I need to do work on a few integrals using the Cauchy-Integral-Formula and I would appreciate your help, since I have just a few difficulties getting it all. $z\in \mathbb{Z}$ here and $z:=x+iy$. $t\in[0,1]$ always:
i) $ \int_\gamma\frac{e^z}{z}+1dz$ for $\gamma(t)=e^{2i \pi t}$
ii) $ \int_\gamma\frac{cos (\pi z)}{z^2+1}dz$ for $\gamma(t)= 3e^{2i \pi t}$
iii) $ \int_\gamma\frac{e^{e^z}}{(z-i)3}dz$ for $\gamma(t)=i+e^{2i \pi t}$
So my biggest problem is how to intgerate the $\gamma(t)$ in my solution. I just started like this:
i) We know that $e^z$ is holomorph, as is $1$, so the sum is also holomorph. For $e^z$ we can use $e^z=e^{x+iy}=e^x e^{iy} = e^x(cos(y)+i(sin(y))$ and the Cauchy Riemann ODEs. $$ \int_\gamma\frac{e^z}{z}+1dz = \int_\gamma\frac{e^z+z}{z}dz $$ So $f(z)=e^z+z$ and $z_0=0$ so $$ \int_\gamma\frac{e^z+z}{z}dz = 2 \pi i f(0) = 2 \pi i (e^0 + 0)= 2 \pi i $$
ii) I am not entirely sure what to do here. Maybe I should use the Partial fraction decomposition, but maybe there is another way? $$ \int_\gamma\frac{cos (\pi z)}{z^2+1}dz= \int_\gamma\frac{cos (\pi z)}{(z+i)(z-i)}dz $$ and $\frac{1}{(z+i)(z-i)} = \frac{A}{(z+i)}+\frac{B}{(z-i)}$ leads to $A=\frac{i}{2}, B=-\frac{i}{2}$. So we would get: $$ \int_\gamma\frac{cos (\pi z)}{(z+i)(z-i)}dz=\int_\gamma\frac{\frac{icos (\pi z)}{2}}{(z+i)}dz - \int_\gamma\frac{\frac{icos (\pi z)}{2}}{(z-i)}dz $$ and then use Cauchy Integral Formula for both integrals? Is there maybe another way?
iii)$$ \int_\gamma\frac{e^{e^z}}{(z-i)3}dz = \int_\gamma\frac{(e^{e^z})^{1/3}}{(z-i)}dz$$ and then use Cauchy-Integral formula? Is there maybe another way here also and more generaly for terms where $(z+m)^n$ is given?
