Proving real and odd function has imaginary and odd Fourier Transform

Cheers, I am trying to prove that a real and odd function/signal has imaginary and odd Fourier Transform. Although it seems fairly easy, I can't find a way to achieve it, and searching online hasn't helped a lot.

Edit:

To prove that it is imaginary I tried:

$$F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t}dt = \\ \int_{-\infty}^{\infty} f(t) [\cos(\omega t) - j\sin(\omega t)]dt = \\ 0 - j\int_{-\infty}^{\infty} f(t)\sin(\omega t)dt$$

and I think this proves that it's imaginary indeed, using the fact that the integral of a even and odd function zero.

How would I go about proving it? Should I use Fourier Transform or could I do something wit the Fourier Series? Thanks a lot for any help!

You've already shown that the real part of $$F(\omega)$$ is zero. And the imaginary part is odd because $$\sin(\omega t)$$ is an odd function, i.e., $$\sin(\omega t)=-\sin(-\omega t)$$. Done.

You should also know that the Fourier transform $$F(\omega)$$ of a real-valued time domain function $$f(t)$$ always satisfies

$$F(\omega)=F^*(-\omega)\tag{1}$$

So in the special case that $$F(\omega)$$ is real-valued, from $$(1)$$ it must be even. And, also from $$(1)$$, if $$F(\omega)$$ is imaginary, its imaginary part must be odd.

A suggestion that is helpful for such parity (even, odd) properties is to remember that every function $$f(t)$$ can be decomposed into an even and an odd part:

$$h(t) = \frac{1}{2}\left(h(t)+h(-t)\right) + \frac{1}{2}\left(h(t)-h(-t)\right)$$

Simply computing the Fourier transform of the odd function $$f(t)-f(-t)$$ and checking the nullity of the real part $$F(\omega)+F(-\omega)$$, along with the conjugation of $$\overline{e^{-j\omega t}}=e^{j\omega t}$$ may yield results without splitting the exponential into sine and cosine.