Hi members,
see attached PDF file for my question
Thank you
The residue of a function f(z) at z=0 is just $\displaystyle \oint_C f(z)\,dz$ where C is a contour around the origin.
If we substitute 2u for z, the contour changes, but that doesn't matter because the integral is the same for all contours around the origin. But $\displaystyle \oint f(z)\,dz$ is twice the size of $\displaystyle \oint f(2u)\,du$ (think back to first-year calculus), so the residue with z is actually twice the residue with u.
If you think of the simplest case, the residue of $\displaystyle \frac{1}{z}$ at $\displaystyle z=0$ is 1, but if you substitute 2u for z there, you get the residue of $\displaystyle \frac{1}{2u}$ at $\displaystyle u=0$, which is $\displaystyle \frac{1}{2}$. The residue is not the same when you change variables.
In your case, you have $\displaystyle \frac{e^\frac{z-1}{2}}{z-1}$ with residue 1 at $\displaystyle z=1$, and $\displaystyle \frac{e^u}{2u}$ with residue $\displaystyle \frac{1}{2}$ at $\displaystyle u=0$.
- Hollywood