Let $0\le r\le 1$ fixed and $0\le \theta< 2\pi$. Let $1<p\le 2$ and $q=p/(p-1).$ We consider the function $$f(\theta)=|1+r\exp(i\theta)|^q+|1-r\exp(i\theta)|^q.$$
Question. Why $$f(\theta)=(1+r^2+2r\cos\theta)^{q/2}-(1+r^2-2r\cos\theta)^{q/2}.$$
I think there is some property hidden on the complex analysis that I don't remember, could someone give me a suggestion?
Put $\text{exp}(i\theta) = \cos(\theta)+i\sin(\theta)$. Then since $\lvert a +ib \rvert = \sqrt{a^2+b^2}$, we see \begin{align*}\lvert 1+r\exp(i\theta)\rvert &= \sqrt{(1+r\cos(\theta))^2 + r^2\sin^2(\theta)}\\ &= \sqrt{1+2r\cos(\theta)+r^2(\cos^2(\theta)+\sin^2(\theta))}\\ &= \sqrt{1+r^2+2r\cos(\theta)} \end{align*} and similarly for the other term, though I don't see where the negative sign in front of the second term came from. Maybe that is a typo.
