Using residue theorem, evaluate the following; $C:\left| {z - 1} \right| = 2$ $$\int_c {{{\tan z} \over z}dz}$$
I want you guys to check my answer.Is it correct? $$\displaylines{ {\mathop{\rm Res}\nolimits} (f(z),0) = \mathop {\lim }\limits_{z \to 0} (z - 0) \times {{\tan z} \over z} \cr = \mathop {\lim }\limits_{z \to 0} \tan z \cr = 0 \cr} $$
The function has a removable singularity at the origin since $$ f(z)=\frac{\tan z}{z} = \frac{\sin z}{z}\frac{1}{\cos z}, $$ and $$ \lim_{z\to 0} \frac{\sin z}{z} = 1. $$ So the singularities you need to worry about are the zeros of $\cos z$, i.e, $z=\pm\,\pi /2$. Then, using the the poles of $\sec z$ are of first order and the positive orientation is counterclockwise: $$ \begin{align} I &= -\left(res_{\frac{\pi}{2}}\,f + res_{-\frac{\pi}{2}}\,f \right)\\ &= -\left(\lim_{z\to\frac{\pi}{2}}\left(z-\frac{\pi}{2}\right)\frac{\tan z}{z} + \lim_{z\to-\frac{\pi}{2}}\left(z+\frac{\pi}{2}\right)\frac{\tan z}{z} \right)\\ &= -\frac{2}{\pi}\left(\lim_{z\to\frac{\pi}{2}}\left(z-\frac{\pi}{2}\right)\sec z + \lim_{z\to-\frac{\pi}{2}}\left(z+\frac{\pi}{2}\right)\sec z \right) \\ &= -\frac{2}{\pi}\left(\lim_{z\to 0}z\sec \left(z+\frac{\pi}{2}\right) + \lim_{z\to 0}z\sec\left(z-\frac{\pi}{2}\right) \right) \\ &= -\frac{2}{\pi}\left(-\lim_{z\to 0}z\csc z+ \lim_{z\to 0}z\csc z\right) \\ &= 0 \end{align} $$
Edit: A simpler way to do this is using that if $g$ is holomorphic in $z_0$ and $f$ has a simple pole in $z_0$, then $res_{z_0} (gf) = g(z_0)\,res_{z_0}f$. Then, observe that $\sec z$ is an even function and use: The Residues of an even function or an odd function on $U$ subset open symmetric