Let $\gamma = \gamma(0;2)$. Is $$\int_{\gamma} \sec ^2z \ \mathrm{d}z$$ equal to $0$?
I'm trying to answer this question using only tools like Cauchy Theorem or the Deformation Theorem since contour integration is treated later in the book I took this exercise from.
So I know that
$$\sec^2z = \frac{1}{\cos^2z}$$ so the points where holomorphy might fail are the zeros of $\cos z$. So
$$\sec^2 z =0 \iff z= \frac{1}{2}(2k+1)\pi$$ for $k \in \mathbb{Z}$. Now, in my path, I have two zeroes, $-\pi/2$ and $\pi/2$, and I don't see how creating a new path around any of those points can help me out here. Any help will be highly appreciated. Thanks in advance!
EDIT: I just realized that Fundamental theorem of calculus applies here to the function $\tan z$, so the integral is indeed zero.
Are you allowed to use path\line integrals?
If yes, then just write the parametrization for $\gamma$: $\gamma(t) = 2 \exp(it).$
Now, $$ \int_\gamma \sec^2(z) \, \rm{d} z = \int_0^{2\pi} \sec^2(2 \exp(it)) \cdot 2i \exp(it) \, \rm{d} t = \left. \tan(2 \exp(it)) \right|_{t=0}^{2\pi} = 0. $$
I guess that what you mean by the Fundamental theorem of calculus (you don't need any complex analysis).