I have the following integral $$ \int _c Re(z) + (1-z)e^z dz $$
where C is the unit circle
So I've separated it in two integrals and if I parametrize the first one, it gives me 0. For the second one I've used integration by parts:
$$\int_c(1-z)e^zdz=(1-z)e^z|_c+\int_ce^zdz$$
I could say maybe that
$$\int_ce^zdz=e^z|_c$$
but what do i do next? How do i find the value of the integrals?
We have \begin{eqnarray} \int_C f' dz &=& \int_0^1 f'(C(t)) C'(t)dt \\ &=& \int_0^1 (f\circ C)'(t)dt \\ &=& (f\circ C) \mid_0^1 \\ &=& f(C(1))-f(C(0)) \end{eqnarray} Hence if $C(0) = C(1)$ then $\int_C f' dz = 0$.
The integration by parts formula is the similar to the usual integration rule: \begin{eqnarray} \int_C (f \cdot g)' dz &=& \int_0^1 (f \cdot g)'(C(t)) C'(t)dt \\ &=& \int_0^1 ((f \cdot g)\circ C)'(t)dt \\ &=& ((f \cdot g)\circ C) \mid_0^1 \\ &=& f(C(1))g(C(1))-f(C(0))g(C(0)) \\ &=& \int_0^1 (f' \cdot g+ f \cdot g')(C(t)) C'(t)dt \\ &=& \int_0^1 (f' \cdot g)(C(t)) C'(t)dt + \int_0^1 (f \cdot g')(C(t)) C'(t)dt \\ &=& \int_C f' \cdot g dz + \int_C f \cdot g'dz \end{eqnarray}
Hence $\int_C f' \cdot g dz = f(C(1))g(C(1))-f(C(0))g(C(0)) - \int_C f \cdot g'dz$.
Now let $f'(z) = e^z$, $g(z) = 1-z$ and note that $C(0) =C(1)$. Then $\int_C f' \cdot g dz = -\int_C f \cdot g' dz = \int_C e^z dz = e^{C(1)}-e^{C(0)} = 0$.