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Suppose we have a line segment joining to complex points $a$ and $b$ in the z-plane, and then, suppose we map the whole plane under an analytic function $f$. Would the shortest curve between $f(b)$ and $f(a)$ be the curve which the line segment is mapped too?

My thoughts:

I thought of an algebraic verification using calculus of variations. We have the arc length functional:

$$ L(\gamma) = \int_{z=a}^{z=b} |f'(\gamma)| |\gamma'(t)| dt$$

And, minimizing the above integral would give the required what is the shortest path.

So, I start by nudging the curve $$L(\gamma + \epsilon \eta) = \int_{z=a}^{z=b} | f'(\gamma) + \epsilon \eta f''(\gamma)| |\gamma'+ \epsilon \eta'| dt= \int_{z=a}^{z=b}| f'(\gamma) \gamma' + \epsilon \left[ f'(\gamma)\eta' + \epsilon \eta f'(\gamma) \gamma'\right] + O(\epsilon^2)| dt$$

I'm not sure how to finish the problem since the modulus is preventing me from using the integration by parts trick which allows us to derive euler lagrange equations in simple cases. One way would to do it be to take real and complex component of function and write the modulus using square roots... but that is even more uglier.

Is there a better way to do this?

Clemens Bartholdy
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  • You're forgetting the step where you have to differentiate w.r.t. $\epsilon$ first. We have to be minimizing the functional, so Euler-Lagrange still applies with $$L(\gamma,\gamma',t) = |f'(\gamma)|\cdot|\gamma'|$$ – Ninad Munshi Dec 01 '21 at 05:14
  • The answer is "no" because of the Riemann mapping theorem: https://en.wikipedia.org/wiki/Riemann_mapping_theorem – markvs Dec 01 '21 at 05:17
  • That seems fishy, $\gamma$ is like a set of two numbers (x,y) coordinates. In the usual case, it was like $y$ which is one number dependent on some other variable x which is integrated over @NinadMunshi, and shouldn't there be an integral somewhere? – Clemens Bartholdy Dec 01 '21 at 05:22
  • Euler Lagrange works in multiple variables as well, just swapping partial derivatives for gradients, so long as those variables are all single variable functions of the same parameter. – Ninad Munshi Dec 01 '21 at 05:24
  • Huh, then could you have a look at this other question that I had posted @NinadMunshi – Clemens Bartholdy Dec 01 '21 at 05:25
  • I don't get how that theorem expands here, and bit confused by the jargon in it. My understanding of complex functions come from Tristan Needham's VCA, it is not a very formal book. – Clemens Bartholdy Dec 01 '21 at 05:27
  • Shortest path is just the line segment right? So you are asking if an analytic function maps line segments to line segments. The answer is of course no. – Arctic Char Dec 01 '21 at 15:46

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