Consider the complex valued function $f(t)$ where $t \in [0, 2\pi)$.

Under what conditions are $f(t)$ and $f'(t)$ orthogonal to each other? I'm defining orthogonal here to be

$$ \int_0^{2\pi} \overline{f(t)} f'(t) dt = 0$$

If $f(t)$ is real valued then this old question suggests a method of proof involving a Fourier decomposition, but I don't fully follow the proof and it's not clear to me if it's applicable to complex functions.

  • 1
    $\begingroup$ Essentially, you’re dealing with integral of the conjugate, that is $\int\overline{z}dz$, which depending on the values assumed at $0$ and $2\pi$, will be probably a change in argument times an area estimation of the interior components of the image curve. See the following post for an example: math.stackexchange.com/questions/445781/… $\endgroup$ – Jack LeGrüß Oct 7 '20 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.