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I'm trying to understand conformal mapping.

A function is conformal if it preserves angles. So $(g \circ \gamma_1)'(z_1) = (g \circ \gamma_2)'(z_1)$

I'm reading a book, and just after the definition, there is information that $\bar{Z}$ is not conformal, as it revers angles and $z^2$ is conformal on $\lbrace Re(z) > 0\rbrace$. I understand why $z^3$ won't be conformal on that set (it reverses angles after "full circle"), but I cannot understand why $z^2$ is.

I see that $z^2 : \lbrace Re(z) > 0\rbrace \to \mathbb{C} \smallsetminus (-\infty, 0]$, so angles are not reversed, but I don't get how they are preserved. It looks for me like every angle is greater. Angle of $\pi/4$ ($\gamma_1=(x, x)$ and $\gamma_2=(x, -x)$ for $x > 0$) is mapped to $\pi/2$.

So, what "preserves angles" means? As I cannot see how $z^2$ may preserves something. (Below picture from Complex Analysis T. Gamelin) enter image description here

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    $\begingroup$ $z^2$ is not conformal at zero and you measure angles around zero in your example; $z^2$ is conformal around any non zero point but you need to have angles centered at such points $\endgroup$
    – Conrad
    Dec 7, 2019 at 4:34
  • $\begingroup$ As Conrad says, $z\mapsto z^2$ is not conformal at $z=0$ since the derivative vanishes at $z=0$. $\endgroup$
    – robjohn ♦
    Dec 7, 2019 at 8:04
  • $\begingroup$ @Conrad I see the root of my problem, thank you! $\endgroup$
    – Tacet
    Dec 7, 2019 at 19:15

1 Answer 1

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Conformal maps preserve angles at a microscopic level; that is, if two curves intersect at an angle $\alpha$ in the source space, they also intersect at the same angle $\alpha$ in the destination space. This means their Jacobian is a constant times a rotation (that is, a radially scaled rotation matrix).

In $\mathbb{C}$, if $f(z)$ differentiable at $z_0$ and $f'(z_0)\ne0$, $f(z)$ is conformal at $z_0$. In fact, near $z_0$, $$ f(z)=f(z_0)+f'(z_0)(z-z_0)+o(z-z_0) $$ so the constant of the radial scale is $|f'(z_0)|$ and the angle of the rotation is $\arg(f(z_0))$.

Here are three examples of conformal mappings of an equilateral triangle in $\mathbb{C}$. At each vertex, there is a small copy of the original triangle so that the angles can more easily be compared.

enter image description here

Here are three examples of conformal mappings of an isoceles right triangle in $\mathbb{C}$. At each vertex, there is a small copy of the original triangle so that the angles can more easily be compared.

enter image description here

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  • $\begingroup$ "radially scaled rotation" - So it does not mean that $\alpha$ does not change, but that $\alpha$ is multiplied by constant? I added picture to my question to better describe my problem. Angle between two visible lines seems to change under $z^2$ map. But I can see that angle is scaled. Am I right? $\endgroup$
    – Tacet
    Dec 7, 2019 at 0:18
  • $\begingroup$ No, $\alpha$ remains unchanged where the mapping is conformal. In $\mathbb{C}$, if $f(z)$ differentiable at $z_0$ and $f'(z_0)\ne0$, $f(z)$ is conformal at $z_0$. $\endgroup$
    – robjohn ♦
    Dec 7, 2019 at 8:01
  • $\begingroup$ I have added small copies of the original triangle at the vertices to show more clearly that the angles are preserved. $\endgroup$
    – robjohn ♦
    Dec 7, 2019 at 9:05
  • $\begingroup$ Thank you for the pictures, they are really helpful. I spent some time on it, and I think I understand it now. $\endgroup$
    – Tacet
    Dec 7, 2019 at 19:19

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