=\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy$$
We can evaluate the integral on the right-hand side by converting to polar coordinates. $x=r\cos\theta$, $y=r\sin\theta,dxdy=rdrd\theta$
$$
J^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta
$$
$$
J^2=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta=\int_{0}^{2\pi}\left[-\frac{1}{2}e^{-r^2}\right]_{0}^{\infty}d\theta
$$
$$
J^2=\int_{0}^{2\pi}\frac{1}{2}d\theta=\pi
$$
$$
J=\sqrt{\pi}
$$
EOP
## Chapter 7 Cauchy's theorem
### Cauchy's theorem (Fundamental theorem of complex function theory)
Let $\gamma$ be a closed curve in $\mathbb{C}$ and let $u$ be an open set _containing $\gamma^*$_. Let $f$ be a holomorphic function on $u$. Then
$$
\int_{\gamma}f(z)dz=0
$$
Note: What "containing $\gamma^*$" means? (Rabbit hole for topologists)
#### Lemma 7.1 (Goursat's lemma)
Cauchy's theorem is true if $\gamma$ is a triangle.
Proof:
We plan to keep shrinking the triangle until $f(\zeta+h)=f(\zeta)+hf'(\zeta)+\mu(h)$ where $\mu(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
Let's start with a triangle $T$ with vertices $z_1,z_2,z_3$.
![carving a triangle](https://notenextra.trance-0.com/Math416/Cauchy_theorem_triangle_carving.png)
We divide $T$ into four smaller triangles by drawing lines from the midpoints of the sides to the opposite vertices.

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@@ -13,4 +13,5 @@ export default {
Math416_L8: "Complex Variables (Lecture 8)",
Math416_L9: "Complex Variables (Lecture 9)",
Math416_L10: "Complex Variables (Lecture 10)",
Math416_L11: "Complex Variables (Lecture 11)",
}

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