💼 Case: Consider the function $f(\omega)=\omega^2=a^2-b^2+2iab$ where $(\omega=a+ib)\in\Z$

To try and visualise it we can set either $a$ or $b$ to a constant $c$

• 🟦 $a=c$ we get $c+ib$ and $f(c+ib)=c^2-b^2+2icb$ (this will be in blue)
• 🟥 $b=c$ we get $a+ic$ and $f(a+ic)=a^2-c^2+2iac$ (this will be in red)

Plotting this we get:

Left is $c+ib$ and $a+ic$, right its the parametric functions $(c^2-b^2,2cb)$ and $(a^2-c^2,2ca)$

We can project it in 3D such that the $z$-axis is $c$ the $y$ axis is imaginary and $x$ is the real axis

https://www.desmos.com/3d/4b999fc662

https://www.desmos.com/3d/404e250f7c

🗒️ Notes:

• The angles between the red and blue is preserved by the mapping
• You can go see the 3D graph of $\omega^2$ where $x=a$, $y=b$ and $z$ is either complex or real here

💼 Case: consider the exponential function $f(\omega)=e^\omega=e^a e^{ib}$

Blue is $a=c$ red is $b=c$ with $c\in[0,4.5]$

Blue is $a=c$ red is $b=c$ with $c\in[0,4.5]$

Plot of $\operatorname{Re}[e^\omega]$ where we see $a$ and $b$ in the axis

🗒️ Note: the function is periodic so generally $b\in[-\pi,\pi]$ to avoid the repetition

Analytic functions and mappings

🧔‍♀️ Theorem: for an analytic function $f(w)$ the lines of $x=$ const and $y=$ const are perpendicular both before and after the mapping except at $f(\omega)=0$

💫 Proof: consider$f(z)=u(x,y)+iv(x,y)$ is analytic, thus it follows Laplace’s equation

$$ \nabla^2u=\nabla^2v=0 \quad \text{where}\quad \nabla=\hat e_x \frac{\partial}{\partial x}+\hat e_y \frac{\partial}{\partial y} $$

• To show that $u$ and $v$ are $\perp$ we show $\nabla u \cdot \nabla v=0$, we can do this as follows:

  $$ \nabla u(x,y)=\frac{\partial u(x,y)}{\partial x}\hat e_x+\frac{\partial u(x,y)}{\partial y}\hat e_y

  $$

• Using Cauchy-Riemann equations:

  $$ \begin{aligned} \nabla v(x,y)&=\frac{\partial v(x,y)}{\partial x}\hat e_x+\frac{\partial v(x,y)}{\partial y}\hat e_y=-\frac{\partial u(x,y)}{\partial y}\hat e_x+\frac{\partial u(x,y)}{\partial x}\hat e_y \\ \nabla u(x,y)\cdot \nabla v(x,y)&=-\frac{\partial u(x,y)}{\partial x}\frac{\partial u(x,y)}{\partial y}+\frac{\partial u(x,y)}{\partial y}\frac{\partial u(x,y)}{\partial x}=0 \end{aligned} $$

Conformal mappings in physics

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e79573b7-3130-4003-9334-a3a98dd09b3d/Conformal_mapping.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e79573b7-3130-4003-9334-a3a98dd09b3d/Conformal_mapping.png" width="40px" /> Conformal mapping: two lines intersecting at $z_0$ in the $z$ plane can be mapped to some $w=f(z)$ plane where they will intersect at $w_0=f(z_0)$ with the same angle

</aside>

💫 Proof: Let $f(z)$ be analytic at $z=z_0$ and let $\omega_0=f(z_0)$. We define $\omega=f(z_0+\Delta z)$

$$ \begin{aligned} \Delta \omega&=\omega-\omega_0 \\ &=f(z_0+\Delta z)-f(z_0) \\ &=\Delta z \left . \frac{\text df}{\text dz} \right |_{z_0}=\Delta zf'(z_0) \end{aligned} $$

ie every small displacement $\Delta z$ from $z_0$ is unfirmly scaled and rotated by $f'(z_0)$ when mapped to the $\omega$-plane; because this transformation is consistent for all directions of $\Delta z$, the angle between any two paths through $z_0$ is preserved in the mapping

💃 Example: suppose $\Delta z=\epsilon e^{i\theta}$ where $\epsilon \ll 1$ and writing $f'(z_0)=Me^{i\alpha}$