# Proving a complex differentiable function is conformal?

For a complex function $f:\Omega \to \mathbb{C}$, which is differentiable at $z_0$ so clearly it satisfies the Cauchy-Riemann equations, how would you prove that this function is conformal at $z_0$? (I.e. that it is angle preserving)?

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Strictly speaking this is only true, if $f'(x_0) \neq 0$. In that case the differential of $f$ in $x_0$ is a nonzero complex linear map, which, as far as I can see, already implies the claim (assuming you know the complex linear maps). – user20266 Mar 26 '12 at 9:23

I gave a non-rigorous answer earlier, and I think it's important to be able to think non-rigorously. I think it's also important to be able to think rigorously, so herewith I submit an attempt.

Say $s_1, s_2 : \mathbb{R} \to \mathbb{C}$ are smooth curves and $s_1(0)= s_2(0) = z_0$.

The angle at which the two curves meet at $z_0$ is $\arg \left(\dfrac{s_1'(0)}{s_2'(0)}\right)$. The angle at which their images under $f$ meet is $$\arg\left(\frac{\left.\frac{d}{dt}\right|_{t=0} f(s_1(t))} {\left.\frac{d}{dt}\right|_{t=0} f(s_1(t))}\right)$$ We would like to be able to say that this is $$\frac{f'(s_1(0)) \; s_1'(0)}{f'(s_2(0)) \; s_2'(0)}.$$ Then cancel $f'(s_1(0))=f'(s_2(0)) = f'(z_0)$ from the top and bottom and---lo and behold---the angle is the same.

You see how this relies on the fact that $f'(z_0)\ne 0$.

Does this proposed chain rule work? We have $\mathbb{R} \overset{s}{\longrightarrow} \mathbb{C} \overset{f}{\longrightarrow}\mathbb{C}$ and we want to be able to say $(f\circ s)' = (f'\circ s)\cdot s'$, when the derivatives of $f\circ s$ and of $s$ are those of complex-valued functions of a real variable, and that of $f$ is that of a complex-valued function of a complex variable. I'm not sure I've seen this before. However, if I'm not mistaken, the proof is actually the same as the proof of the chain rule for composition of functions of real variables or that for composition of functions of complex variables. Except maybe one needs to mention that if $$\frac{f(z_0+k)- f(z_0)}{k} \to f'(z_0)\text{ as }k\to 0,$$ where $k\to 0$ means $k$ is approaches $0$ within $\mathbb{C}$, then in particular this implies $$\frac{f(s(0+h)-f(s(0))}{s(0+h)-s(0)} \to f'(z_0)\text{ as }h\to 0,$$ where $h\to 0$ means $h$ approaches $0$ within $\mathbb{R}$. The fact that differentiable complex-valued functions $s$ of a real variable are continuous would need to get cited at the appropriate place.

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It's not conformal at points where $f'$ is $0$. At other points, I'd start by thinking about the fact that if $w=f(z)$ then $dw = f'(z)\;dz$, and the mapping $dz \mapsto f'(z_0)\;dz$ is a multiplication by a nonzero complex number $f'(z_0)$. Every multiplication $z\mapsto az$ by a nonzero complex number $a$ amounts to rotating by $\arg a$ and dilating by $|a|$. If you rotate and dilate, you don't change angles.

After that, I'd start thinking about how to make all this logically rigorous.

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