I have a question regarding complex-valued functions as integrand. If $$f : \mathbb{C} \rightarrow \mathbb{C}$$ is even function, then $$\int_{C(0,r)}f=0$$ for all $r>0$.
I tried to split the integral because the function is even, but the problem was when I came across $e^{it}$. And I wasn´t whether I can split the integral?
Why don't you think about it like this?
$\displaystyle \int_{C(0,r)}f(z)\,dz=\lim_{n \to \infty}\sum_{j=1}^{n}f\left(re^{\frac{2\pi ij}{n}}\right)\cdot rie^{\frac{2\pi ij}{n}}$, which is basically your $f(z)\,dz$.
Now, why is this special at all? Each $f(z_1)\,(dz)_1$ is canceled out by the $f(z_2)(dz)_2$ on the other end of the circle. You can see this by substituting $j_2=j+\frac{n}{2}$ for even $n$. The summands will be exact opposites due to the second factor.
Here's a more geometric, intuitive example. $f(z)\,dz$ means the value of $f(z)$ multiplied by $dz$, which is the direction vector corresponding to the curve. On opposite sides of the circle, the $dz$'s are exact opposites, while the $f(z)$'s will be the same, since $f(z)=f(-z)$ in an even function. If you think about the limit definition of an integral and this property, it'll make sense how everything cancels out and the integral comes out to $0$.