All,
According to the residue theorem, from the following integral, we can get
$$ \begin{split} \int_c t^k \cdot \frac{t+\zeta}{t-\zeta} \frac{dt}{t} &=4\pi i \zeta^k \end{split} $$ where $k \geq 1$ and $c$ is a unit circle. Also, $\zeta^k = e^{k \theta i} = \cos k \theta + i \sin k \theta$
My question is, what would be the result when $k=0?$
I have derived the integral below but still not sure if this is correct.
$$ \begin{split} \int_c t^k \cdot \frac{t+\zeta}{t-\zeta} \frac{dt}{t} = \int_c \frac{t+\zeta}{t-\zeta} \frac{dt}{t} \end{split} $$
by using partial fraction expansion of the integrand, of the form as follow:
$$\frac{t+\zeta}{(t-\zeta)t} = \frac{A}{t-\zeta}+\frac{B}{t}$$
Multiplying thorough and compare the terms in the numerator, we get,
$$t+\zeta=At+B(t-\zeta)=(A+B)t-B\zeta$$
which can be solved for $A=2$ and $B=-1$, hence
$$\frac{t+\zeta}{(t-\zeta)t}=\frac{2}{t-\zeta}+\frac{1}{t}$$
using this last form, we can break the initial integrand becomes,
$$ \begin{split} \int_c \frac{t+\zeta}{t-\zeta} \frac{dt}{t} \end{split} = \int_c \frac{2}{t-\zeta}dt-\int_c \frac{dt}{t} = 2\pi i(2)-2\pi i = 2\pi i $$
is this correct?