• 🧠 Memory: Recap on integration

• We can sum over a curve in complex space using

  $$ \begin{aligned} S_n&=\sum^n_{k=1}(u_k+i v_k)(\Delta x_k+i\Delta y_k)\\ &=\sum^n_{k=1}(u_k\Delta x_k -v_k \Delta y_k)+i(v_k\Delta x_k+u_k\Delta y_k) \end{aligned} $$

• In the limit of $n\to \infin$

  $$ \begin{aligned} \int_Cf(z)\,\text dz&=\lim_{n\to\infin}S_n \\ &=\int_Cu(x,y)\,\text dx-v(x,y)\,\text dy +i \int_C v(x,y)\,\text dx+u(x,y)\,\text dy \end{aligned} $$

  🗒️ Note: these are now dependent line integral that relate to the path $C$

• If we define $\text d \vec r=(\text dx,\text dy)$ then

  $$ \boxed{\int_Cf(z)\,\text dz=\int_C(u,-v)\cdot \text d\vec r+i \int_C (v,u)\cdot \text d \vec r} $$

💼 Case: Consider a path $C$ from $a$ to $b$ that is analytic over the range $[a,b]$

• The reversed path will run from $b \to a$

  $$ \int^b_af(z)\,\text dz=-\int^a_bf(z)\,\text dz \qquad \int_Cf(z)\,\text dz=-\int_{-C}f(z)\,\text dz $$

• If $s$ is a point on $C$ between $a$ and $b$, then we can split the integral

  $$ \begin{aligned} \left . \int^b_af(z)\,\text dz=\int^s_af(z)\,\text dz+\int^b_sf(z)\,\text dz \;\right | \; \int_{C}f(z)\,\text dz=\int_{C_1}f(z)\,\text dz+\int_{C_2}f(z)\,\text dz \end{aligned} $$

  where $C=C_1+C_2$

• We can also write a difference

$$ \begin{aligned} \int_{C_1}f(z)\,\text dz=\int_{C}f(z)\,\text dz-\int_{C_2}f(z)\,\text dz \end{aligned} $$

where $C_1=C-C_2$

🗒️ Note: if $(a=b)$, $C$ closed, then $\oint_C$, if non-self intersection = Jordan curve, traversed clockwise

Estimation lemma (helper theorem)

🧔‍♀️ Theorem: If $M\ge|f(z)|$ for $|f(z)|$ along $C$ then

$$ \left | \int_Cf(z)\,\text dz \right |\le ML $$

where $L$ is the length of the path

💫 Proof: Starting with $S_n=\sum^n_{k=1}f(\xi_k)\Delta k$, taking the absolute value

$$ |S_n|\le \sum^n_{k=1}|f(\xi_k)||\Delta z_k|\le M\sum^n_{k=1}|\Delta z_k| $$

where we used $|z_1+z_1|\le |z_1|+|z_2|$.

$n\to\infin$: $|S_n|\to |\int_Cf(z)|$ and $\sum^n_{k=1}|\Delta z_k|\to L$. Thus we showed $\left | \int_Cf(z)\,\text dz \right |\le ML$

💃 Example: lets try integrating $\int^{i+1}_0 z^2 \,\text dz$, where $z^2=f(z)=x^2-y^2+2ixy$

1. 🐤 Start by integrating along $y=x^2$ we have $\text dy=2x\, \text dx$

• On the path $y=x^2$, $f(z)$ reduces to

  $$ f(z)=x^2-x^4+2ix^3=u+iv $$

• The integral becomes

  $$ \begin{aligned} \int_C f(z)\,\text dz&=\int_Cu\,\text dx-v\,\text dy+i\int_C u\,\text dy+v\,\text dx\\ &=\int^1_0(x^2-x^4)\,\text dx-\int^1_0 2x^3\,\text dy+i\int^1_0(x^2-x^4)\,\text dy+i\int^1_02x^3\,\text dx

  \end{aligned} $$

  Changing $\text dy\to 2x\,\text dx$ we get

  $$ \begin{aligned} \int_C f(z)\,\text dz&=\int_0^1 (x^2-x^4)\,\text dx+i\int_0^1 (4x^3-2x)\,\text dx \\ &=\frac 13-1+i\left ( 1-\frac 26 \right ) =\frac 23(-1+i)

  \end{aligned} $$

This is the first path from $0$ to $1+i$

This is the second path from $0$ to $1+i$

1. 🐥 Along $C'=C'_1+C'_2$ where $C_1'$ is $y=0$ for $x\in[0,1]$ and $C'_2$ is $x=1$ for $y\in[0,1]$

• On $C_1'$ we have $y=0$, $\text dy=0$, $u=x^2$ and $v=0$ so the integral is

  $$ \int_{C'_1}f(z)\,\text dz=\int_0^1u\,\text dx=\frac 13 $$

• On $C'_2$ we have $x=1$, $\text dx=0$, $u=1-y^2$ and $v=2y$ so the integral is

  $$ \begin{aligned} \int_{C_2'}f(z)\,\text dz&=\int^1_0 (-v+iu)\,\text dy \\ &=-\int^1_0 2y\,\text dy+i\int^1_0 (1-y^2)\,\text dy \\ &=-1+i\left ( 1-\frac 13 \right ) \end{aligned} $$

• Putting together we get

  $$ \begin{aligned} \int_{C'}f(z)\,\text dz&=\int_{C_1'}f(z)\,\text dz+\int_{C_2'}f(z)\,\text dz=\frac{1}{3}-1+i\frac{2}{3} \\ &=\frac{2}{3}(-1+i)

  \end{aligned} $$