I would like to find Laurent series for function Log(z)/(z-i) around i, where Log(i) = $\pi$i/2
I thought that I should start with using Taylor series for Log(1+z) what gives:
$$ Log(z)/(z-i) = \sum_{n=1}^{∞} \frac{(-1)^{n+1}}{n} * \frac{(z-1)^n }{z-i} $$
but I have no idea how to use it to expand around point i. Is it even a good start?
Thank you for any suggestions!
Since $\log(z)$is analytic in a neighborhood around $i$, we can apply a general Taylor Series expansion $\log(z) = \sum_{n = 0}^{\infty} \frac{\log^{(n)}(i)}{n!} (z-i)^{n} = \frac{\pi}{2} i + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot i^{n}}(z-i)^{n}$. With some calculation, one obtains that: $$ \frac{\log(z)}{(z-i)} = \frac{\frac{\pi}{2}i}{z-i} + \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{(n+1) \cdot i^{n+1}}(z-i)^{n} $$