I am trying to calculate the residue of the function $f(z) = \frac{\mathrm{e}^z}{z^2}$ in the point $z=0$, straight from the definition.
From the residual formula, we know that the answer has to be: $$ \underset{z=0}{\textrm{Res}} \,\frac{\mathrm{i}^z}{z^2}=(\mathrm{e}^z)'(0)=\mathrm{i}^0 = 1. $$ The definition of the residue in the point $a$ is: $$ \underset{z=a}{\textrm{Res}}\,f(z)=\frac1{2\pi\mathrm{i}}\int_{C(a,\epsilon)^+}f(z)\,\mathrm{d}z. $$ I use $\epsilon=1$, Therefore, I need to calculate the following integral by hand: $$ \int_{C(0,1)^+}\frac{\mathrm{e}^z}{z^2}\,\mathrm{d}z. $$ That's where I am stuck. Can we use a complex anti-deriviative perhaps? I now tried to parameterize it using $\gamma:t\mapsto \mathrm{e}^{\mathrm{i}t}$, for $t\in(0,2\pi)$. This then gives by definition the integral $$ \int_{C(0,1)^+} \frac{\mathrm{e}^z}{z^2}\,\mathrm{d}z=\int_0^{2\pi} f(\gamma(t))\gamma'(t)=\int_0^{2\pi}\frac{\mathrm{e}^{\mathrm{e}^{\mathrm{i}t}}}{\mathrm{e}^{2\mathrm{i}t}}\mathrm{ie}^{\mathrm{i}t}\,\mathrm{d}t=\int_0^{2\pi}\frac{\mathrm{ie}^{\mathrm{e}^{\mathrm{i}t}}}{\mathrm{e}^{\mathrm{i}t}}\,\mathrm{d}t. $$ I don't know how to integrate this.
Depending on your definition, but my definition says that the residus it the 1st coefficient of the singular part. Since $$e^z=1+z+o(z),$$ you get $$\frac{e^z}{z^2}=\frac{1}{z^2}+\frac{1}{z}+o(1).$$ And so, the residue arise naturally.