I have to calculate the Residue at $z=i$ of this function: $$ {e^{iz}} \over {\sqrt z (2ln(z)-i\pi )} $$
I know it's a 1st order pole and I'd like to use the traditional formula with the limit of: $$ (z-i) {e^{iz}} \over {\sqrt z (2ln(z)-i\pi )} $$ as z->i
I am having problems calculating this limit. I can I handle residues with trascendent functions?
You have $$ \lim_{z\to i} \frac{ (z-i) e^{i z}}{\sqrt{z}{ (2 \ln z- i\pi)}} = \frac{ e^{-1}}{\sqrt{i}} \left( \lim_{z\to i} \frac{2 \ln z -i\pi} {z-i} \right)^{-1} = \frac{ e^{-1}}{\sqrt{i}} \left( 2 \frac{d}{dz} \ln (z=i) \right)^{-1} =\frac{ e^{-1} i}{2 \sqrt{i}} = \frac{ e^{-1} e^{i \pi/4}}{2 } ,$$ by the definition of the complex derivative and $(d/dz)\ln z=1/z$.