To compute $\int_{C(0,1)}Re(z)dz $
Can one do the following ;
$Re(z)=cos(\theta) $ and a parametrisation of $C(0,1)$ is $e^{i\theta},\theta \in[0,2\pi]$
So $\int_{C(0,1)}Re(z)dz=\int_{0}^{2\pi}cos(\theta)ie^{i\theta}d\theta=i[sin(z)e^{i\theta]^{2\pi}_0}=0$ ?
Your reasoning is correct. However, you should be careful when evaluating the integral: $$I=\int_C \mathrm{Re}(z)\ dz = i \int_0^{2\pi} \cos\theta \cdot e^{i\theta}\ d\theta. $$ Then you may integrate by parts, but it is quicker to split up $e^{i\theta}$ using Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$. Hence $$ I = i\int_0^{2\pi} \cos^2\theta\ d\theta - \int_0^{2\pi} \cos\theta\sin\theta\ d\theta.$$ You may see that the second integral vanishes. For the first one, you may exploit the trig identity $$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$ to get rid of the square, and then conclude easily.