Evaluate following complex line integral.Let $c=\{z|\max\{|\text{Re}(z)|,|\text{Im}(z)|\}=1\}$ be the square with orientation $+1$. Calculate $$\int_c \frac{z\ dz}{\cos(z)-1}$$ with $f(z)=\frac{z}{\cos(z)-1}$
Answer: The integral has a singularity by $z_0=0$, hence $z_0\notin c$ we could integrate $f(z)$ over the following segments:$$(i):-t+i,\quad t\in [-1,1]$$$$(ii):-1+i(2-t),\quad t\in [1,3]$$$$(iii):t-4-i,\quad t\in [3,5]$$$$(iv):1+i(t-6),\quad t\in [5,7]$$ Is this the right way? I integrate then $f(z)$ over $(i),(iii)$ and get $$\int_{(i)}f(z)dz=\int_{(ii)}f(z)dz=\int_{-1}^{1}\frac{k}{cos(k)-1}dz$$ Should I do so on?
No need to integrate over the boundary of the square. The singularities of the meromorphic function $\frac{z}{\cos(z)-1}$ are located at $2\pi\mathbb{Z}$, so the only singularity inside the square is at the origin.
In a neighbourhood of the origin we have $\cos(z)=1-\frac{z^2}{2}+o(z^3)$, hence: $$ \text{Res}\left(f(z),z=0\right) = -2 \tag{1}$$ and by the residue theorem: $$ \int_c f(z)\,dz = 2\pi i\cdot (-2) = \color{red}{-4\pi i}.\tag{2}$$